Rd Sharma XII Vol 2 2020 for Class 12 Science Maths Chapter 4 - Algebra Of Vectors

Share with your friends

Rd Sharma XII Vol 2 2020 Solutions for Class 12 Science Maths Chapter 4 Algebra Of Vectors are provided here with simple step-by-step explanations. These solutions for Algebra Of Vectors are extremely popular among Class 12 Science students for Maths Algebra Of Vectors Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma XII Vol 2 2020 Book of Class 12 Science Maths Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma XII Vol 2 2020 Solutions. All Rd Sharma XII Vol 2 2020 Solutions for class Class 12 Science Maths are prepared by experts and are 100% accurate.

Page No 22.17:

Question 1:

If P, Q and R are three collinear points such that $\vec{P Q} = \vec{a}$ and $\vec{Q R} = \vec{b}$ . Find the vector $\vec{P R}$ .

Answer:

Given: $P, Q$ and $R$ are collinear such that $\vec{P Q} = \vec{a}$ and $\vec{Q R} = \vec{b}$ . Then,
$\vec{P Q} + \vec{Q R} = \vec{P R}$
$\Rightarrow \vec{P R} = \vec{a} + \vec{b}$

Page No 22.17:

Question 2:

Give a condition that three vectors $\vec{a}$ , $\vec{b}$ and $\vec{c}$ form the three sides of a triangle. What are the other possibilities?

Answer:

Let $A B C$ be a triangle such that $\vec{B C} = \vec{a}$ , $\vec{A B} = \vec{c}$ and $\vec{C A} = \vec{b}$ . Then,
$\vec{a} + \vec{b} + \vec{c} = \vec{B C} + \vec{C A} + \vec{A B}$
$\Rightarrow \vec{a} + \vec{b} + \vec{c} = \vec{B A} + \vec{A B}$ [∵ $\vec{B C} + \vec{C A} = \vec{B A}$ ]
$\Rightarrow \vec{a} + \vec{b} + \vec{c} = \vec{B B}$ [ Using triangle law]
$\Rightarrow \vec{a} + \vec{b} + \vec{c} = \vec{0}$ [ By definition of null vector]

Other possibilities are
$(i) \vec{c} + \vec{a} = \vec{b}$
$(ii) \vec{a} + \vec{b} = \vec{c}$
$(iii) \vec{b} + \vec{c} = \vec{a}$

Page No 22.17:

Question 3:

If $\vec{a}$ and $\vec{b}$ are two non-collinear vectors having the same initial point. What are the vectors represented by $\vec{a}$ + $\vec{b}$ and $\vec{a}$ − $\vec{b}$ .

Answer:

Given: $\vec{a}, \vec{b}$ are two non-collinear vectors having same initial points. Complete the parallelogram $A B C D$ such that $\vec{A B} = \vec{a}$ and $\vec{B C} = \vec{b} .$
In $△ A B C$ ,
$\vec{A B} + \vec{B C} = \vec{A C} \Rightarrow \vec{a} + \vec{b} = \vec{A C}$
In $△ A B D$ ,
$\vec{A D} + \vec{D B} = \vec{A B} \Rightarrow \vec{b} + \vec{D B} = \vec{a} \Rightarrow \vec{D B} = \vec{a} - \vec{b}$

Therefore, $\vec{A C}$ and $\vec{D B}$ are the diagonals of a parallelogram whose adjacent sides are $\vec{a}$ and $\vec{b}$ respectively.

Page No 22.17:

Question 4:

If $\vec{a}$ is a vector and m is a scalar such that m $\vec{a}$ = $\vec{0}$ , then what are the alternatives for m and $\vec{a}$ ?

Answer:

Given: $\vec{a}$ is a vector and $m$ is a scalar such that, $m \vec{a} = \vec{0}$
Then either $m = 0$ or, $\vec{a} = \vec{0}$

Page No 22.17:

Question 5:

If $\vec{a,} \vec{b}$ are two vectors, then write the truth value of the following statements:
(i) $\vec{a} = - \vec{b} \Rightarrow |\vec{a}| = |\vec{b}|$

(ii) $|\vec{a}| = |\vec{b}| \Rightarrow \vec{a} = \pm \vec{b}$

(iii) $|\vec{a}| = |\vec{b}| \Rightarrow \vec{a} = \vec{b}$

Answer:

(i) True.
$\vec{a} = - \vec{b}$
Taking modulus on both sides of the equation, we get,
$|\vec{a}| = |- \vec{b}|$
$\Rightarrow |\vec{a}| = |\vec{b}|$ [ ∵ $|- \vec{b}| = |\vec{b}|$ ]

(ii) False.
We cannot say $|\vec{a}| = |\vec{b}| \Rightarrow \vec{a} = \pm \vec{b}$
Consider an example,
$\vec{a} = i + \sqrt{3} j and \vec{b} = \sqrt{2} i + \sqrt{2} j |\vec{a}| = \sqrt{1^{2} + {(\sqrt{3})}^{2}} = 2 and |\vec{b}| = \sqrt{{(\sqrt{2})}^{2} + {(\sqrt{2})}^{2}} = 2 Thus, |\vec{a}| = |\vec{b}| but \vec{a} \neq \pm \vec{b}$

(iii) False.
We cannot say $|\vec{a}| = |\vec{b}| \Rightarrow \vec{a} = \vec{b}$
Consider an example,
$\vec{a} = i + \sqrt{3} j and \vec{b} = \sqrt{2} i + \sqrt{2} j |\vec{a}| = \sqrt{1^{2} + {(\sqrt{3})}^{2}} = 2 and |\vec{b}| = \sqrt{{(\sqrt{2})}^{2} + {(\sqrt{2})}^{2}} = 2 Thus, |\vec{a}| = |\vec{b}| but \vec{a} \neq \vec{b}$

Page No 22.17:

Question 6:

ABCD is a quadrilateral. Find the sum the vectors $\vec{B A}, \vec{B C}, \vec{C D}$ and $\vec{D A}$ .

Answer:

Given: $A B C D$ is a quadrilateral.
We need to find the sum of $\vec{B A} + \vec{B C} + \vec{C D} + \vec{D A} .$
Consider,
$\vec{B A} + \vec{B C} + \vec{C D} + \vec{D A} = (\vec{B A} + \vec{D A}) + (\vec{B C} + \vec{C D})$
$= (\vec{B D} + 2 \vec{D A}) + \vec{B D}$ [∵ $\vec{B D} + \vec{D A} = \vec{B A}$ and $\vec{B C} + \vec{C D} = \vec{B D}$ ]
$= 2 (\vec{B D} + \vec{D A}) = 2 \vec{B A}$

Page No 22.17:

Question 7:

ABCDE is a pentagon, prove that
(i) $\vec{A B} + \vec{B C} + \vec{C D} + \vec{D E} + \vec{E A} = \vec{0}$

(ii) $\vec{A B} + \vec{A E} + \vec{B C} + \vec{D C} + \vec{E D} + \vec{A C} = 3 \vec{A C}$

Answer:

Given: $A B C D E$ is a pentagon.
(i) To Prove: $\vec{A B} + \vec{B C} + \vec{C D} + \vec{D E} + \vec{E A} = \vec{0} .$
Proof: We have,
$LHS = \vec{A B} + \vec{B C} + \vec{C D} + \vec{D E} + \vec{E A}$
   $= \vec{A C} + \vec{C D} + \vec{D A}$ [ ∵ $\vec{A B} + \vec{B C} = \vec{A C}$ and $\vec{D E} + \vec{E A} = \vec{A D}$ ]
   $= \vec{A D} + \vec{D A}$ [ ∵ $\vec{A C} + \vec{C D} = \vec{A D}$ ]
$= \vec{0}$ = RHS
(ii) To Prove: $\vec{A B} + \vec{A E} + \vec{B C} + \vec{D C} + \vec{E D} + \vec{A C} = 3 \vec{A C} .$
Proof: We have,
   $LHS = \vec{A B} + \vec{A E} + \vec{B C} + \vec{D C} + \vec{E D} + \vec{A C}$
    $= (\vec{A B} + \vec{B C}) + (\vec{A E} + \vec{E D}) + \vec{D C} + \vec{A C}$
    $= \vec{A C} + \vec{A D} + \vec{D C} + \vec{A C}$      [∵ $\vec{A B} + \vec{B C} = \vec{A C}$ and $\vec{A E} + \vec{E D} = \vec{A D}$ ]
= $\vec{A C} + \vec{A D} + \vec{A C} - \vec{A D} + \vec{A C}$ [∵ $\vec{A D} + \vec{D C} = \vec{A C} \Rightarrow \vec{D C} = \vec{A C} - \vec{A D}$ ]
   $= 3 \vec{A C} = RHS$
Hence proved.

Page No 22.17:

Question 8:

Prove that the sum of all vectors drawn from the centre of a regular octagon to its vertices is the zero vector.

Answer:

Given: A regular octagon of eight sides with centre O.
To show: $\vec{O A} + \vec{O B} + \vec{O C} + \vec{O D} + \vec{O E} + \vec{O F} + \vec{O G} + \vec{O H} = \vec{0} .$
Proof: We know centre of the regular octagon bisects all the diagonals passing through it.
$\vec{O A} = - \vec{O E}, \vec{O B} = - \vec{O F}, \vec{O D} = - \vec{O H}$ and $\vec{O C} = - \vec{O G} .$
$\Rightarrow \vec{O A} + \vec{O E} = \vec{0}, \vec{O B} + \vec{O F} = \vec{0}, \vec{O D} + \vec{O H} = \vec{0}$ and $\vec{O C} + \vec{O G} = \vec{0} .$ ............(i)
Now,
$\vec{O A} + \vec{O B} + \vec{O C} + \vec{O D} + \vec{O E} + \vec{O F} + \vec{O G} + \vec{O H} = (\vec{O A} + \vec{O E}) + (\vec{O B} + \vec{O F}) + (\vec{O C} + \vec{O G}) + (\vec{O D} + \vec{O H}) = \vec{0} + \vec{0} + \vec{0} + \vec{0} = \vec{0}$
Hence proved.

Page No 22.17:

Question 9:

If P is a point and ABCD is a quadrilateral and $\vec{A P} + \vec{P B} + \vec{P D} = \vec{P C}$ , show that ABCD is a parallelogram.

Answer:

Given: $A B C D$ is a quadrilateral such that $\vec{A P} + \vec{P B} + \vec{P D} = \vec{P C} .$
To show: $A B C D$ is a parallelogram.
Proof: Consider,
$\vec{A P} + \vec{P B} + \vec{P D} = \vec{P C} \Rightarrow \vec{A P} + \vec{P B} = \vec{P C} - \vec{P D}$
$\Rightarrow \vec{A B} = \vec{D C}$ [ ∵ $\vec{A P} + \vec{P B} = \vec{A B}$ and $\vec{P D} + \vec{D C} = \vec{P C}$ ]
Again,
$\vec{A P} + \vec{P B} + \vec{P D} = \vec{P C} \Rightarrow \vec{A P} + \vec{P D} = \vec{P C} - \vec{P B}$
$\Rightarrow \vec{A D} = \vec{B C}$ [ ∵ $\vec{A P} + \vec{P D} = \vec{A D}$ and $\vec{P B} + \vec{B C} = \vec{P C}$ ]

Since, opposite sides of the quadrilateral are equal and parallel.
Hence, $A B C D$ is a parallelogram.

Page No 22.17:

Question 10:

Five forces $\vec{A B,} \vec{A C,} \vec{A D,} \vec{A E}$ and $\vec{A F}$ act at the vertex of a regular hexagon ABCDEF. Prove that the resultant is
6 $\vec{A O,}$ where O is the centre of hexagon.

Answer:

$\vec{A B} + \vec{A C} + \vec{A D} + \vec{A E} + \vec{A F}$

Consider ∆ADE,

$\begin{array}{l} \vec{A D} + \vec{D E} + \vec{E A} = 0 \\ \vec{A D} + \vec{D E} = \vec{A E} \\ 2 \vec{A O} + (- \vec{A B}) = \vec{A E} [\vec{A D} = 2 \vec{A O} and E D | | A B \vec{D E} = - \vec{A B}] \\ ∴ \vec{A E} + \vec{A B} = 2 \vec{A O} . . . . . (1) \end{array}$

Now, consider ∆ADC

$\vec{A C} + \vec{C D} + \vec{D A} = 0 \vec{A C} + \vec{C D} = \vec{A D} [∵ \vec{C D} = \vec{A F}] \vec{A C} + \vec{A F} = 2 \vec{A O} . . . . . (2)$

Using (1) and (2),

$\vec{A B} + \vec{A E} + \vec{A C} + \vec{A F} + \vec{A D} 2 \vec{A O} + 2 \vec{A O} + 2 \vec{A O} = 6 \vec{A O}$

Page No 22.23:

Question 1:

Find the position vector of a point R which divides the line joining the two points P and Q with position vectors $\vec{OP} = 2 \vec{a} + \vec{b}$ and $\vec{OQ} = \vec{a} - 2 \vec{b}$ , respectively in the ratio 1 : 2 internally and externally. [NCERT EXEMPLAR]

Answer:

It is given that P and Q are two points with position vectors $\vec{OP} = 2 \vec{a} + \vec{b}$ and $\vec{OQ} = \vec{a} - 2 \vec{b}$ , respectively.

When R divides PQ internally in the ratio 1 : 2, then

Position vector of R = $\frac{1 \times (\vec{a} - 2 \vec{b}) + 2 \times (2 \vec{a} + \vec{b})}{1 + 2} = \frac{5 \vec{a}}{3}$

When R divides PQ externally in the ratio 1 : 2, then

Position vector of R = $\frac{1 \times (\vec{a} - 2 \vec{b}) - 2 \times (2 \vec{a} + \vec{b})}{1 - 2} = \frac{- 3 \vec{a} - 4 \vec{b}}{- 1} = 3 \vec{a} + 4 \vec{b}$

Page No 22.24:

Question 2:

X and Y are two points with position vectors $3 \vec{a} + \vec{b} and \vec{a} - 3 \vec{b}$ respectively. Write down the position vector of a point Z which divides the line segment XY in the ration 2:1 externally.

Answer:

Given:
X and Y are two points with position vectors $3 \vec{a} + \vec{b} and \vec{a} - 3 \vec{b}$ , respectively.
Z divides the line segment XY in the ratio 2 : 1 externally

We know,
The position vector of R which divided P and Q externally in the ratio m : n is given by $\vec{O R} = \frac{m \vec{b} - n \vec{a}}{m - n}$ , where position vector of P and Q are $\vec{a} and \vec{b}$ , respectively.

Thus,
$\vec{O Z} = \frac{2 (\vec{a} - 3 \vec{b}) - 1 (3 \vec{a} + \vec{b})}{2 - 1} \Rightarrow \vec{O Z} = \frac{2 \vec{a} - 6 \vec{b} - 3 \vec{a} - \vec{b}}{1} \Rightarrow \vec{O Z} = - \vec{a} - 7 \vec{b}$

Hence, the position vector of a point Z is $- \vec{a} - 7 \vec{b} .$

Page No 22.24:

Question 3:

Let $\vec{a,} \vec{b,} \vec{c,} \vec{d}$ be the position vectors of the four distinct points A, B, C, D. If $\vec{b} - \vec{a} = \vec{c} - \vec{d}$ , then show that ABCD is a parallelogram.

Answer:

Given: $\vec{a}, \vec{b}, \vec{c}$ and $\vec{d}$ are the position vectors of the four distinct points $A, B, C$ and $D$ .
Also, we have, $\vec{b} - \vec{a} = \vec{c} - \vec{d} .$
$\Rightarrow \vec{A B} = \vec{D C}$
Again,
$\vec{b} - \vec{a} = \vec{c} - \vec{d} \Rightarrow \vec{b} - \vec{c} = \vec{a} - \vec{d} \Rightarrow \vec{C B} = \vec{D A}$
Consequently, $A B ∥ D C, C B ∥ D A$ and $\vec{A B} = \vec{D C}, \vec{C B} = \vec{D A}$ . Thus two of its opposite sides are equal and parallel.
Hence, $A B C D$ is a parallelogram.

Page No 22.24:

Question 4:

If $\vec{a,} \vec{b}$ are the position vectors of A, B respectively, find the position vector of a point C in AB produced such that AC = 3 AB and that a point D in BA produced such that BD = 2BA.

Answer:

Let the position vectors of C and D are $\vec{c}$ and $\vec{d}$ respectively. We have,
$A C = 3 A B . \Rightarrow A C = 3 (A C - B C) . \Rightarrow 2 A C = 3 B C .$
$\Rightarrow \frac{A C}{B C} = \frac{3}{2} .$
So C divides AB in the ratio of $3 : 2$ externally.
$\vec{c} = \frac{2 \vec{a} - 3 \vec{b}}{2 - 3} = 3 \vec{b} - 2 \vec{a} .$

Position vector of point C is $3 \vec{b} - 2 \vec{a}$
Moreover,
$B D = 2 B A . \Rightarrow B D = 2 (B D - A D) . \Rightarrow B D = 2 A D .$
$\Rightarrow \frac{B D}{A D} = \frac{2}{1}$ .
∴ $\vec{d} = \frac{\vec{b} - 2 \vec{a}}{1 - 2} = 2 \vec{a} - \vec{b} .$
Position vector of point D is $2 \vec{a} - \vec{b}$

Page No 22.24:

Question 5:

Show that the four points A, B, C, D with position vectors $\vec{a,} \vec{b,} \vec{c,} \vec{d}$ respectively such that $3 \vec{a} - 2 \vec{b} + 5 \vec{c} - 6 \vec{d} = 0,$ are coplanar. Also, find the position vector of the point of intersection of the line segments AC and BD.

Answer:

Let $A C$ and $B D$ intersects at a point $P$ .
We have,
$3 \vec{a} - 2 \overset{⇀}{b} + 5 \overset{⇀}{c} - 6 \overset{⇀}{d} = \vec{0} . \Rightarrow 3 \vec{a} + 5 \vec{c} = 2 \vec{b} + 6 \vec{d} Since sum of coefficients on both sides of the above equation is 8 . so we divide the equation on both sides by 8 . \Rightarrow \frac{3 \vec{a} + 5 \vec{c}}{8} = \frac{2 \vec{b} + 6 \vec{d}}{8} \Rightarrow \frac{3 \vec{a} + 5 \vec{c}}{3 + 5} = \frac{2 \vec{b} + 6 \vec{d}}{2 + 6}$

Therefore, $P$ divides $A C$ in the ratio of $3 : 5$ and $P$ divides $B D$ in the ratio of 2:6.
Therefore, position vector of the point of intersection of AC and BD will be
$\frac{3 \vec{a} + 5 \vec{c}}{8} = \frac{2 \vec{b} + 6 \vec{d}}{8}$

Page No 22.24:

Question 6:

Show that the four points P, Q, R, S with position vectors $\vec{p}$ , $\vec{q}$ , $\vec{r}$ , $\vec{s}$ respectively such that 5 $\vec{p}$ −2 $\vec{q}$ +6 $\vec{r}$ −9 $\vec{s}$ = $\vec{0}$ , are coplanar. Also, find the position vector of the point of intersection of the line segments PR and QS.

Answer:

Let the point of intersection of the line segments $P R$ and $Q S$ is A. Then
$5 \vec{p} - 2 \vec{q} + 6 \vec{r} - 9 \vec{s} = \vec{0} . \Rightarrow 5 \vec{p} + 6 \vec{r} = 2 \vec{q} + 9 \vec{s} the sum of the coefficients on both the sides of the above equation is 11 . So, we divide the given equation with 11 . \Rightarrow \frac{5 \vec{p} + 6 \vec{r}}{11} = \frac{2 \vec{q} + 9 \vec{s}}{11}$

$\frac{5 \vec{p} + 6 \vec{r}}{5 + 6} = \frac{2 \vec{q} + 9 \vec{s}}{2 + 9}$

Therefore, A divides $P R$ in the ratio of $5 : 6$ and $Q S$ in the ratio of $2 : 9$ .
The position vector of the point of intersection of the line segment is $\frac{5 \vec{p} + 6 \vec{r}}{11}, \frac{2 \vec{q} + 9 \vec{s}}{11}$ .

Page No 22.24:

Question 7:

The vertices A, B, C of triangle ABC have respectively position vectors $\vec{a}$ , $\vec{b}$ , $\vec{c}$ with respect to a given origin O. Show that the point D where the bisector of ∠A meets BC has position vector $\vec{d} = \frac{β \vec{b} + γ \vec{c}}{β + γ}, where β = |\vec{c} - \vec{a}| and, γ = |\vec{a} - \vec{b}|$ .
Hence, deduce that the incentre I has position vector $\frac{α \vec{a} + β \vec{b} + γ \vec{c}}{α + β + γ}, where α = |\vec{b} - \vec{c}|$ .

Answer:

Let the position vectors of A, B and C with respect to some origin, O be $\vec{a}, \vec{b} and \vec{c}$ respectively.

Let D be the point on BC where bisectors of ∠A meets.
Let $\vec{d}$ be the position vector of D which divides CB internally in the ratio β and γ, where $β = |\vec{AC}| and γ = |\vec{AB}|$
Thus, $β = |\vec{c} - \vec{a}| and γ = |\vec{b} - \vec{a}|$
By section formula, the position vector of D is given by
$\vec{OD} = \frac{β \vec{b} + γ \vec{c}}{β + γ} Let α = |\vec{b} - \vec{c}|$

Incentre is the concurrent point of angle bisectors and incentre divides the line AD in the ratio ∝: β + γ.

So, the position vector of incentre is given as,

$\frac{α \vec{a} + (\frac{β \vec{b} + γ \vec{c}}{β + γ}) (β + γ)}{α + β + γ} = \frac{α \vec{a} + β \vec{b} + γ \vec{c}}{α + β + γ}$

Page No 22.36:

Question 1:

If O is a point in space, ABC is a triangle and D, E, F are the mid-points of the sides BC, CA and AB respectively of the triangle, prove that $\vec{O A} + \vec{O B} + \vec{O C} = \vec{O D} + \vec{O E} + \vec{O F}$ .

Answer:

Let D, E and F are the midpoints of BC, CA and AB respectively.
Therefore,
$\frac{\vec{O B} + \vec{O C}}{2} = \vec{O D}$
$\vec{O B} + \vec{O C} = 2 \vec{O D} . . . . . (1) S i m i l a r l y, \vec{O C} + \vec{O A} = 2 \vec{O E} . . . . . (2) \vec{O A} + \vec{O B} = 2 \vec{O F} . . . . . (3)$

Adding (1), (2) and (3). We get,

$2 (\vec{O A} + \vec{O B} + \vec{O C}) = 2 (\vec{O D} + \vec{O E} + \vec{O F}) . \Rightarrow \vec{O A} + \vec{O B} + \vec{O C} = \vec{O D} + \vec{O E} + \vec{O F} .$
Hence Proved.

Page No 22.36:

Question 2:

Show that the sum of three vectors determined by the medians of a triangle directed from the vertices is zero.

Answer:

Let $\vec{a}, \vec{b}$ and $\vec{c}$ are the position vectors of the vertices $A, B$ and $C$ respectively.
Then we know that the position vector of the centroid $O$ of the triangle is $\frac{\vec{a} + \vec{b} + \vec{c}}{3}$ .
Therefore sum of the three vectors $\vec{O A}, \vec{O B}$ and $\vec{O C}$ is
$\vec{O A} + \vec{O B} + \vec{O C} = \vec{a} - (\frac{\vec{a} + \vec{b} + \vec{c}}{3}) + \vec{b} - (\frac{\vec{a} + \vec{b} + \vec{c}}{3}) + \vec{c} - (\frac{\vec{a} + \vec{b} + \vec{c}}{3}) = (\vec{a} + \vec{b} + \vec{c}) - 3 (\frac{\vec{a} + \vec{b} + \vec{c}}{3}) = \vec{0}$
Hence, Sum of the three vectors determined by the medians of a triangle directed from the vertices is zero.

Page No 22.37:

Question 3:

ABCD is a parallelogram and P is the point of intersection of its diagonals. If O is the origin of reference, show that $\vec{O A} + \vec{O B} + \vec{O C} + \vec{O D} = 4 \vec{O P}$ .

Answer:

Given a parallelogram ABCD and P is the point of intersection of its diagonals. We know the diagonals of a parallelogram, bisect each other. Therefore,
$\frac{\vec{O A} + \vec{O C}}{2} = \vec{O P} \vec{O A} + \vec{O C} = 2 \vec{O P} . . . . . (1) a n d \frac{\vec{O B} + \vec{O D}}{2} = \vec{O P} \vec{O B} + \vec{O D} = 2 \vec{O P} . . . . . (2)$
Adding (1) and (2), We get,
$\vec{O A} + \vec{O B} + \vec{O C} + \vec{O D} = 4 \vec{O P}$

Page No 22.37:

Question 4:

Show that the line segments joining the mid-points of opposite sides of a quadrilateral bisects each other.

Answer:

Let ABCD is the quadrilateral and P, Q, R, S are mid points of the sides AB, BC, CD, DA respectively.

Join DB to form triangle ABD.

$\frac{AS}{SD} = \frac{AP}{PB} \Rightarrow SP | | DB and SP = \frac{1}{2} DB$

In triangle BCD

$\frac{CR}{RD} = \frac{CQ}{QB} \Rightarrow RQ | | DB and RQ = \frac{1}{2} DB$

In quadrilateral PQRS,
SP = RQ and SP || RQ
∴ PQRS is a parallelogram.

Diagonals of a parallelogram bisect each other.

∴ PR and QS bisect each other.

Page No 22.37:

Question 5:

ABCD are four points in a plane and Q is the point of intersection of the lines joining the mid-points of AB and CD; BC and AD. Show that $\vec{P A} + \vec{P B} + \vec{P C} + \vec{P D} = 4 \vec{P Q}$ , where P is any point.

Answer:

Let E, F, G and H are the midpoints of the sides AB, BC, CD and DA respectively of say quadrilateral ABCD. By geometry of the figure formed by joining the midpoints E, F, G and H will be a parallelogram. Hence its diagonals will bisect each other, say at Q.
Now, F is the midpoint of BC.
$\frac{\overset{⇀}{P B} + \overset{⇀}{P C}}{2} = \overset{⇀}{P F} ∴ \overset{⇀}{P B} + \overset{⇀}{P C} = 2 \overset{⇀}{P F} . . . . . (1)$
And, H is the midpoint of AD.
$\frac{\overset{⇀}{P A} + \overset{⇀}{P D}}{2} = \overset{⇀}{P H} ∴ \overset{⇀}{P A} + \overset{⇀}{P D} = 2 \overset{⇀}{P H} . . . . . (2)$
Adding (1) and (2). We get,
$\overset{⇀}{P A} + \overset{⇀}{P B} + \overset{⇀}{P C} + \overset{⇀}{P D} = 2 (\overset{⇀}{P F} + \overset{⇀}{P H}) = 2 (2 \overset{⇀}{P Q}) = 4 \overset{⇀}{P Q} .$

Page No 22.37:

Question 6:

Prove by vector method that the internal bisectors of the angles of a triangle are concurrent.

Answer:

Let ABC be a triangle and $\vec{α}, \vec{β}, \vec{γ}$ be the position vectors of the vertices A, B and C respectively. Let AD, BE and CF be the internal bisectors of $∠ A, ∠ B$ and $∠ C$ respectively.
We know that D divides BC in the ratio of AB : AC that is c : b.
Then,
P.V. of D is $\frac{c \vec{γ} + b \vec{β}}{c + b}$ .
P.V. of E is $\frac{c \vec{γ} + a \vec{α}}{c + a}$ .
and P.V. of F is $\frac{a \vec{α} + b \vec{β}}{a + b}$ .

The point dividing AD in the ratio $b + c : a$ is $\frac{a \vec{α} + b \vec{β} + c \vec{γ}}{a + b + c}$ .
The point dividing BE in the ratio of $a + c : b$ is $\frac{a \vec{α} + b \vec{β} + c \vec{γ}}{a + b + c}$ .
The point dividing CF in the ratio of $a + b : c$ is $\frac{a \vec{α} + b \vec{β} + c \vec{γ}}{a + b + c}$ .

Since the point $\frac{a \vec{α} + b \vec{β} + c \vec{γ}}{a + b + c}$ lies on all the three internal bisectors AD, BE and CF.
Hence the internal bisectors are concurrent .

Page No 22.4:

Question 1:

Represent the following graphically:
(i) a displacement of 40 km, 30° east of north
(ii) a displacement of 50 km south-east
(iii) a displacement of 70 km, 40° north of west.

Answer:

(i) The vector $\vec{O P}$ represents the required displacement vector.
(ii) The vector $\vec{O Q}$ represents the required vector.
(iii) The vector $\vec{O R}$ represents the required vector.

Page No 22.4:

Question 2:

Classify the following measures as scalars and vectors:
(i) 15 kg
(ii) 20 kg weight
(iii) 45°
(iv) 10 meters south-east
(v) 50 m/sec²

Answer:

The quantities which have only magnitude and which are not related to any
fixed direction in space are called scaler quantities or simply scalars.
The quantities which have both magnitude and direction are called vector quantities or simply vectors.

(i) Mass - Scalar
(ii) Weight(Force) - Vector
(iii) Angle - Scalar
(iv) Directed Disptance- Vector
(v) Magnitude of acceleration - Scalar

Page No 22.4:

Question 3:

Classify the following as scalars and vector quantities:
(i) Time period
(ii) Distance
(iii) displacement
(iv) Force
(v) Work
(vi) Velocity
(vii) Acceleration

Answer:

The quantities which have only magnitude and which are not related to any
fixed direction in space are called scaler quantities or simply scalars.
The quantities which have both magnitude and direction are called vector quantities or simply vectors.

(i) Scalar
(ii) Scalar
(iii) Vector
(iv)Vector
(v) Scalar
(vi) Vector
(vii) Vector

Page No 22.4:

Question 4:

In Figure ABCD is a regular hexagon, which vectors are:
(i) Collinear
(ii) Equal
(iii) Coinitial
(iv) Collinear but not equal.
Figure

Answer:

(i) Vectors having the same or parallel supports are called collinear vector.
In the given figure the collinear vectors are
$\vec{a}, \vec{d}; \vec{x}, \vec{z}, \vec{b}; \vec{c}, \vec{y}$

(ii) Vectors having the same magnitude and direction are called equal vector.
In the given figure the equal vectors are
$\vec{b,} \vec{x}; \vec{c}, \vec{y}; \vec{a}, \vec{d}$

(iii) Vectors having the same initial point are called co-initial vector.
In the given figure the co-initial vectors are
$\vec{a}, \vec{y}, \vec{z}$

(iv) The vectors which are collinear but not equal are $\vec{b}, \vec{z}; \vec{x}, \vec{z}$

Page No 22.4:

Question 5:

Answer the following as true or false:
(i) $\vec{a}$ and $\vec{a}$ are collinear.
(ii) Two collinear vectors are always equal in magnitude.
(iii) Zero vector is unique.
(iv) Two vectors having same magnitude are collinear.
(v) Two collinear vectors having the same magnitude are equal.

Answer:

(i) True, As vectors having the same and parallel support are collinear.
(ii) False, Collinear vectors are parallel vector not equal vectors.
(iii) False.
(iv) False, Collinear vectors may not have a same magnitude.
(v) False, As two collinear vectors are equal only if they have same length and same sense.

Page No 22.42:

Question 1:

If the position vector of a point (−4, −3) be $\vec{a,}$ find $|\vec{a}|$ .

Answer:

Given a point $(- 4, - 3)$ such that its position vector $\vec{a}$ is given by
$\vec{a} = - 4 \overset{\land}{i} - 3 \overset{\land}{j}$
Then,
$|\vec{a}| = \sqrt{{(- 4)}^{2} + {(- 3)}^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5$

Page No 22.42:

Question 2:

If the position vector $\vec{a}$ of a point (12, n) is such that $|\vec{a}|$ = 13, find the value (s) of n.

Answer:

Given a position vector $\vec{a}$ of a point $(12, n)$ such that,
$\vec{a} = 12 \overset{\land}{i} + n \overset{\land}{j}$
Then,
$|\vec{a}| = \sqrt{12^{2} + n^{2}}$
Also , $|\vec{a}| = 13$ (given)
Thus, we get,

$\sqrt{12^{2} + n^{2}} = 13 \Rightarrow 12^{2} + n^{2} = 169 \Rightarrow n^{2} = 169 - 144 \Rightarrow n^{2} = 25 \Rightarrow n = \pm 5$

Page No 22.42:

Question 3:

Find a vector of magnitude 4 units which is parallel to the vector $\sqrt{3} \hat{i} + \hat{j}$ .

Answer:

Let $\vec{a} = \sqrt{3} \hat{i} + \hat{j}$
Then, $|\vec{a}| = \sqrt{{(\sqrt{3})}^{2} + 1} = \sqrt{3 + 1} = \sqrt{4} = 2$
A unit vector parallel to $\vec{a}$ = $\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{1}{2} (\sqrt{3} \hat{i} + \hat{j})$
Hence, Required vector = $4 \hat{a} = 4 \times \frac{1}{2} (\sqrt{3} \hat{i} + \hat{j}) = 2 \sqrt{3} \hat{i} + 2 \hat{j}$

Page No 22.42:

Question 4:

Express $\vec{A B}$ in terms of unit vectors $\hat{i}$ and $\hat{j}$ , when the points are:
(i) A (4, −1), B (1, 3)
(ii) A (−6, 3), B (−2, −5)
Find $|\vec{A} B|$ in each case.

Answer:

(i) Given: $A (4, - 1)$ and $B (1, 3)$
Then the position vector $\vec{A B}$ is given by
$\vec{A B}$ = Position vector of $B$ $-$ Position vector of $A$
$= (\overset{\land}{i} + 3 \overset{\land}{j}) - (4 \overset{\land}{i} - \overset{\land}{j}) = \overset{\land}{i} + 3 \overset{\land}{j} - 4 \overset{\land}{i} + \overset{\land}{j} = - 3 \overset{\land}{i} + 4 \overset{\land}{j}$
So, $|\vec{A B}| = \sqrt{{(- 3)}^{2} + 4^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5$

(ii) Given: $A (- 6, 3)$ and $B (- 2, - 5)$
Then, the position vector $\vec{A B}$ is given by
$\vec{A B} =$ Position vector of $B$ $-$ Position vector of $A$
$= (- 2 \overset{\land}{i} - 5 \overset{\land}{j}) - (- 6 \overset{\land}{i} + 3 \overset{\land}{j}) = - 2 \overset{\land}{i} - 5 \overset{\land}{j} + 6 \overset{\land}{i} - 3 \overset{\land}{j} = 4 \overset{\land}{i} - 8 \overset{\land}{j}$
So, $|\vec{A B}| = \sqrt{4^{2} + {(- 8)}^{2}} = \sqrt{16 + 64} = \sqrt{80} = 4 \sqrt{5}$

Page No 22.42:

Question 5:

Find the coordinates of the tip of the position vector which is equivalent to $\vec{A} B$ , where the coordinates of A and B are (−1, 3) and (−2, 1) respectively.

Answer:

Let $O$ be the origin. Let $P (x, y)$ be the required point. Then, $\vec{P}$ is the tip of the position vector $\vec{O P}$ of the point $P$ .
We have,
$\vec{O P} = x \overset{⏜}{i} + y \overset{⏜}{j} .$
and, $\vec{A B} =$ Position vector of $B$ $-$ Position vector of $A$ .
$= (- 2 \overset{⏜}{i} + \overset{⏜}{j}) - (- \overset{⏜}{i} + 3 \overset{⏜}{j}) = - 2 \overset{⏜}{i} + \overset{⏜}{j} + \overset{⏜}{i} - 3 \overset{⏜}{j} = - \overset{⏜}{i} - 2 \overset{⏜}{j}$

Given that $\vec{O P} = \vec{A B}$
So, $x \overset{⏜}{i} + y \overset{⏜}{j} = - \overset{⏜}{i} - 2 \overset{⏜}{j} \Leftrightarrow x = - 1, y = - 2$
Hence, coordinated of the required point is $(- 1 . - 2)$

Page No 22.43:

Question 6:

ABCD is a parallelogram. If the coordinates of A, B, C are (−2, −1), (3, 0) and (1, −2) respectively, find the coordinates of D.

Answer:

Let the coordinates of $D$ is $(x, y)$ .
Since, $A B C D$ is a parallelogram.
∴ $A B = D C$
We have,

$\vec{A B} = \vec{D C} \Rightarrow 3 \overset{⏜}{i} - (- 2 \overset{⏜}{i} - \overset{⏜}{j}) = (\overset{⏜}{i} - 2 \overset{⏜}{j}) - (x \overset{⏜}{i} + y \overset{⏜}{j}) \Rightarrow 5 \overset{⏜}{i} + \overset{⏜}{j} = \overset{⏜}{i} (1 - x) + \overset{⏜}{j} (- 2 - y) \Rightarrow 1 - x = 5 and 1 = - 2 - y \Rightarrow x = - 4 and y = - 3$
Hence, the coordinates of $D$ is $(- 4, - 3)$

Page No 22.43:

Question 7:

If the position vectors of the points A (3, 4), B (5, −6) and C (4, −1) are $\vec{a,}$ $\vec{b,}$ $\vec{c}$ respectively, compute $\vec{a} + 2 \vec{b} - 3 \vec{c}$ .

Answer:

Let $\vec{a}, \vec{b}, \vec{c}$ are the position vectors of the points $A (3, 4)$ , $B (5, - 6)$ and $C (4, - 1)$ .
Then,
$\vec{a} = 3 \overset{⏜}{i} + 4 \overset{⏜}{j}$
   $\vec{b} = 5 \overset{⏜}{i} - 6 \overset{⏜}{j}$
   $\vec{c} = 4 \overset{⏜}{i} - \overset{⏜}{j}$
Therefore,
   $\vec{a} + 2 \vec{b} - 3 \vec{c} = 3 \overset{⏜}{i} + 4 \overset{⏜}{j} + 2 (5 \overset{⏜}{i} - 6 \overset{⏜}{j}) - 3 (4 \overset{⏜}{i} - \overset{⏜}{j}) = 3 \overset{⏜}{i} + 4 \overset{⏜}{j} + 10 \overset{⏜}{i} - 12 \overset{⏜}{j} - 12 \overset{⏜}{i} + 3 \overset{⏜}{j} = \overset{⏜}{i} - 5 \overset{⏜}{j}$

Page No 22.43:

Question 8:

If $\vec{a}$ be the position vector whose tip is (5, −3), find the coordinates of a point B such that $\vec{A B} =$ $\vec{a}$ , the coordinates of A being (4, −1).

Answer:

Let $O$ be the origin and let $P (5, - 3)$ be the tip of the position vector $\vec{a}$ . Then, $\vec{a} = \vec{O P} = 5 \hat{i} - 3 \hat{j} .$ Let the coordinate of $B$ be $(x, y)$ and $A$ has coordinates $(4, - 1)$ .
Therefore,
$\vec{A B}$ = Position vector of $B$ $-$ Position vector of $A$
$= (x \hat{i} + y \hat{j}) - (4 \hat{i} - \hat{j}) = (x - 4) \hat{i} + (y + 1) \bar{j}$
Now,
$\vec{A B} = \vec{a} \Rightarrow (x - 4) \hat{i} + (y + 1) \hat{j} = 5 \hat{i} - 3 \hat{j} \Rightarrow x - 4 = 5 and y + 1 = - 3 \Rightarrow x = 9 and y = - 4$
Hence, the coordinates of $B$ are $(9, - 4)$ .

Page No 22.43:

Question 9:

Show that the points 2 $\hat{i}$ , − $\hat{i}$ − 4 $\hat{j}$ and − $\hat{i}$ + 4 $\hat{j}$ form an isosceles triangle.

Answer:

Given:- The points $A, B, C$ with position vectors $\vec{a}, \vec{b}, \vec{c}$ respectively.
Also, $\vec{a} = 2 \hat{i}$
$\vec{b} = - \hat{i} - 4 \hat{j}$
$\vec{c} = - \hat{i} + 4 \hat{j}$
Then,
$\vec{A B} = \vec{b} - \vec{a} \Rightarrow \vec{A B} = (- \hat{i} - 4 \hat{j}) - 2 \hat{i} \Rightarrow \vec{A B} = - 3 \hat{i} - 4 \hat{j} Now, |\vec{A B}| = \sqrt{{(- 3)}^{2} + {(- 4)}^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5 \vec{B C} = \vec{c} - \vec{b} \Rightarrow \vec{B C} = (- \hat{i} + 4 \hat{j}) - (- \hat{i} - 4 \hat{j}) \Rightarrow \vec{B C} = - \hat{i} + 4 \hat{j} + \hat{i} + 4 \hat{j} \Rightarrow \vec{B C} = 8 \hat{j}$
and

$\vec{A C} = \vec{c} - \vec{a} \Rightarrow \vec{A C} = (- \hat{i} + 4 \hat{j}) - 2 \hat{i} \Rightarrow \vec{A C} = - 3 \hat{i} + 4 \hat{j} Now, |\vec{A C}| = \sqrt{{(- 3)}^{2} + {(4)}^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5$

Since, the magnitude of AB and AC is equal.
Hence, the points 2 $\hat{i}$ , − $\hat{i}$ − 4 $\hat{j}$ and − $\hat{i}$ + 4 $\hat{j}$ form an isosceles triangle.

Page No 22.43:

Question 10:

Find a unit vector parallel to the vector $\hat{i} + \sqrt{3} \hat{j}$ .

Answer:

Let $\vec{a} = \hat{i} + \sqrt{3} \hat{j}$
Then, $|\vec{a}| = \sqrt{1^{2} + (\sqrt{3})^{2}} = \sqrt{1 + 3} = \sqrt{4} = 2$
Unit vector parallel to $\vec{a}$ = $\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{1}{2} (\hat{i} + \sqrt{3} \hat{j}) = \frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j}$

Page No 22.43:

Question 11:

The position vectors of points $A, B$ and $C$ are $λ \hat{i} +$ 3 $\hat{j}$ , 12 $\hat{i} + μ$ $\hat{j}$ and 11 $\hat{i} -$ 3 $\hat{j}$ respectively. If C divides the line segment joining A and B in the ratio 3:1, find the values of $λ$ and $μ$

Answer:

The position vectors of points A, B and C are $λ \hat{i} +$ 3 $\hat{j}$ , 12 $\hat{i} + μ$ $\hat{j}$ and 11 $\hat{i} -$ 3 $\hat{j}$ , respectively.

It is given that, C divides the line segment joining A and B in the ratio 3 : 1.

$11 \hat{i} - 3 \hat{j} = \frac{3 \times (12 \hat{i} + μ \hat{j}) + 1 \times (λ \hat{i} + 3 \hat{j})}{3 + 1} \Rightarrow 11 \hat{i} - 3 \hat{j} = \frac{(36 + λ) \hat{i} + (3 μ + 3) \hat{j}}{4} \Rightarrow 44 \hat{i} - 12 \hat{j} = (36 + λ) \hat{i} + (3 μ + 3) \hat{j}$
Equating the corresponding components, we get

$36 + λ = 44$

$\Rightarrow λ = 44 - 36 = 8$

and

$3 μ + 3 = - 12$

$\Rightarrow 3 μ = - 12 - 3 = - 15$

$\Rightarrow μ = - 5$

Thus, the values of $λ$ and $μ$ are 8 and −5, respectively.

Page No 22.43:

Question 12:

Find the components along the coordinate axes of the position vector of each of the following points:
(i) P(3, 2)
(ii) Q(–5, 1)
(iii) R(–11, –9)
(iv) S(4, –3)

Answer:

(i) Let O be the origin.
    The position vector of point P(3,2), $\vec{O P} = 3 \hat{i} + 2 \hat{j}$
    Component of $\vec{O P}$ along x-axis = a vector of magnitude 3 having its direction along the positive direction of x-axis.
    Component of $\vec{O P}$ along y-axis = a vector of magnitude 2 having its direction along the positive direction of y-axis.

(ii) The position vector of point Q(-5,1), $\vec{O Q} = - 5 \hat{i} + \hat{j}$
    Component of $\vec{O Q}$ along x-axis = a vector of magnitude 5 having its direction along the negative direction of x-axis.
    Component of $\vec{O Q}$ along y-axis = a vector of magnitude 1 having its direction along the positive direction of y-axis.

(iii) The position vector of point R(-11,-9), $\vec{O R} = - 11 \hat{i} - 9 \hat{j}$
     Component of $\vec{O R}$ along x-axis = a vector of magnitude 11 having its direction along the negative direction of x-axis.
     Component of $\vec{O R}$ along y-axis = a vector of magnitude 9 having its direction along the negative direction of y-axis.

(iv) The position vector of point S(4,-3), $\vec{O S} = 4 \hat{i} - 3 \hat{j}$
     Component of $\vec{O S}$ along x-axis = a vector of magnitude 4 having its direction along the positive direction of x-axis.
     Component of $\vec{O S}$ along y-axis = a vector of magnitude 3 having its direction along the negative direction of y-axis.

Page No 22.48:

Question 1:

Find the magnitude of the vector $\vec{a} = 2 \hat{i} + 3 \hat{j} - 6 \hat{k} .$

Answer:

Given: $\vec{a} = 2 \hat{i} + 3 \hat{j} - 6 \hat{k}$
∴ Magnitude of the vector = $|\vec{a}| = \sqrt{2^{2} + 3^{2} + {(- 6)}^{2}} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$

Page No 22.48:

Question 2:

Find the unit vector in the direction of $3 \hat{i} + 4 \hat{j} - 12 \hat{k} .$

Answer:

Let $\vec{a} = 3 \hat{i} + 4 \hat{j} - 12 \hat{k}$
Then, $|\vec{a}| = \sqrt{3^{2} + 4^{2} + {(- 12)}^{2}} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$
So, a unit vector in the direction of $\vec{a}$ is given by
$\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{1}{13} (3 \hat{i} + 4 \hat{j} - 12 \hat{k}) = \frac{3}{13} \hat{i} + \frac{4}{13} \hat{j} - \frac{12}{13} \hat{k}$

Page No 22.48:

Question 3:

Find a unit vector in the direction of the resultant of the vectors $\hat{i} - \hat{j} + 3 \hat{k}, 2 \hat{i} + \hat{j} - 2 \hat{k} and \hat{i} + 2 \hat{j} - 2 \hat{k} .$

Answer:

Given: $\vec{a} = \hat{i} - \hat{j} + 3 \hat{k}, \vec{b} = 2 \hat{i} + \hat{j} - 2 \hat{k}$ and $\vec{c} = \hat{i} + 2 \hat{j} - 2 \hat{k}$ are the position vectors.
Then, Resultant of the vectors = $\vec{a} + \vec{b} + \vec{c}$
$= \hat{i} - \hat{j} + 3 \hat{k} + 2 \hat{i} + \hat{j} - 2 \hat{k} + \hat{i} + 2 \hat{j} - 2 \hat{k} = 4 \hat{i} + 2 \hat{j} - \hat{k}$

So, $|\vec{a} + \vec{b} + \vec{c}| = \sqrt{4^{2} + 2^{2} + 1^{2}} = \sqrt{16 + 4 + 1} = \sqrt{21}$
∴ Unit vector in the direction of the resultant vector = $\frac{\vec{a} + \vec{b} + \vec{c}}{|\vec{a} + \vec{b} + \vec{c}|} = \frac{1}{\sqrt{21}} (4 \hat{i} + 2 \hat{j} - \hat{k})$

Page No 22.49:

Question 4:

The adjacent sides of a parallelogram are represented by the vectors $\vec{a} = \hat{i} + \hat{j} - \hat{k} and \vec{b} = - 2 \hat{i} + \hat{j} + 2 \hat{k} .$ Find unit vectors parallel to the diagonals of the parallelogram.

Answer:

$\begin{array}{l} \vec{a} = \hat{i} + \hat{j} - \hat{k} \\ \vec{b} = - 2 \hat{i} + \hat{j} + 2 \hat{k} \end{array}$

$\begin{array}{ccl} \vec{A C} & = & \vec{a} + \vec{b} \\ = & - \hat{i} + 2 \hat{j} + \hat{k} \end{array}$

$Let the unit vector along the diagonals A C and B D of the parallelogram be \hat{A C} and \hat{B D} . \Rightarrow \hat{A C} = \frac{- \hat{i} + 2 \hat{j} + \hat{k}}{\sqrt{6}}$

$\begin{matrix} Similarly, \vec{B D} & = & \vec{b} - \vec{a} \\ = & - 3 \hat{i} + 3 \hat{k} \end{matrix}$

$\Rightarrow \hat{B D} = \frac{- 3 \hat{i} + 3 \hat{k}}{3 \sqrt{2}} = \frac{- \hat{i} + k}{\sqrt{2}}$

Page No 22.49:

Question 5:

$If \vec{a} = 3 \hat{i} - \hat{j} - 4 \hat{k}, \vec{b} = - 2 \hat{i} + 4 \hat{j} - 3 \hat{k} and \vec{c} = \hat{i} + 2 \hat{j} - \hat{k}, find |3 \vec{a} - 2 \vec{b} + 4 \vec{c}| .$

Answer:

Given: $\vec{a} = 3 \hat{i} - \hat{j} - 4 \hat{k}, \vec{b} = - 2 \hat{i} + 4 \hat{j} - 3 \hat{k}$ and $\vec{c} = \hat{i} + 2 \hat{j} - \hat{k} .$

$Now, 3 \vec{a} - 2 \vec{b} + 4 \vec{c} = 3 (3 \hat{i} - \hat{j} - 4 \hat{k}) - 2 (- 2 \hat{i} + 4 \hat{j} - 3 \hat{k}) + 4 (\hat{i} + 2 \hat{j} - \hat{k}) = 9 \hat{i} - 3 \hat{j} - 12 \hat{k} + 4 \hat{i} - 8 \hat{j} + 6 \hat{k} + 4 \hat{i} + 8 \hat{j} - 4 \hat{k} = 17 \hat{i} - 3 \hat{j} - 10 \hat{k}$
Hence, $|3 \vec{a} - 2 \vec{b} + 4 \vec{c}| = \sqrt{17^{2} + {(- 3)}^{2} + {(- 10)}^{2}} = \sqrt{289 + 9 + 100} = \sqrt{398}$

Page No 22.49:

Question 6:

If $\vec{P Q} = 3 \hat{i} + 2 \hat{j} - \hat{k}$ and the coordinates of P are (1, −1, 2), find the coordinates of Q.

Answer:

Given: $\vec{P Q} = 3 \hat{i} + 2 \hat{j} - \hat{k} .$ Let the position vector of $P$ $(1, - 1, 2)$ is $\vec{p}$ such that $\vec{p} = \hat{i} - \hat{j} + 2 \hat{k}$ and the position vector of $Q (x, y, z)$ is $\vec{q}$ such that $\vec{q} = x \hat{i} + y \hat{j} + z \hat{k} .$
Therefore,
$∴ \vec{P Q} = \vec{q} - \vec{p} \Rightarrow 3 \hat{i} + 2 \hat{j} - \hat{k} = (x \hat{i} + y \hat{j} + z \hat{k}) - (\hat{i} - \hat{j} + 2 \hat{k}) \Rightarrow 3 \hat{i} + 2 \hat{j} - \hat{k} = (x - 1) \hat{i} + (y + 1) \hat{j} + (z - 2) \hat{k} \Rightarrow x - 1 = 3, y + 1 = 2, z - 2 = - 1 \Rightarrow x = 4, y = 1, z = 1$
Hence, the coordinates of $Q$ are $(4, 1, 1)$

Page No 22.49:

Question 7:

Prove that the points $\hat{i} - \hat{j}, 4 \hat{i} + 3 \hat{j} + \hat{k} and 2 \hat{i} - 4 \hat{j} + 5 \hat{k}$ are the vertices of a right-angled triangle.

Answer:

Given the points $\hat{i} - \hat{j}, 4 \hat{i} + 3 \hat{j} + \hat{k}$ and $2 \hat{i} - 4 \hat{j} + 5 \hat{k}$ Are A, B and C respectively.
Then,
$\vec{A B} = 4 \hat{i} + 3 \hat{j} + \hat{k} - \hat{i} + \hat{j} = 3 \hat{i} + 4 \hat{j} + \hat{k} . \vec{B C} = 2 \hat{i} - 4 \hat{j} + 5 \hat{k} - 4 \hat{i} - 3 \hat{j} - \hat{k} = - 2 \hat{i} - 7 \hat{j} + 4 \hat{k} . \vec{C A} = \hat{i} - \hat{j} - 2 \hat{i} + 4 \hat{j} - 5 \hat{k} = - \hat{i} + 3 \hat{j} - 5 \hat{k} .$
$\vec{A B} + \vec{B C} + \vec{C A} = 3 \hat{i} + 4 \hat{j} + \hat{k} - 2 \hat{i} - 7 \hat{j} + 4 \hat{k} - \hat{i} + 3 \hat{j} - 5 \hat{k} = \vec{0} .$
The given points forms a vertices of a triangle.
Now,
$|\vec{A B}| = \sqrt{9 + 16 + 1} = \sqrt{26} . |\vec{B C}| = \sqrt{4 + 49 + 16} = \sqrt{69} . |\vec{C A}| = \sqrt{1 + 9 + 25} = \sqrt{35} .$
${|\vec{A B}|}^{2} + {|\vec{C A}|}^{2} = 26 + 35 = 61$ ≠ ${|\vec{B C}|}^{2}$
The given triangle is not right-angled.

Page No 22.49:

Question 8:

If the vertices of a triangle are the points with position vectors $a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}, b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}, c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k},$ what are the vectors determined by its sides? Find the length of these vectors.

Answer:

Given the vertices of a triangle A, B and C with position vectors $a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}, b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$ and $c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k}$ respectively. Then,
$\vec{A B} = (b_{1} - a_{1}) \hat{i} + (b_{2} - a_{2}) \hat{j} + (b_{3} - a_{3}) \hat{k} . \vec{B C} = (c_{1} - b_{1}) \hat{i} + (c_{2} - b_{2}) \hat{j} + (b_{3} - a_{3}) \hat{k} . \vec{C A} = (a_{1} - c_{1}) \hat{i} + (a_{2} - c_{2}) \hat{j} + (a_{3} - c_{3}) \hat{k} .$
Therefore, the length of these vectors are:

$|\vec{A B}| = \sqrt{(b_{1} - a_{1})^{2} + (b_{2} - a_{2})^{2} + (b_{3} - a_{3})^{2}} . |\vec{B C}| = \sqrt{(c_{1} - b_{1})^{2} + (c_{2} - b_{2})^{2} + (c_{3} - b_{3})^{2}} . |\vec{C A}| = \sqrt{(a_{1} - c_{1})^{2} + (a_{2} - c_{2})^{2} + (a_{3} - c_{3})^{2}} .$

Page No 22.49:

Question 9:

Find the vector from the origin O to the centroid of the triangle whose vertices are (1, −1, 2), (2, 1, 3) and (−1, 2, −1).

Answer:

Given the vertices of the triangle $(1, - 1, 2), (2, 1, 3)$ and $(- 1, 2, - 1)$ . Then,
Position vectors are
$\vec{a} = \hat{i} - \hat{j} + 2 \hat{k} . \vec{b} = 2 \hat{i} + \hat{j} + 3 \hat{k} . \vec{c} = - \hat{i} + 2 \hat{j} - \hat{k} .$
The centroid of a triangle is given by $\frac{\vec{a} + \vec{b} + \vec{c}}{3}$
So, $\frac{\vec{a} + \vec{b} + \vec{c}}{3} = \frac{\hat{i} - \hat{j} + 2 \hat{k} + 2 \hat{i} + \hat{j} + 3 \hat{k} - \hat{i} + 2 \hat{j} - \hat{k}}{3} = \frac{2 \hat{i} + 2 \hat{j} + 4 \hat{k}}{3} = \frac{2}{3} \hat{i} + \frac{2}{3} \hat{j} + \frac{4}{3} \hat{k}$ .

Page No 22.49:

Question 10:

Find the position vector of a point R which divides the line segment joining points $P (\hat{i} + 2 \hat{j} + \hat{k}) and Q (- \hat{i} + \hat{j} + \hat{k})$ in the ratio 2:1.
(i) internally
(ii) externally

Answer:

(i) Given: R divides the line segment joining the points $P (\hat{i} + 2 \hat{j} + \hat{k}), Q (- \hat{i} + \hat{j} + \hat{k})$ in the ratio 2 : 1 internally.
Therefore. position vector of R = $\frac{2 (- \hat{i} + \hat{j} + \hat{k}) + 1 (\hat{i} + 2 \hat{j} + \hat{k})}{2 + 1}$
= $\frac{1}{3} (- \hat{i} + 4 \hat{j} + 3 \hat{k})$

(ii) Given: R divides the line segment joining the points $P (\hat{i} + 2 \hat{j} + \hat{k}), Q (- \hat{i} + \hat{j} + \hat{k})$ in the ratio 2 : 1 externally.
Therefore. position vector of R = $\frac{2 (- \hat{i} + \hat{j} + \hat{k}) - 1 (\hat{i} + 2 \hat{j} + \hat{k})}{2 - 1}$
= $- 3 \hat{i} + \hat{k}$

Page No 22.49:

Question 11:

Find the position vector of the mid-point of the vector joining the points $P (2 \hat{i} - 3 \hat{j} + 4 \hat{k}) and Q (4 \hat{i} + \hat{j} - 2 \hat{k}) .$

Answer:

Given: $P (2 \hat{i} - 3 \hat{j} + 4 \hat{k})$ and $Q (4 \hat{i} + \hat{j} - 2 k)$
The position vector of the midpoint of the vector
joining these points = $\frac{Position vector of P + Position vector of Q}{2}$
$= \frac{(2 \hat{i} - 3 \hat{j} + 4 \hat{k}) + (4 \hat{i} + \hat{j} - 2 \hat{k})}{2} = \frac{6 \hat{i} - 2 \hat{j} + 2 \hat{k}}{2} = 3 \hat{i} - \hat{j} + \hat{k}$

Page No 22.49:

Question 12:

Find the unit vector in the direction of vector $\vec{P Q},$ where P and Q are the points (1, 2, 3) and (4, 5, 6).

Answer:

Let $\vec{a}$ and $\vec{b}$ are the position vectors of the points $P (1, 2, 3)$ and $Q (4, 5, 6)$
Then,
$\vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k} \vec{b} = 4 \hat{i} + 5 \hat{j} + 6 \hat{k}$
So,
$\vec{P Q} = \vec{b} - \vec{a} = 4 \hat{i} + 5 \hat{j} + 6 \hat{k} - \hat{i} - 2 \hat{j} - 3 \hat{k} = 3 \hat{i} + 3 \hat{j} + 3 \hat{k}$

Now, $|\vec{P Q}| = \sqrt{3^{2} + 3^{2} + 3^{2}} = \sqrt{9 + 9 + 9} = 3 \sqrt{3}$
Therefore, Unit vector parallel to $\vec{P Q}$ = $\frac{\vec{P Q}}{|P Q|} = \frac{1}{3 \sqrt{3}} (3 \hat{i} + 3 \hat{j} + 3 \hat{k}) = \frac{1}{\sqrt{3}} (\hat{i} + \hat{j} + \hat{k})$

Page No 22.49:

Question 13:

Show that the points $A (2 \hat{i} - \hat{j} + \hat{k}), B (\hat{i} - 3 \hat{j} - 5 \hat{k}), C (3 \hat{i} - 4 \hat{j} - 4 \hat{k})$ are the vertices of a right angled triangle.

Answer:

Given the points $A (2 \hat{i} - \hat{j} + \hat{k}), B (\hat{i} - 3 \hat{j} - 5 \hat{k})$ and $C (3 \hat{i} - 4 \hat{j} - 4 \hat{k}) .$
Then,
$\vec{A B} =$ Position vector of $B$ $-$ Position vector of A
   $= \hat{i} - 3 \hat{j} - 5 \hat{k} - (2 \hat{i} - \hat{j} + \hat{k}) = \hat{i} - 3 \hat{j} - 5 \hat{k} - 2 \hat{i} + \hat{j} - \hat{k} = - \hat{i} - 2 \hat{j} - 6 \hat{k}$

$\vec{B C} =$ Position vector of $C$ $-$ Position vector of $B$
   $= 3 \hat{i} - 4 \hat{j} - 4 \hat{k} - (\hat{i} - 3 \hat{j} - 5 \hat{k}) = 3 \hat{i} - 4 \hat{j} - 4 \hat{k} - \hat{i} + 3 \hat{j} + 5 \hat{k} = 2 \hat{i} - \hat{j} + \hat{k}$

$\vec{C A} =$ Position vector of $A$ $-$ Position vector of $C$
   $= 2 \hat{i} - \hat{j} + \hat{k} - (3 \hat{i} - 4 \hat{j} - 4 \hat{k}) = 2 \hat{i} - \hat{j} + \hat{k} - 3 \hat{i} + 4 \hat{j} + 4 \hat{k} = - \hat{i} + 3 \hat{j} + 5 \hat{k}$
Clearly, $\vec{A B} + \vec{B C} + \vec{C A} = \vec{0}$

$Now, \vec{|A B|} = \sqrt{{(- 1)}^{2} + {(- 2)}^{2} + {(- 6)}^{2}} = \sqrt{1 + 4 + 36} = \sqrt{41} \vec{|B C|} = \sqrt{{(2)}^{2} + {(- 1)}^{2} + {(1)}^{2}} = \sqrt{4 + 1 + 1} = \sqrt{6} \vec{|C A|} = \sqrt{{(- 1)}^{2} + {(3)}^{2} + {(5)}^{2}} = \sqrt{1 + 9 + 25} = \sqrt{35} Clearly, {\vec{|A B|}}^{2} = {\vec{|B C|}}^{2} + {\vec{|C A|}}^{2} \Rightarrow A B^{2} = B C^{2} + C A^{2}$

So, $A, B, C$ forms a right angled triangle.

Page No 22.49:

Question 14:

Find the position vector of the mid-point of the vector joining the points P (2, 3, 4) and Q (4, 1, −2).

Answer:

Let $\vec{p}, \vec{q}$ be the position vectors of the points $P (2, 3, 4), Q (4, 1, - 2)$
Then,
$\vec{p} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$ and $\vec{q} = 4 \hat{i} + \hat{j} - 2 \hat{k}$
Therefore, the position vector of the midpoint of the given points is $\frac{\vec{p} + \vec{q}}{2}$
∴ $\frac{\vec{p} + \vec{q}}{2} = \frac{(2 \hat{i} + 3 \hat{j} + 4 \hat{k}) + (4 \hat{i} + \hat{j} - 2 \hat{k})}{2} = 3 \hat{i} + 2 \hat{j} + \hat{k}$

Page No 22.49:

Question 15:

Find the value of x for which $x (\hat{i} + \hat{j} + \hat{k})$ is a unit vector.

Answer:

We have, $x (\hat{i} + \hat{j} + \hat{k})$ is a unit vector.
$∴ \sqrt{x^{2} + x^{2} + x^{2}} = 1 \Rightarrow \sqrt{3} |x| = 1 \Rightarrow |x| = \frac{1}{\sqrt{3}} \Rightarrow x = \pm \frac{1}{\sqrt{3}}$

Page No 22.49:

Question 16:

If $\vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = 2 \hat{i} - \hat{j} + 3 \hat{k} and \vec{c} = \hat{i} - 2 \hat{j} + \hat{k},$ find a unit vector parallel to $2 \vec{a} - \vec{b} + 3 \vec{c .}$

Answer:

We have, $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ , $\vec{b} = 2 \hat{i} - \hat{j} + 3 \hat{k}$ and $\vec{c} = \hat{i} - 2 \hat{j} + \hat{k}$
∴ $2 \vec{a} - \vec{b} + 3 \vec{c} = 2 (\hat{i} + \hat{j} + \hat{k}) - (2 \hat{i} - \hat{j} + 3 \hat{k}) + 3 (\hat{i} - 2 \hat{j} + \hat{k}) = 3 \hat{i} - 3 \hat{j} + 2 \hat{k} .$

A unit vector parallel to $2 \vec{a} - \vec{b} + 3 \vec{c}$ is given by $\frac{2 \vec{a} - \vec{b} + 3 \vec{c}}{|2 \vec{a} - \vec{b} + 3 \vec{c}|}$ $= \frac{(3 \hat{i} - 3 \hat{j} + 2 \hat{k})}{\sqrt{3^{2} + {(- 3)}^{2} + 2^{2}}}$
$= \frac{(3 \hat{i} - 3 \hat{j} + 2 \hat{k})}{\sqrt{22}} = \frac{3}{\sqrt{22}} \hat{i} - \frac{3}{\sqrt{22}} \hat{j} + \frac{2}{\sqrt{22}} \hat{k}$

Page No 22.49:

Question 17:

If $\vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = 4 \hat{i} - 2 \hat{j} + 3 \hat{k} and \vec{c} = \hat{i} - 2 \hat{j} + \hat{k},$ find a vector of magnitude 6 units which is parallel to the vector $2 \vec{a} - \vec{b} + 3 \vec{c .}$

Answer:

We have, $\vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = 4 \hat{i} - 2 \hat{j} + 3 \hat{k}$ and $\vec{c} = \hat{i} - 2 \hat{j} + \hat{k} .$
Then,
$2 \vec{a} - \vec{b} + 3 \vec{c} = 2 (\hat{i} + \hat{j} + \hat{k}) - (4 \hat{i} - 2 \hat{j} + 3 \hat{k}) + 3 (\hat{i} - 2 \hat{j} + \hat{k}) = \hat{i} - 2 \hat{j} + 2 \hat{k} .$
∴ A unit vector parallel to $2 \vec{a} - \vec{b} + 3 \vec{c}$ is $\frac{2 \vec{a} - \vec{b} + 3 \vec{c}}{|2 \vec{a} - \vec{b} + 3 \vec{c}|}$ $= \frac{(\hat{i} - 2 \hat{j} + 2 \hat{k})}{\sqrt{1^{2} + {(- 2)}^{2} + 2^{2}}}$

$= \frac{(\hat{i} - 2 \hat{j} + 2 \hat{k})}{\sqrt{9}} = \frac{(\hat{i} - 2 \hat{j} + 2 \hat{k})}{3}$
Hence, Required vector = $\frac{6}{3} (\hat{i} - 2 \hat{j} + 2 \hat{k}) = 2 \hat{i} - 4 \hat{j} + 4 \hat{k} .$

Page No 22.49:

Question 18:

Find a vector of magnitude of 5 units parallel to the resultant of the vectors $\vec{a} = 2 \hat{i} + 3 \hat{j} - \hat{k} and \vec{b} = \hat{i} - 2 \hat{j} + \hat{k .}$

Answer:

Given the position vectors $\vec{a} = 2 \hat{i} + 3 \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - 2 \hat{j} + \hat{k}$
∴ Resultant Vector = $\vec{a} + \vec{b} = 2 \hat{i} + 3 \hat{j} - \hat{k} + \hat{i} - 2 \hat{j} + \hat{k} = 3 \hat{i} + \hat{j}$
So, a unit vector parallel to the resultant vector is $\frac{3 \hat{i} + \hat{j}}{|3 \hat{i} + \hat{j}|} = \frac{3 \hat{i} + \hat{j}}{\sqrt{3^{2} + 1^{2}}} = \frac{3 \hat{i} + \hat{j}}{\sqrt{10}}$
Hence, required vector = $5 \times \frac{(3 \hat{i} + \hat{j})}{\sqrt{10}} = \sqrt{\frac{5}{2}} (3 \hat{i} + \hat{j})$

Page No 22.49:

Question 19:

The two vectors $\hat{j} + \hat{k}$ and $3 \hat{i} - \hat{j} + 4 \hat{k}$ represents the sides $\vec{AB}$ and $\vec{AC}$ respectively of a triangle ABC. Find the length of the median through A. [CBSE 2015]

Answer:

Disclaimer: The question has been solved by taking the vector $\vec{AB}$ as $\hat{j} + \hat{k}$ .

In ∆ABC, $\vec{AB} = \hat{j} + \hat{k}$ and $\vec{AC} = 3 \hat{i} - \hat{j} + 4 \hat{k}$ .

Let the position vector of A be (0, 0, 0). Then, the position vectors of B and C are (0, 1, 1) and (3, −1, 4), respectively.

Suppose D be the mid-point of the line segment joining the points B(0, 1, 1) and C(3, −1, 4).

∴ Position vector of D $= \frac{(\hat{j} + \hat{k}) + (3 \hat{i} - \hat{j} + 4 \hat{k})}{2} = \frac{3 \hat{i} + 5 \hat{k}}{2} = \frac{3}{2} \hat{i} + \frac{5}{2} \hat{k}$

Now,

Length of the median, AD = $|\vec{AD}| = |(\frac{3}{2} \hat{i} + \frac{5}{2} \hat{k}) - (0 \hat{i} + 0 \hat{j} + 0 \hat{k})| = |\frac{3}{2} \hat{i} + \frac{5}{2} \hat{k}| = \sqrt{{(\frac{3}{2})}^{2} + 0^{2} + {(\frac{5}{2})}^{2}} = \sqrt{\frac{34}{4}} = \sqrt{\frac{17}{2}}$ units

Page No 22.50:

Question 1:

Show that the points A, B, C with position vectors $\vec{a} - 2 \vec{b} + 3 \vec{c}, 2 \vec{a} + 3 \vec{b} - 4 \vec{c}$ and $- 7 \vec{b} + 10 \vec{c}$ are collinear.

Answer:

We have, A, B ,C with position vectors $\vec{a} - 2 \vec{b} + 3 \vec{c}, 2 \vec{a} + 3 \vec{b} - 4 \vec{c}, - 7 \vec{b} + 10 \vec{c}$ Then,
$\vec{A B} =$ Position Vector of B $-$ Position Vector of A
$= 2 \vec{a} + 3 \vec{b} - 4 \vec{c} - \vec{a} + 2 \vec{b} - 3 \vec{c} = \vec{a} + 5 \vec{b} - 7 \vec{c}$

$\vec{B C} =$ Position Vector of C $-$ Position Vector of B
$= - 7 \vec{b} + 10 \vec{c} - 2 \vec{a} - 3 \vec{b} + 4 \vec{c} = - 2 \vec{a} - 10 \vec{b} + 14 \vec{c} = - 2 (\vec{a} + 5 \vec{b} - 7 \vec{c})$
∴ $\vec{B C} = - 2 \vec{A B}$
Hence, $\vec{A B} and \vec{B C}$ are parallel vectors.
But B is a point common to them.
So, $\vec{A B} and \vec{B C}$ are collinear.
Hence, points A, B and C are collinear.

Page No 22.50:

Question 2:

If $\vec{a}$ , $\vec{b}$ , $\vec{c}$ are non-coplanar vectors, prove that the points having the following position vectors are collinear:
(i) $\vec{a,} \vec{b,} 3 \vec{a} - 2 \vec{b}$

(ii) $\vec{a} + \vec{b} + \vec{c}, 4 \vec{a} + 3 \vec{b}, 10 \vec{a} + 7 \vec{b} - 2 \vec{c}$

Answer:

(i) Given: $\vec{a}, \vec{b}, \vec{c}$ are non coplanar vectors.
Let the points be $A, B, C$ respectively with position vectors $\vec{a}, \vec{b,} 3 \vec{a} - 2 \vec{b} .$ Then,
$\vec{A B} =$ Position vector of B $-$ Position vector of A
   $= \vec{b} - \vec{a}$

$\vec{B C} =$ Position vector of C $-$ Position vector of B
$= 3 \vec{a} - 2 \vec{b} - \vec{b} = 3 \vec{a} - 3 \vec{b} = - 3 (\vec{b} - \vec{a})$
∴ $\vec{B C} = - 3 \vec{A B}$
$So, \vec{A B} and \vec{B C}$ are parallel vectors.
But $B$ is a point common to them.
Hence, $A, B$ and $C$ are collinear.

(ii) Given $\vec{a}, \vec{b}, \vec{c}$ are non coplanar vectors.
Let the points be $A, B, C$ respectively with the position vectors $\vec{a} + \vec{b} + \vec{c}, 4 \vec{a} + 3 \vec{b}, 10 \vec{a} + 7 \vec{b} - 2 \vec{c} .$ Then,
$\vec{A B} =$ Position vector of B $-$ Position vector of A
   $= 4 \vec{a} + 3 \vec{b} - \vec{a} - \vec{b} - \vec{c} = 3 \vec{a} + 2 \vec{b} - \vec{c}$

$\vec{B C} =$ Position vector of C $-$ Position vector of B
   $= 10 \vec{a} + 7 \vec{b} - 2 \vec{c} - 4 \vec{a} - 3 \vec{b} = 6 \vec{a} + 4 \vec{b} - 2 \vec{c} = 2 (3 \vec{a} + 2 \vec{b} - \vec{c})$

∴ $\vec{B C} = 2 \vec{A B}$
$So, \vec{A B} and \vec{B C}$ are parallel vectors.
But $B$ is a point common to them.
So, $\vec{A B}$ and $\vec{B C}$ are collinear.
Hence, $A, B$ and $C$ are collinear.

Page No 22.50:

Question 3:

Prove that the points having position vectors $\hat{i} + 2 \hat{j} + 3 \hat{k}, 3 \hat{i} + 4 \hat{j} + 7 \hat{k}, - 3 \hat{i} - 2 \hat{i} - 5 \hat{k}$ are collinear.

Answer:

Let $A, B, C$ be the points with position vectors $\hat{i} + 2 \hat{j} + 3 \hat{k}, 3 \hat{i} + 4 \hat{j} + 7 \hat{k}, - 3 \hat{i} - 2 \hat{j} - 5 \hat{k} .$ Then,
$\vec{A B} =$ Position vector of B $-$ Position vector of A
$= 3 \hat{i} + 4 \hat{j} + 7 \hat{k} - \hat{i} - 2 \hat{j} - 3 \hat{k} = 2 \hat{i} + 2 \hat{j} + 4 \hat{k}$

$\vec{B C} =$ Position vector of C $-$ Position vector of B
$= - 3 \hat{i} - 2 \hat{j} - 5 \hat{k} - 3 \hat{i} - 4 \hat{j} - 7 \hat{k} = - 6 \hat{i} - 6 \hat{j} - 12 \hat{k} = - 3 (2 \hat{i} + 2 \hat{j} + 4 \hat{k})$
∴ $\vec{B C} = - 3 \vec{A B}$
$So, \vec{A B} and \vec{B C}$ are parallel vectors.
But $B$ is a point common to them.
So, $\vec{A B}$ and $\vec{B C}$ are collinear.
Hence, $A, B, C$ are collinear.

Page No 22.50:

Question 4:

If the points with position vectors $10 \hat{i} + 3 \hat{j}, 12 \hat{i} - 5 \hat{j} and a \hat{i} + 11 \hat{j}$ are collinear, find the value of a.

Answer:

Let $A, B, C$ be the points with position vectors $10 \hat{i} + 3 \hat{j}, 12 \hat{i} - 5 \hat{j}, a \hat{i} + 11 \hat{j}$ . Then,
$\vec{A B} =$ Position vector of B $-$ Position vector of A
$= 12 \hat{i} - 5 \hat{j} - 10 \hat{i} - 3 \hat{j} = 2 \hat{i} - 8 \hat{j}$

$\vec{B C} =$ Position vector of C $-$ Position vector of B
$= a \hat{i} + 11 \hat{j} - 12 \hat{i} + 5 \hat{j} = (a - 12) \hat{i} + 16 \hat{j}$

Since, $A, B$ and $C$ are collinear.
∴ $\vec{A B} = λ \vec{B C} .$
$\Rightarrow 2 \hat{i} - 8 \hat{j} = λ (a - 12) \hat{i} + λ 16 \hat{j} \Rightarrow 2 = λ (a - 12), - 8 = λ 16 \Rightarrow 2 = λ (a - 12), λ = - \frac{1}{2} \Rightarrow 2 = - \frac{1}{2} (a - 12) \Rightarrow - a + 12 = 4 \Rightarrow a = 8$

Page No 22.51:

Question 5:

If $\vec{a,} \vec{b}$ are two non-collinear vectors, prove that the points with position vectors $\vec{a} + \vec{b,} \vec{a} - \vec{b} and \vec{a} + λ \vec{b}$ are collinear for all real values of λ.

Answer:

Given: $\vec{a}, \vec{b}$ are non collinear vectors.
Let the position vectors of points A, B and C be $\vec{a} + \vec{b}, \vec{a} - \vec{b}, \vec{a} + λ \hat{b}$ respectively.
Then,
$\vec{A B} =$ P.V. of B − P.V. of A.
$= \vec{a} - \vec{b} - \vec{a} - \vec{b} . = - 2 \vec{b} .$

$\vec{B C} =$ P.V. of C − P.V. of B.
$= \vec{a} + λ \vec{b} - \vec{a} + \vec{b} . = \vec{b} (λ - 1) .$

$\vec{C A} =$ P.V. of A − P.V. of C.
$= \vec{a} + \vec{b} - \vec{a} - λ \vec{b} . = \vec{b} (1 - λ) .$
Now, the position vectors are collinear if and only if $\vec{A B}$ and $\vec{C A}$ is some multiple of $\vec{B C}$ .
So,
$\vec{A B} = β \overset{⇀}{B C} \Rightarrow - 2 b = β \overset{⇀}{b} (λ - 1) \Rightarrow - 2 = β (λ - 1) \Rightarrow β = - \frac{2}{λ - 1}$

and $\vec{B C} = - \vec{C A} .$
Hence, for real values of $λ$ , the given position vectors are parallel.

Page No 22.51:

Question 6:

If $\vec{A O} + \vec{O B} = \vec{B O} + \vec{O C},$ prove that A, B, C are collinear points.

Answer:

We have,
$\vec{A O} + \vec{O B} = \vec{B O} + \vec{O C} . \Rightarrow \vec{A O} - \vec{B O} = \vec{O C} - \vec{O B} . \Rightarrow \vec{O B} - \vec{O A} = \vec{O C} - \vec{O B} . \Rightarrow \vec{A B} = \vec{B C} .$
Hence A, B and C are collinear points.

Page No 22.51:

Question 7:

Show that the vectors $2 \hat{i} - 3 \hat{j} + 4 \hat{k} and - 4 \hat{i} + 6 \hat{j} - 8 \hat{k}$ are collinear.

Answer:

Given the position vectors $2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ and $- 4 \hat{i} + 6 \hat{j} - 8 \hat{k}$
Let $\vec{a} = 2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ and $\vec{b} = - 4 \hat{i} + 6 \hat{j} - 8 \hat{k}$
Then,
$\vec{b} = - 4 \hat{i} + 6 \hat{j} - 8 \hat{k} = - 2 (2 \hat{i} - 3 \hat{j} + 4 \hat{k}) = - 2 \vec{a}$
Hence, $\vec{a}, \vec{b}$ are collinear.

Page No 22.51:

Question 8:

If the points A(m, −1), B(2, 1) and C(4, 5) are collinear, find the value of m.

Answer:

The given points are A(m, −1), B(2, 1) and C(4, 5).

Now,

$\vec{AB} = (2 \hat{i} + \hat{j}) - (m \hat{i} - \hat{j}) = (2 - m) \hat{i} + 2 \hat{j}$

$\vec{AC} = (4 \hat{i} + 5 \hat{j}) - (m \hat{i} - \hat{j}) = (4 - m) \hat{i} + 6 \hat{j}$

If A, B, C are collinear, then

$\vec{AB} = λ \vec{AC} \Rightarrow (2 - m) \hat{i} + 2 \hat{j} = λ [(4 - m) \hat{i} + 6 \hat{j}] \Rightarrow 2 - m = λ (4 - m) and 2 = 6 λ (Equating coefficients of \hat{i} and \hat{j}) \Rightarrow λ = \frac{1}{3}$
and
$2 - m = \frac{1}{3} (4 - m) \Rightarrow 6 - 3 m = 4 - m \Rightarrow 2 m = 2 \Rightarrow m = 1$

Thus, the value of m is 1.

Page No 22.51:

Question 9:

Show that the points (3, 4), (−5, 16) and (5, 1) are collinear.

Answer:

Let the given points be A(3, 4), B(−5, 16) and C(5, 1).

Now,

$\vec{AB} = [(- 5 \hat{i} + 16 \hat{j}) - (3 \hat{i} + 4 \hat{j})] = - 8 \hat{i} + 12 \hat{j}$

$\vec{AC} = [(5 \hat{i} + \hat{j}) - (3 \hat{i} + 4 \hat{j})] = 2 \hat{i} - 3 \hat{j}$

Clearly, $\vec{AB} = - 4 \vec{AC}$

Therefore, $\vec{AB}$ and $\vec{AC}$ are parallel vectors. But, A is a common point of $\vec{AB}$ and $\vec{AC}$ .

Hence, the given points (3, 4), (−5, 16) and (5, 1) are collinear.

Page No 22.51:

Question 10:

If the vectors $\vec{a} = 2 \hat{i} - 3 \hat{j}$ and $\vec{b} = - 6 \hat{i} + m \hat{j}$ are collinear, find the value of m.

Answer:

It is given that the vectors $\vec{a} = 2 \hat{i} - 3 \hat{j}$ and $\vec{b} = - 6 \hat{i} + m \hat{j}$ are collinear.

$∴ \vec{b} = λ \vec{a}$ for some scalar λ

$\Rightarrow - 6 \hat{i} + m \hat{j} = λ (2 \hat{i} - 3 \hat{j}) \Rightarrow - 6 \hat{i} + m \hat{j} = 2 λ \hat{i} - 3 λ \hat{j} \Rightarrow - 6 = 2 λ and m = - 3 λ \Rightarrow m = - 3 \times (\frac{- 6}{2}) = 9$

Thus, the value of m is 9.

Page No 22.51:

Question 11:

Show that the points A (1, −2, −8), B (5, 0, −2) and C (11, 3, 7) are collinear, and find the ratio in which B divides AC.

Answer:

Given points $A (1, - 2, - 8), B (5, 0, - 2), C (11, 3, 7)$ .
Therefore, $\vec{A B} = 5 \hat{i} + 0 \hat{j} - 2 \hat{k} - \hat{i} + 2 \hat{j} + 8 \hat{k} = 4 \hat{i} + 2 \hat{j} + 6 \hat{k}$
$\vec{B C} = 11 \hat{i} + 3 \hat{j} + 7 \hat{k} - 5 \hat{i} + 2 \hat{k} = 6 \hat{i} + 3 \hat{j} + 9 \hat{k}$
and, $\vec{A C} = 11 \hat{i} + 3 \hat{j} + 7 \hat{k} - \hat{i} + 2 \hat{j} + 8 \hat{k} = 10 \hat{i} + 5 \hat{j} + 15 \hat{k}$
Clearly, $\vec{A B} + \vec{B C} = \vec{A C}$
Hence $A, B, C$ are collinear.
Suppose $B$ divides in the ratio AC in the ratio $λ : 1$ . Then the position vector $B$ is
$(\frac{11 λ + 1}{λ + 1}) \hat{i} + (\frac{3 λ - 2}{λ + 1}) \hat{j} + (\frac{7 λ - 8}{λ + 1}) \hat{k}$
But the position vector of $B$ is $5 \hat{i} + 0 \hat{j} - 2 \hat{k} .$

$\frac{11 λ + 1}{λ + 1} = 5, \frac{3 λ - 2}{λ + 1} = 0, \frac{7 λ - 8}{λ + 1} = - 2 \Rightarrow 11 λ + 1 = 5 λ + 5, 3 λ - 2 = 0, 7 λ - 8 = - 2 λ - 2 \Rightarrow 6 λ = 4, 3 λ = 2, 9 λ = 6 \Rightarrow λ = \frac{2}{3}, λ = \frac{2}{3}, λ = \frac{2}{3}$

Page No 22.51:

Question 12:

Using vectors show that the points A (−2, 3, 5), B (7, 0, −1) C (−3, −2, −5) and D (3, 4, 7) are such that AB and CD intersect at the point P (1, 2, 3).

Answer:

We have,
$\vec{AP} = position vector of P - position vector of A \Rightarrow \vec{AP} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) - (- 2 \hat{i} + 3 \hat{j} + 5 \hat{k}) = 3 \hat{i} - \hat{j} - 2 \hat{k} \vec{PB} = position vector of B - position vector of P \Rightarrow \vec{PB} = (7 \hat{i} - 0 \hat{j} - \hat{k}) - (\hat{i} + 2 \hat{j} + 3 \hat{k}) = 6 \hat{i} - 2 \hat{j} - 4 \hat{k} Since \vec{PB} = 2 \vec{AP} . So, vectors \vec{P B} a n d \vec{A P} are collinear . But P is a point common to \vec{PB} and \vec{AP} .$

Hence, P, A, B are collinear points.

$Now, \vec{CP} = (- 3 \hat{i} - 2 \hat{j} - 5 \hat{k}) - (\hat{i} + 2 \hat{j} + 3 \hat{k}) = (- 4 \hat{i} - 4 \hat{j} - 8 \hat{k}) \vec{PD} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) - (3 \hat{i} + 4 \hat{j} + 7 \hat{k}) = (- 2 \hat{i} - 2 \hat{j} - 4 \hat{k}) Thus, \vec{CP} = 2 \vec{PD} . So the vectors \vec{CP} and \vec{PD} are collinear . But P is a common point to \vec{CP} and \vec{PD}$
Hence, C,P,D are collinear points.
Thus A, B, C, D and P are points such that A,P,B and C,P,D are two sets of collinear points.
Hence, AB and CD intersect at point P.

Page No 22.51:

Question 13:

Using vectors, find the value of λ such that the points (λ, −10, 3), (1, −1, 3) and (3, 5, 3) are collinear. [NCERT EXEMPLAR]

Answer:

Let the given points be A(λ, −10, 3), B(1, −1, 3) and C(3, 5, 3).

$\vec{AB} = (\hat{i} - \hat{j} + 3 \hat{k}) - (λ \hat{i} - 10 \hat{j} + 3 \hat{k}) = (1 - λ) \hat{i} + 9 \hat{j}$

$\vec{AC} = (3 \hat{i} + 5 \hat{j} + 3 \hat{k}) - (λ \hat{i} - 10 \hat{j} + 3 \hat{k}) = (3 - λ) \hat{i} + 15 \hat{j}$

If the points A, B, C are collinear, then

$\vec{AB} = k \vec{AC}$ for some scalar k

$\Rightarrow (1 - λ) \hat{i} + 9 \hat{j} = k [(3 - λ) \hat{i} + 15 \hat{j}] \Rightarrow 1 - λ = k (3 - λ) and 9 = 15 k (Equating coefficients of \hat{i} and \hat{j}) \Rightarrow 1 - λ = \frac{3}{5} (3 - λ) \Rightarrow 5 - 5 λ = 9 - 3 λ \Rightarrow 2 λ = - 4 \Rightarrow λ = - 2$

Thus, the value of λ is −2.

Page No 22.65:

Question 1:

Show that the points whose position vectors are as given below are collinear:
(i) $2 \hat{i} + \hat{j} - \hat{k}, 3 \hat{i} - 2 \hat{j} + \hat{k} and \hat{i} + 4 \hat{j} - 3 \hat{k}$

(ii) $3 \hat{i} - 2 \hat{j} + 4 \hat{k}, \hat{i} + \hat{j} + \hat{k} and - \hat{i} + 4 \hat{j} - 2 \hat{k}$

Answer:

(i) Let the points be $A, B$ and $C$ with position vectors $2 \hat{i} + \hat{j} - \hat{k}, 3 \hat{i} - 2 \hat{j} + \hat{k}$ and $\hat{i} + 4 \hat{j} - 3 \hat{k} .$ Then,
   $\vec{A B} =$ Position vector of B $-$ Position vector of A
$= 3 \hat{i} - 2 \hat{j} + \hat{k} - 2 \hat{i} - \hat{j} + \hat{k} = \hat{i} - 3 \hat{j} + 2 \hat{k}$

$\vec{B C} =$ Position vector of C $-$ Position vector of B
$= \hat{i} + 4 \hat{j} - 3 \hat{k} - 3 \hat{i} + 2 \hat{j} - \hat{k} = - 2 \hat{i} + 6 \hat{j} - 4 \hat{k} = - 2 (\hat{i} - 3 \hat{j} + 2 \hat{k})$

$∴ \vec{A B} = - 2 \vec{B C}$ .
$So, \vec{A B}$ and $\vec{B C}$ are parallel vectors. But B is a point common to them.
Hence, $A, B$ and C are collinear.

(ii) Let the points be $A, B$ and $C$ with position vectors $3 \hat{i} - 2 \hat{j} + 4 \hat{k}, \hat{i} + \hat{j} + \hat{k}$ and $- \hat{i} + 4 \hat{j} - 2 \hat{k}$ respectively. Then,
$\vec{A B} =$ Position vector of B $-$ Position vector of A
   $= \hat{i} + \hat{j} + \hat{k} - 3 \hat{i} + 2 \hat{j} - 4 \hat{k} = - 2 \hat{i} + 3 \hat{j} - 3 \hat{k}$

$\vec{B C} =$ Position vector of C $-$ Position vector of B
   $= - \hat{i} + 4 \hat{j} - 2 \hat{k} - \hat{i} - \hat{j} - \hat{k} = - 2 \hat{i} + 3 \hat{j} - 3 \hat{k}$

$∴ \vec{A B} = \vec{B C}$
$So, \vec{A B}$ and $\vec{B C}$ are parallel vectors.But $B$ is a point common to them.
Hence, $A, B$ and $C$ are collinear.

Page No 22.65:

Question 2:

Using vector method, prove that the following points are collinear:
(i) A (6, −7, −1), B (2, −3, 1) and C (4, −5, 0)
(ii) A (2, −1, 3), B (4, 3, 1) and C (3, 1, 2)
(iii) A (1, 2, 7), B (2, 6, 3) and C (3, 10, −1)
(iv) A (−3, −2, −5), B (1, 2, 3) and C (3, 4, 7)

Answer:

(i) Given the points $A (6, - 7, - 1), B (2, - 3, 1)$ and $C (4, - 5, 0)$ . Then,
   $\vec{A B} =$ Position vector of B $-$ Position vector of A
   $= 2 \hat{i} - 3 \hat{j} + \hat{k} - 6 \hat{i} + 7 \hat{j} + \hat{k} = - 4 \hat{i} + 4 \hat{j} + 2 \hat{k} = - 2 (2 \hat{i} - 2 \hat{j} - \hat{k})$

$\vec{B C} =$ Position vector of C $-$ Position vector of B
   $= 4 \hat{i} - 5 \hat{j} - 2 \hat{i} + 3 \hat{j} - \hat{k} = 2 \hat{i} - 2 \hat{j} - \hat{k}$

$∴ \vec{A B} = - 2 \vec{B C}$
$So, \vec{A B}, \vec{B C}$ are parallel vectors. But $B$ is a point common to them.
Hence, the given points $A, B$ and $C$ are collinear.

(ii) Given the points $A (2, - 1, 3), B (4, 3, 1)$ and $C (3, 1, 2)$ . Then,
   $\vec{A B} =$ Position vector of B $-$ Position vector of A
$= 4 \hat{i} + 3 \hat{j} + \hat{k} - 2 \hat{i} + \hat{j} - 3 \hat{k} = 2 \hat{i} + 4 \hat{j} - 2 \hat{k} = - 2 (- \hat{i} - 2 \hat{j} + \hat{k})$

$\vec{B C} =$ Position vector of C $-$ Position vector of B
$= 3 \hat{i} + \hat{j} + 2 \hat{k} - 4 \hat{i} - 3 \hat{j} - \hat{k} = - \hat{i} - 2 \hat{j} + \hat{k}$

   $∴ \vec{A B} = - 2 \vec{B C}$
$So, \vec{A B}, \vec{B C}$ are parallel vectors. But $B$ is a point common to them.
Hence, The given points $A, B$ and $C$ are collinear.

(iii) Given the points $A (1, 2, 7), B (2, 6, 3)$ and $C (3, 10, - 1)$ . Then,
$\vec{A B} =$ Position vector of B $-$ Position vector of A
$= 2 \hat{i} + 6 \hat{j} + 3 \hat{k} - \hat{i} - 2 \hat{j} - 7 \hat{k} = \hat{i} + 4 \hat{j} - 4 \hat{k}$

$\vec{B C} =$ Position vector of C $-$ Position vector of B.
   $= 3 \hat{i} + 10 \hat{j} - \hat{k} - 2 \hat{i} - 6 \hat{j} - 3 \hat{k} = \hat{i} + 4 \hat{j} - 4 \hat{k}$

$∴ \vec{A B} = \vec{B C}$
$So, \vec{A B}, \vec{B C}$ are parallel vectors. But $B$ is a point common to them.
Hence, the given points $A, B$ and $C$ are collinear.

(iv) Given the points $A (- 3, - 2, - 5), B (1, 2, 3)$ and $C (3, 4, 7)$ . Then,
$\vec{A B} =$ Position vector of B $-$ Position vector of A
$= \hat{i} + 2 \hat{j} + 3 \hat{k} + 3 \hat{i} + 2 \hat{j} + 5 \hat{k} = 4 \hat{i} + 4 \hat{j} + 8 \hat{k} = 2 (2 \hat{i} + 2 \hat{j} + 4 \hat{k})$

$\vec{B C} =$ Position vector of C $-$ Position vector of B
$= 3 \hat{i} + 4 \hat{j} + 7 \hat{k} - \hat{i} - 2 \hat{j} - 3 \hat{k} = 2 \hat{i} + 2 \hat{j} + 4 \hat{k}$

$∴ \vec{A B} = 2 \vec{B C}$
$So, \vec{A B}, \vec{B C}$ are parallel vectors. But $B$ is a point common to them.
Hence, the given points $A, B$ and $C$ are collinear.

Page No 22.65:

Question 3:

If $\vec{a}$ , $\vec{b}$ , $\vec{c}$ are non-zero, non-coplanar vectors, prove that the following vectors are coplanar:
(i) $5 \vec{a} + 6 \vec{b} + 7 \vec{c,} 7 \vec{a} - 8 \vec{b} + 9 \vec{c} and 3 \vec{a} + 20 \vec{b} + 5 \vec{c}$

(ii) $\vec{a} - 2 \vec{b} + 3 \vec{c}, - 3 \vec{b} + 5 \vec{c} and - 2 \vec{a} + 3 \vec{b} - 4 \vec{c}$

Answer:

(i) The three vectors are coplanar if one of them is expressible as a linear combination of the other two . Let
$5 \vec{a} + 6 \vec{b} + 7 \vec{c} = x (7 \vec{a} - 8 \vec{b} + 9 \overset{⇀}{c}) + y (3 \vec{a} + 20 \vec{b} + 5 \vec{c}) . = \vec{a} (7 x + 3 y) + \vec{b} (- 8 x + 20 y) + \vec{c} (9 x + 5 y) .$
$\Rightarrow 7 x + 3 y = 5, - 8 x + 20 y = 6 and 9 x + 5 y = 7 .$
Solving first two of these equations, we get $x = \frac{1}{2}, y = \frac{1}{2}$ . Clearly, these values of x and y satisfies the third equation.
Hence, the given vectors are coplanar.

(ii) The three vectors are coplanar if one of them is expressible as a linear combination of the other two. Let
$\vec{a} - 2 \vec{b} + 3 \vec{c} = x (- 3 \vec{b} + 5 \vec{c}) + y (- 2 \vec{a} + 3 \vec{b} - 4 \vec{c}) . = \vec{a} (- 2 y) + \vec{b} (- 3 x + 3 y) + \vec{c} (5 x - 4 y) .$
$\Rightarrow - 2 y = 1, - 3 x + 3 y = - 2 and 5 x - 4 y = 3$
Solving first two of these equations, we get $x = \frac{1}{6}, y = - \frac{1}{2}$ .
These values of x and y does not satisfy the third equation.
Hence, the given vectors are not coplanar.

Page No 22.65:

Question 4:

Show that the four points having position vectors $6 \hat{i} - 7 \hat{j}, 16 \hat{i} - 19 \hat{j} - 4 \hat{k}, 3 \hat{j} - 6 \hat{k}, 2 \hat{i} - 5 \hat{j} + 10 \hat{k}$ are coplanar.

Answer:

Let the given four points be $P, Q, R$ and $S$ respectively. Three points are coplanar if the vectors $\vec{P Q}, \vec{P R}$ and $\vec{P S}$ are coplanar. These vectors are coplanar iff one of them can be expressed as a linear combination of the other two. So, let
$\vec{P Q} = x \vec{P R} + y \vec{P S} . 10 \hat{i} - 12 \hat{j} - 4 \hat{k} = x (- 6 \hat{i} + 10 \hat{j} - 6 \hat{k}) + y (- 4 \hat{i} + 2 \hat{j} + 10 \hat{k}) . \Rightarrow 10 \hat{i} - 12 \hat{j} - 4 \hat{k} = \hat{i} (- 6 x - 4 y) + \hat{j} (10 x + 2 y) + \hat{k} (- 6 x + 10 y) .$
$\Rightarrow - 6 x - 4 y = 10, 10 x + 2 y = - 12$ and $- 6 x + 10 y = - 4 .$ [ Equating coefficients of $\hat{i}, \hat{j}, \hat{k}$ on both sides]
Solving the first of these three equations, we get $x = - 1$ and $y = - 1$ . These values also satisfy the third equation.
Hence, the given four points are coplanar.

Page No 22.65:

Question 5:

Prove that the following vectors are coplanar:
(i) $2 \hat{i} - \hat{j} + \hat{k}, \hat{i} - 3 \hat{j} - 5 \hat{k} and 3 \hat{i} - 4 \hat{j} - 4 \hat{k}$

(ii) $\hat{i} + \hat{j} + \hat{k}, 2 \hat{i} + 3 \hat{j} - \hat{k} and - \hat{i} - 2 \hat{j} + 2 \hat{k}$

Answer:

(i) Given the vectors $P (2 \hat{i} - \hat{j} + \hat{k}), Q (\hat{i} - 3 \hat{j} - 5 \hat{k})$ and $R (3 \hat{i} - 4 \hat{j} - 4 \hat{k})$ .
We know the three vectors are coplanar if one of them is expressible as a linear combination of the other two. Let,
$2 \hat{i} - \hat{j} + \hat{k} = x (\hat{i} - 3 \hat{j} - 5 \hat{k}) + y (3 \hat{i} - 4 \hat{j} - 4 \hat{k}) . = \hat{i} (x + 3 y) + \hat{j} (- 3 x - 4 y) + \hat{k} (- 5 x - 4 y) .$
$\Rightarrow x + 3 y = 2, - 3 x - 4 y = - 1, - 5 x - 4 y = 1$ [Equating the coefficients of $\hat{i}, \hat{j}, \hat{k}$ respectively]
Solving first two of these equation, we get $x = - 1, y = 1$ . Clearly these two values satisfy the third equation.
Hence, the given vectors are coplanar.

(ii) Given the vectors $P (\hat{i} + \hat{j} + \hat{k}), Q (2 \hat{i} + 3 \hat{j} - \hat{k})$ and $R (- \hat{i} - 2 \hat{j} + 2 \hat{k})$ .
We know the three vectors are coplanar if one of them is expressible as a linear combination of the other two. Let,
$\hat{i} + \hat{j} + \hat{k} = x (2 \hat{i} + 3 \hat{j} - \hat{k}) + y (- \hat{i} - 2 \hat{j} + 2 \hat{k}) . = \hat{i} (2 x - y) + \hat{j} (3 x - 2 y) + \hat{k} (- x + 2 y) .$
$\Rightarrow 2 x - y = 1, 3 x - 2 y = 1, - x + 2 y = 1$ [ Equating the coefficients of $\hat{i}, \hat{j}, \hat{k}$ respectvely]
Solving first two of these equation , we get $x = 1, y = 1$ . Clearly these two values satisfy the third equation.
Hence, the given vectors are coplanar.

Page No 22.65:

Question 6:

Prove that the following vectors are non-coplanar:
(i) $3 \hat{i} + \hat{j} - \hat{k}, 2 \hat{i} - \hat{j} + 7 \hat{k} and 7 \hat{i} - \hat{j} + 23 \hat{k}$

(ii) $\hat{i} + 2 \hat{j} + 3 \hat{k}, 2 \hat{i} + \hat{j} + 3 \hat{k} and \hat{i} + \hat{j} + \hat{k}$

Answer:

(i) Let if possible the given vectors are coplanar. Then one of the given vector is expressible in terms of the other two.
We have,
$3 \hat{i} + \hat{j} - \hat{k} = x (2 \hat{i} - \hat{j} + 7 \hat{k}) + y (7 \hat{i} - \hat{j} + 23 \hat{k}) . = \hat{i} (2 x + 7 y) + \hat{j} (- x - y) + \hat{k} (7 x + 23 y) . \Rightarrow 2 x + 7 y = 3, x + y = - 1, 7 x + 23 y = - 1 . By solving the first two equations, we get \Rightarrow x = - 2, y = 1 .$
Clearly these values of x and y does not satisfy the third equation.
Hence the given vectors are non-coplanar.

(ii) Let if possible the given vectors are coplanar. Then one of the given vector is expressible in terms of the other two.
We have,
$\hat{i} + 2 \hat{j} + 3 \hat{k} = x (2 \hat{i} + \hat{j} + 3 \hat{k}) + y (\hat{i} + \hat{j} + \hat{k}) . = \hat{i} (2 x + y) + \hat{j} (x + y) + \hat{k} (3 x + y) . \Rightarrow 2 x + y = 1, x + y = 2, 3 x + y = 3 . By solving the first two equation, we get \Rightarrow x = - 1, y = 3 .$
Clearly these values of x and y does not satisfy the third equation.
Hence the given vectors are non-coplanar.

Page No 22.66:

Question 7:

If $\vec{a}$ , $\vec{b}$ , $\vec{c}$ are non-coplanar vectors, prove that the following vectors are non-coplanar:
(i) $2 \vec{a} - \vec{b} + 3 \vec{c}, \vec{a} + \vec{b} - 2 \vec{c} and \vec{a} + \vec{b} - 3 \vec{c}$

(ii) $\vec{a} + 2 \vec{b} + 3 \vec{c}, 2 \vec{a} + \vec{b} + 3 \vec{c} and \vec{a} + \vec{b} + \vec{c}$

Answer:

(i) Let if possible the following vectors are coplanar. Then one of the vector is expressible in terms of the other two.
We have,
$2 \vec{a} - \vec{b} + 3 \vec{c} = x (\vec{a} + \vec{b} - 2 \vec{c}) y (\vec{a} + \vec{b} - 3 \vec{c}) . = \vec{a} (x + y) + \vec{b} (x + y) + \vec{c} (- 2 x - 3 y) . \Rightarrow x + y = 2, x + y = - 1, - 2 x - 3 y = 3 .$
which is not true, as $x + y = 2$ ≠ $- 1$ . Hence the given vectors are non-coplanar.

(ii) Let if possible the following vector are coplanar. Then one of the vector is expressible in terms of the other two.
We have,
$\vec{a} + 2 \vec{b} + 3 \vec{c} = x (2 \vec{a} + \vec{b} + 3 \vec{c}) + y (\vec{a} + \vec{b} + \vec{c)} . = \vec{a} (2 x + y) + \vec{b} (x + y) + \vec{c} (3 x + y) . \Rightarrow 2 x + y = 1, x + y = 2, 3 x + y = 3 .$
On solving the first two equations we get $x = - 1, y = 3$ . Clearly the values of $x, y$ does not satisfy the third equation.
Hence the given vectors are non-coplanar.

Page No 22.66:

Question 8:

Show that the vectors $\vec{a,} \vec{b,} \vec{c}$ given by $\vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k}, \vec{b} = 2 \hat{i} + \hat{j} + 3 \hat{k} and \vec{c} = \hat{i} + \hat{j} + \hat{k}$ are non-coplanar. Express vector $\vec{d} = 2 \hat{i} - \hat{j} - 3 \hat{k}$ as a linear combination of the vectors $\vec{a,} \vec{b} and \vec{c} .$

Answer:

Let the given vectors $\vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k}, \vec{b} = 2 \hat{i} + \hat{j} + 3 \hat{k}$ and $\vec{c} = \hat{i} + \dot{\hat{j} + \dot{\hat{k}}}$ are coplanar. Then one of the vector is expressible as a linear combination of the other two. Let,
$\hat{i} + 2 \hat{j} + 3 \hat{k} = x (2 \hat{i} + \hat{j} + 3 \hat{k}) + y (\hat{i} + \hat{j} + \hat{k}) . = \hat{i} (2 x + y) + \hat{j} (x + y) + \hat{k} (3 x + y) .$
$\Rightarrow 2 x + y = 1, x + y = 2, 3 x + y = 3 .$
On solving the first two equations we get $x = - 1, y = 3$ . Clearly the values of $x, y$ does not satisfy the third equation.
Hence the given vectors are non-coplanar.
Now, $\vec{d} = 2 \hat{i} - \hat{j} - 3 \hat{k}$ which can be expressed as
$2 \hat{i} - \hat{j} - 3 \hat{k} = x (\hat{i} + 2 \hat{j} + 3 \hat{k}) + y (2 \hat{i} + \hat{j} + 3 \hat{k}) + z (\hat{i} + \hat{j} + \hat{k}) .$
= $\hat{i} (x + 2 y + z) + \hat{j} (2 x + y + z) + \hat{k} (3 x + 3 y + z)$ .

$\Rightarrow x + 2 y + z = 2, 2 x + y + z = - 1, 3 x + 3 y + z = - 3 . \Rightarrow x = - \frac{8}{3}, y = \frac{1}{3}, z = 4$
Hence $\vec{d}$ is expressible as the linear combination of $\vec{a}, \vec{b}$ and $\vec{c}$ .

Page No 22.66:

Question 9:

Prove that a necessary and sufficient condition for three vectors $\vec{a}$ , $\vec{b}$ and $\vec{c}$ to be coplanar is that there exist scalars l, m, n not all zero simultaneously such that $l \vec{a} + m \vec{b} + n \vec{c} = \vec{0} .$

Answer:

$Necessary Condition : F i r s t l e t \vec{a}, \vec{b}, \vec{c} be three coplanar vectors . Then one of them is expressable as a linear combination of the other two . Let \vec{c} = x \vec{a} + y \vec{b} for some scalars x, y . Then, \vec{c} = x \vec{a} + y \vec{b} for some scalars x, y \Rightarrow l \vec{a} + m \vec{b} + n \vec{c} = 0, w h e r e l = x, m = y, n = - 1 Thus, i f \vec{a}, \vec{b}, \vec{c} are coplanar vectors, then there exists scalars l, m, n s u c h t h a t l \vec{a} + m \vec{b} + n \vec{c} = 0 w h e r e l, m, n are all non zero simultaneously .$

$Sufficient Condition : L e t \vec{a}, \vec{b}, \vec{c} be three vectors such that there exists scalars l, m, n n o t a l l z e r o s i m u l a t a n e o u s l y satisfying l \vec{a} + m \vec{b} + n \vec{c} = \vec{0} . We have tp prove that \vec{a}, \vec{b}, \vec{c} are coplanar vectors . Now, l \vec{a} + m \vec{b} + n \vec{c} = \vec{0} \Rightarrow n \vec{c} = - l \vec{a} - m \vec{b} \Rightarrow \vec{c} = (- \frac{1}{n}) \vec{a} + (- \frac{m}{n}) \vec{b} \Rightarrow \vec{c} is a linear combination of \vec{a} a n d \vec{b} . Hence \vec{a}, \vec{b}, \vec{c} are coplanar vectors .$

Page No 22.66:

Question 10:

Show that the four points A, B, C and D with position vectors $\vec{a}$ , $\vec{b}$ , $\vec{c}$ and $\vec{d}$ respectively are coplanar if and only if $3 \vec{a} - 2 \vec{b} + \vec{c} - 2 \vec{d} = \vec{0} .$

Answer:

Necessary Condition: Firstly , let $\vec{a}, \vec{b}, \vec{c}$ are coplanar vectors. Then, one of them is expressible as a linear combination of the other two. Let $\vec{c} = x \vec{a} + y \vec{b}$ for some scalars $x, y .$ Then,
$\vec{c} = x \vec{a} + y \vec{b}$ for some scalars $x, y$ .
$\Rightarrow l \vec{a} + m \vec{b} + n \vec{c} = \vec{0},$ where $l = x, m = y, n = - 1$ .
Thus, if $\vec{a}, \vec{b}, \vec{c}$ are coplanar vectors, then there exists a scalars $l, m, n$ not all zero simultaneously satisfying $l \vec{a} + m \vec{b} + n \vec{c} = \vec{0}$ where $l, m, n$ are not all zero simultaneously.

Sufficient Condition: Let $\vec{a}, \vec{b}, \vec{c}$ are three scalars such that there exists scalars $l, m, n$ not all zero simultaneously satisfying $l \vec{a} + m \vec{b} + n \vec{c} = \vec{0}$ . We have to prove that $\vec{a}, \vec{b}, \vec{c}$ are coplanar vectors.
Now,
$l \vec{a} + m \vec{b} + n \vec{c} = \vec{0 .} \Rightarrow n \vec{c} = - l \vec{a} - m \vec{b} . \Rightarrow \vec{c} = (\frac{- l}{n}) \vec{a} + (\frac{- m}{n}) \vec{b} .$
$\Rightarrow \vec{c}$ is a linear combination of $\vec{a}$ and $\vec{b}$ .
$\Rightarrow \vec{c}$ lies in a plane $\vec{a}$ and $\vec{b}$ .
Hence, $\vec{a}, \vec{b}, \vec{c}$ are coplanar vectors.

Page No 22.73:

Question 1:

Can a vector have direction angles 45°, 60°, 120°?

Answer:

Yes,
Let a vector makes an angle $α = 45 °, β = 60 °, γ = 120 °$ with $O X, O Y, O Z$ respectively. Let $l, m, n$ be the direction cosines of the vector. Then,
$l = \cos 45 ° = \frac{1}{\sqrt{2}}, m = \cos 60 ° = \frac{1}{2}, n = \cos 120 ° = - \frac{1}{2}$
So,
$l^{2} + m^{2} + n^{2} = \frac{1}{2} + \frac{1}{4} + \frac{1}{4} = 1$
Since, the vector has direction cosines such that $l^{2} + m^{2} + n^{2} = 1$
Hence, a vector can have direction angles $45 °, 60 °, 120 °$

Page No 22.73:

Question 2:

Prove that 1, 1, 1 cannot be direction cosines of a straight line.

Answer:

Let $1, 1, 1$ be the direction cosines of a straight line. Then
$1^{2} + 1^{2} + 1^{2} = 3$ ≠ 1
Since direction cosines of a line which makes equal angle with the axes must satisfy $l^{2} + m^{2} + n^{2} = 1$ .
Hence $1, 1, 1$ cannot be the direction cosines of a straight line.

Page No 22.73:

Question 3:

A vector makes an angle of $\frac{π}{4}$ with each of x-axis and y-axis. Find the angle made by it with the z-axis.

Answer:

Let the vector $\vec{O P}$ makes an angle $α = 45 ° and β = 45 °$ with $O X, O Y$ respectively. Suppose $\vec{O P}$ is inclined at angle $γ$ to $O Z$ .
Let $l, m, n$ be the direction cosines of $\vec{O P}$ . Then,
$l = \cos \frac{π}{4} = \frac{1}{\sqrt{2}}$ , $m = \cos \frac{π}{4} = \frac{1}{\sqrt{2}}$ and $n = \cos γ$
Now, we have,

$l^{2} + m^{2} + n^{2} = 1 \Rightarrow \frac{1}{2} + \frac{1}{2} + n^{2} = 1 \Rightarrow n^{2} = 0 \Rightarrow n = 0 \Rightarrow \cos γ = \cos \frac{π}{2} \Rightarrow γ = \frac{π}{2}$
Hence, the angle made by it with the $z -$ axis is $\frac{π}{2}$ .

Page No 22.73:

Question 4:

A vector $\vec{r}$ is inclined at equal acute angles to x-axis, y-axis and z-axis. If $|\vec{r}|$ = 6 units, find $\vec{r}$ .

Answer:

Suppose, vector $\vec{r}$ makes an angle $α$ with each of the axis $O X, O Y$ and $O Z$ . Then, its direction cosines are $l = \cos α, m = \cos α$ and $n = \cos α$ i.e. $l = m = n .$

$Now, l^{2} + m^{2} + n^{2} = 1 \Rightarrow l^{2} + l^{2} + l^{2} = 1 \Rightarrow 3 l^{2} = 1 \Rightarrow l^{2} = \frac{1}{3} \Rightarrow l = \pm \frac{1}{\sqrt{3}} Since, \vec{r} makes acute angle with the axis . Hence, we take only positive value .$
Therefore,

$\vec{r} = |\vec{r}| (l \hat{i} + m \hat{j} + n \hat{k}) \vec{r} = 6 (\frac{1}{\sqrt{3}} \hat{i} + \frac{1}{\sqrt{3}} \hat{j} + \frac{1}{\sqrt{3}} \hat{k}) = 2 \sqrt{3} (\hat{i} + \hat{j} + \hat{k})$

Page No 22.73:

Question 5:

A vector $\vec{r}$ is inclined to x-axis at 45° and y-axis at 60°. If $|\vec{r}|$ = 8 units, find $\vec{r}$ .

Answer:

Here, $\vec{r}$ makes an angle $45 °$ with $O X$ and $60 °$ with $O Y$ . So,
$l = \cos 45 ° = \frac{1}{\sqrt{2}} and m = \cos 60 ° = \frac{1}{2}$
$Now, l^{2} + m^{2} + n^{2} = 1 \Rightarrow \frac{1}{2} + \frac{1}{4} + n^{2} = 1 \Rightarrow n^{2} = \frac{1}{4} \Rightarrow n = \pm \frac{1}{2}$
Therefore,
$\vec{r} = |\vec{r}| (l \hat{i} + m \hat{j} + n \hat{k}) = 8 (\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{2} \hat{j} \pm \frac{1}{2} \hat{k}) = 4 (\sqrt{2} \hat{i} + \hat{j} \pm \hat{k})$

Page No 22.73:

Question 6:

Find the direction cosines of the following vectors:
(i) $2 \hat{i} + 2 \hat{j} - \hat{k}$

(ii) $6 \hat{i} - 2 \hat{j} - 3 \hat{k}$

(iii) $3 \hat{i} - 4 \hat{k}$

Answer:

(i) We have, $2 \hat{i} + 2 \hat{j} - \hat{k}$
The direction cosines are $\frac{2}{\sqrt{2^{2} + 2^{2} + {(- 1)}^{2}}}, \frac{2}{\sqrt{2^{2} + 2^{2} + {(- 1)}^{2}}}, \frac{- 1}{\sqrt{2^{2} + 2^{2} + {(- 1)}^{2}}}$ or, $\frac{2}{3}, \frac{2}{3}, \frac{- 1}{3}$

(ii) We have, $6 \hat{i} - 2 \hat{j} - 3 \hat{k}$
The direction cosines are $\frac{6}{\sqrt{6^{2} + {(- 2)}^{2} + {(- 3)}^{2}}}, \frac{- 2}{\sqrt{6^{2} + {(- 2)}^{2} + {(- 3)}^{2}}}, \frac{- 3}{\sqrt{6^{2} + {(- 2)}^{2} + {(- 3)}^{2}}}$ or, $\frac{6}{7}, \frac{- 2}{7}, \frac{- 3}{7}$

(iii) We have, $3 \hat{i} - 4 \hat{k}$
The direction cosines are $\frac{3}{\sqrt{3^{2} + 0 + {(- 4)}^{2}}}, \frac{0}{\sqrt{3^{2} + 0 + {(- 4)}^{2}}}, \frac{- 4}{\sqrt{3^{2} + 0 + {(- 4)}^{2}}}$ or, $\frac{3}{5}, 0, \frac{- 4}{5}$

Page No 22.73:

Question 7:

Find the angles at which the following vectors are inclined to each of the coordinate axes:
(i) $\hat{i} - \hat{j} + \hat{k}$

(ii) $\hat{j} - \hat{k}$

(iii) $4 \hat{i} + 8 \hat{j} + \hat{k}$

Answer:

(i) Let $\vec{r}$ be the given vector, and let it make an angle $α, β, γ$ with $O X, O Y, O Z$ respectively. Then, its direction cosines are $\cos α, \cos β, \cos γ$ . So, direction ratios of $\vec{r}$ = $\hat{i} - \hat{j} + \hat{k}$ are proportional to $1, - 1, 1$ . Therefore,
Direction cosine of $\vec{r}$ are $\frac{1}{\sqrt{1^{2} + {(- 1)}^{2} + 1^{2}}}, \frac{- 1}{\sqrt{1^{2} + {(- 1)}^{2} + 1^{2}}}, \frac{1}{\sqrt{1^{2} + {(- 1)}^{2} + 1^{2}}}$
or, $\frac{1}{\sqrt{3}}, \frac{- 1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$ .
∴ $\cos α = \frac{1}{\sqrt{3}}, \cos β = \frac{- 1}{\sqrt{3}}, \cos γ = \frac{1}{\sqrt{3}}$
$α = \cos^{- 1} (\frac{1}{\sqrt{3}}), β = \cos^{- 1} (\frac{- 1}{\sqrt{3}}), γ = \cos^{- 1} (\frac{1}{\sqrt{3}})$

(ii) Let $\vec{r}$ be the given vector, and let it make an angles $α, β, γ$ with $O X, O Y, O Z$ respectively. Then, its direction cosines are $\cos α, \cos β, \cos γ$ . So, direction ratios of $\vec{r}$ $= \hat{j} - \hat{k}$ are proportional to $0, 1, - 1$ . Therefore, direction cosines of $\vec{r}$ are
$\frac{0}{\sqrt{0 + 1^{2} + {(- 1)}^{2}}}, \frac{1}{\sqrt{0 + 1^{2} + {(- 1)}^{2}}}, \frac{- 1}{\sqrt{0 + 1^{2} + {(- 1)}^{2}}}$ or, $0, \frac{1}{\sqrt{2}}, \frac{- 1}{\sqrt{2}}$

∴ $\cos α = 0, \cos β = \frac{1}{\sqrt{2}}, \cos γ = \frac{- 1}{\sqrt{2}}$
$\Rightarrow α = \cos^{- 1} (0), β = \cos^{- 1} (\frac{1}{\sqrt{2}}), γ = \cos^{- 1} (\frac{- 1}{\sqrt{2}}) \Rightarrow α = \frac{π}{2}, β = \frac{π}{4}, γ = \frac{3 π}{4}$

(iii) Let $\vec{r}$ be the given vector, and let it make an angle $α, β, γ$ with $O X, O Y, O Z$ respectively. Then, its direction cosines are $\cos α, \cos β, \cos γ$ . So, direction ratio of $\vec{r}$ $= 4 \hat{i} + 8 \hat{j} + \hat{k}$ are proportional to $4, 8, 1$
Therefore, direction ratio of $\vec{r}$ are
$\frac{4}{\sqrt{4^{2} + 8^{2} + 1^{2}}}, \frac{8}{\sqrt{4^{2} + 8^{2} + 1^{2}}}, \frac{1}{\sqrt{4^{2} + 8^{2} + 1^{2}}}$ or, $\frac{4}{9}, \frac{8}{9}, \frac{1}{9}$ .

∴ $α = \cos^{- 1} (\frac{4}{9}), β = \cos^{- 1} (\frac{8}{9}), γ = \cos^{- 1} (\frac{1}{9}) .$

Page No 22.73:

Question 8:

Show that the vector $\hat{i} + \hat{j} + \hat{k}$ is equally inclined with the axes OX, OY and OZ.

Answer:

Let $\vec{r} = \hat{i} + \hat{j} + \hat{k}$ and it make an angle $α, β, γ$ with $O X, O Y, O Z$ respectively.Then its direction cosines are $\cos α, \cos β$ and $\cos γ$ . So, Direction ratios of $\vec{r} = \hat{i} + \hat{j} + \hat{k}$ are proportional $(1, 1, 1)$ . Therefore, direction cosines of $\vec{r}$ are
$\frac{1}{\sqrt{1^{2} + 1^{2} + 1^{2}}}, \frac{1}{\sqrt{1^{2} + 1^{2} + 1^{2}}}, \frac{1}{\sqrt{1^{2} + 1^{2} + 1^{2}}}$ or, $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$ .
Thus,
$\cos α = \frac{1}{\sqrt{3}}, \cos β = \frac{1}{\sqrt{3}}$ and $\cos γ = \frac{1}{\sqrt{3}}$ .
$\Rightarrow α = β = γ$ .
Hence, all are equally inclined with the coordinate axis.

Page No 22.73:

Question 9:

Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} .$

Answer:

Suppose the vector makes equal angle $α$ with the coordinate axis.
Then, its direction cosines are $l = \cos α, m = \cos α, n = \cos α$ . Therefore, $l = m = n$ .
$l^{2} + m^{2} + n^{2} = 1 \Rightarrow l^{2} + l^{2} + l^{2} = 1 \Rightarrow 3 l^{2} = 1 \Rightarrow l^{2} = \frac{1}{3} \Rightarrow l = \frac{1}{\sqrt{3}}$
Hence, direction cosines are $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$ .

Page No 22.73:

Question 10:

If a unit vector $\vec{a}$ makes an angle $\frac{π}{3}$ with $\hat{i}, \frac{π}{4}$ with $\hat{j}$ and an acute angle θ with $\hat{k}$ , then find θ and hence, the components of $\vec{a}$ .

Answer:

The Direction cosines of vector $\vec{a}$ are $l = \cos (\frac{π}{3}) = \frac{1}{2}, m = \cos (\frac{π}{4}) = \frac{1}{\sqrt{2}}, n = \cos θ$

Therefore,
$l^{2} + m^{2} + n^{2} = 1 \Rightarrow \frac{1}{4} + \frac{1}{2} + n^{2} = 1 \Rightarrow n^{2} = 1 - \frac{3}{4} \Rightarrow n^{2} = \frac{1}{4} \Rightarrow n = \frac{1}{2} [∵ \vec{a} makes acute angle with \hat{k}]$
$\Rightarrow \cos θ = \frac{1}{2} \Rightarrow θ = \frac{π}{3}$
Since, $\vec{a}$ is the unit vector.
$∴ \vec{a} = l \hat{i} + m \hat{j} + n \hat{k} \Rightarrow \vec{a} = \frac{1}{2} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} + \frac{1}{2} \hat{k}$
Hence, components of $\vec{a}$ are $\frac{1}{2} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} + \frac{1}{2} \hat{k}$

Page No 22.74:

Question 11:

Find a vector $\vec{r}$ of magnitude $3 \sqrt{2}$ units which makes an angle of $\frac{π}{4}$ and $\frac{π}{2}$ with y and z-axes respectively. [NCERT EXEMPLAR]

Answer:

Suppose vector $\vec{r}$ makes an angle α with the x-axis.

Let l, m, n be the direction cosines of $\vec{r}$ . Then,

$l = \cos α, m = \cos \frac{π}{4} = \frac{1}{\sqrt{2}}, n = \cos \frac{π}{2} = 0$

Now,

$l^{2} + m^{2} + n^{2} = 1 \Rightarrow \cos^{2} α + \frac{1}{2} + 0 = 1 \Rightarrow \cos^{2} α = 1 - \frac{1}{2} = \frac{1}{2} \Rightarrow \cos α = \pm \frac{1}{\sqrt{2}}$

We know that

$\vec{r} = |\vec{r}| (l \hat{i} + m \hat{j} + n \hat{k}) ∴ \vec{r} = 3 \sqrt{2} (\pm \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} j + 0 \hat{k}) (|\vec{r}| = 3 \sqrt{2}) \Rightarrow \vec{r} = \pm 3 \hat{i} + 3 \hat{j}$

Page No 22.74:

Question 12:

A vector $\vec{r}$ is inclined at equal angles to the three axes. If the magnitude of $\vec{r}$ is $2 \sqrt{3}$ , find $\vec{r}$ . [NCERT EXEMPLAR]

Answer:

Let l, m, n be the direction cosines of $\vec{r}$ .

Now, $\vec{r}$ is inclined at equal angles to the three axes.

$∴ l = m = n [α = β = γ \Rightarrow \cos α = \cos β = \cos γ] So, l^{2} + m^{2} + n^{2} = 1 \Rightarrow 3 l^{2} = 1 \Rightarrow l = \pm \frac{1}{\sqrt{3}} \Rightarrow l = m = n = \pm \frac{1}{\sqrt{3}}$
We know that

$\vec{r} = |\vec{r}| (l \hat{i} + m \hat{j} + n \hat{k}) \Rightarrow \vec{r} = 2 \sqrt{3} (\pm \frac{1}{\sqrt{3}} \hat{i} \pm \frac{1}{\sqrt{3}} \hat{j} \pm \frac{1}{\sqrt{3}} \hat{k}) \Rightarrow \vec{r} = 2 (\pm \hat{i} \pm \hat{j} \pm \hat{k})$

Page No 22.75:

Question 1:

If in a ∆ABC, A = (0, 0), B = (3, 3 $\sqrt{3}$ ), C = (−3 $\sqrt{3}$ , 3), then the vector of magnitude 2 $\sqrt{2}$ units directed along AO, where O is the circumcentre of ∆ABC is
(a) $(1 - \sqrt{3}) \hat{i} + (1 + \sqrt{3}) \hat{j}$

(b) $(1 + \sqrt{3}) \hat{i} + (1 - \sqrt{3}) \hat{j}$

(c) $(1 + \sqrt{3}) \hat{i} + (\sqrt{3} - 1) \hat{j}$

(d) none of these

Answer:

(a) $(1 - \sqrt{3}) \hat{i} + (1 + \sqrt{3}) \hat{j}$

$\begin{array}{l} | \vec{A O} | = 2 \sqrt{2} \\ | \vec{A O} | = | \vec{B O} | = | \vec{C O} | = 2 \sqrt{2} = R \\ Let the position vector of O b e x \hat{i} + y \hat{j} . \\ | \vec{A O} | = \sqrt{x^{2} + y^{2}} \\ ∴ x^{2} + y^{2} = 8 . . . . . (1) \\ Also, | \vec{B O} | = | \vec{C O} | \\ \sqrt{{(x - 3)}^{2} + {(y - 3 \sqrt{3})}^{2}} = \sqrt{{(x + 3 \sqrt{3})}^{2} + {(y - 3)}^{2}} \\ x^{2} - 6 x + 9 + y^{2} - 6 \sqrt{3} y + 27 = x^{2} + 6 \sqrt{3} x + 27 + y^{2} - 6 y + 9 \\ y (6 - 6 \sqrt{3}) = x (6 \sqrt{3} + 6) \\ y = \frac{x (1 + \sqrt{3})}{(1 - \sqrt{3})} . . . . . (2) \end{array}$

Substituting y from (2) in (1) we get,

$\begin{array}{l} {(1 - \sqrt{3})}^{2} x^{2} + {(1 + \sqrt{3})}^{2} x^{2} = 8 {(1 - \sqrt{3})}^{2} \\ x^{2} \times 8 = 8 {(1 - \sqrt{3})}^{2} \\ x = 1 - \sqrt{3} \\ y = 1 + \sqrt{3} \\ ∴ The position vector of O is (1 - \sqrt{3}) \hat{i} + (1 + \sqrt{3}) \hat{j .} \\ \vec{AO} = (1 - \sqrt{3}) \hat{i} + (1 + \sqrt{3}) \hat{j} \end{array}$

Page No 22.75:

Question 2:

If $\vec{a}, \vec{b}$ are the vectors forming consecutive sides of a regular hexagon ABCDEF, then the vector representing side CD is
(a) $\vec{a} + \vec{b}$

(b) $\vec{a} - \vec{b}$

(c) $\vec{b} - \vec{a}$

(d) $- (\vec{a} + \vec{b})$

Answer:

(c) $\vec{b} - \vec{a}$
Let $A B C D E F$ be a regular hexagon such that $\vec{A B} = \vec{a}$ and $\vec{B C} = \vec{b} .$ We know, $A D$ is parallel to $B C$ such that $A D = 2 B C$ .
∴ $\vec{A D} = 2 \vec{B C} = 2 \vec{b}$
In $△ A B C$ , we have
$\vec{A B} + \vec{B C} = \vec{A C} \Rightarrow \vec{a} + \vec{b} = \vec{A C}$

In $△ A C D$ , we have
$\vec{A C} + \vec{C D} = \vec{A D} \Rightarrow \vec{C D} = \vec{A D} - \vec{A C} \Rightarrow \vec{C D} = 2 \vec{b} - (\vec{a} + \vec{b}) \Rightarrow \vec{C D} = \vec{b} - \vec{a}$

Page No 22.75:

Question 3:

Forces 3 O $\vec{A}$ , 5 O $\vec{B}$ act along OA and OB. If their resultant passes through C on AB, then

(a) C is a mid-point of AB
(b) C divides AB in the ratio 2 : 1
(c) 3 AC = 5 CB
(d) 2 AC = 3 CB

Answer:

(c) 3 AC = 5 CB

Draw ON, the perpendicular to the line AB

Let $\vec{i}$ be the unit vector along ON
The resultant force $\vec{R} = 3 \vec{O A} + 5 \vec{O B} . . . . . (1)$

The angles between $\vec{i}$ and the forces $\vec{R}, 3 \vec{O A}, 5 \vec{O B}$ are ∠CON, ∠AON, ∠BON respectively.

$\vec{R} \cdot \vec{i} = 3 \vec{O A} \cdot \vec{i} + 5 \vec{O B} \cdot \vec{i}$

⇒ R⋅1⋅ cos ∠CON = 3 $\vec{O A}$ ⋅1⋅cos∠AON + 5 $\vec{O B}$ ⋅1⋅cos∠BON

$R \cdot \frac{O N}{O C} = 3 O A \times \frac{O N}{O A} + 5 O B \frac{O N}{O B} \frac{R}{O C} = (3 + 5)$
R = 8 $\vec{O C}$

We know that,

$\vec{O A} = \vec{O C} + \vec{C A} \Rightarrow 3 \vec{O A} = 3 \vec{O C} + 3 \vec{C A} . . . . . (i) \vec{O B} = \vec{O C} + \vec{C B} \Rightarrow 5 \vec{O B} = 5 \vec{O C} + 5 \vec{C B} . . . . . (ii)$

on adding (i) and (ii) we get,

$\begin{matrix} \begin{matrix} 3 \vec{O A} & + & 5 \vec{O B} \end{matrix} & = & 8 \vec{O C} & + & 3 \vec{CA} & + & 5 \vec{CB} \\ \vec{R} & = & 8 \vec{O C} & + & 3 \vec{CA} & + & 5 \vec{CB} \\ 8 \vec{O C} & = & 8 \vec{O C} & + & 3 \vec{CA} & + & 5 \vec{CB} \end{matrix} |3 \vec{AC}| = |5 \vec{CB}| \Rightarrow 3 AC = 5 CB$

Page No 22.75:

Question 4:

If $\vec{a}, \vec{b}, \vec{c}$ are three non-zero vectors, no two of which are collinear and the vector $\vec{a} + \vec{b}$ is collinear with $\vec{c}, \vec{b} + \vec{c}$ is collinear with $\vec{a},$ then $\vec{a} + \vec{b} + \vec{c} =$
(a) $\vec{a}$

(b) $\vec{b}$

(c) $\vec{c}$

(d) none of these

Answer:

(d) None of these

$\vec{a} + \vec{b}$ is collinear with $\vec{c}$

$∴ \vec{a} + \vec{b} = x \vec{c} . . . . . (1)$

where x is scalar and x ≠ 0.

$\vec{b} + \vec{c}$ is collinear with $\vec{a}$

$\vec{b} + \vec{c} = y \vec{a} . . . . . (2)$

y is scalar and y ≠ 0

Substracting (2) from (1) we get,

$\begin{array}{l} \vec{a} - \vec{c} = x \vec{c} - y \vec{a} \\ \vec{a} (1 + y) = (1 + x) \vec{c} \end{array}$

As given $\vec{a}, \vec{c}$ are not collinear,

∴ 1 + y = 0 and 1 + x = 0

y = −1 and x = −1

Putting value of x in equation (1)

$\begin{array}{l} \vec{a} + \vec{b} = - \vec{c} \\ \vec{a} + \vec{b} + \vec{c} = 0 \end{array}$

Page No 22.75:

Question 5:

If points A (60 $\hat{i}$ + 3 $\hat{j}$ ), B (40 $\hat{i}$ − 8 $\hat{j}$ ) and C (a $\hat{i}$ − 52 $\hat{j}$ ) are collinear, then a is equal to
(a) 40
(b) −40
(c) 20
(d) −20

Answer:

(b) −40
Given: Three points $A (60 \hat{i} + 3 \hat{j}), B (40 \hat{i} - 8 \hat{j})$ and $C (a \hat{i} - 52 \hat{j})$ are collinear. Then, $\vec{A B} = λ \vec{B C} .$
We have,
$\vec{A B} = (40 \hat{i} - 8 \hat{j}) - (60 \hat{i} + 3 \hat{j}) = - 20 \hat{i} - 11 \hat{j}$
$\vec{B C} = (a \hat{i} - 52 \hat{j}) - (40 \hat{i} - 8 \hat{j}) = (a - 40) \hat{i} - 44 \hat{j}$
Therefore,
$\vec{A B} = λ \vec{B C} \Rightarrow - 20 \hat{i} - 11 \hat{j} = λ (a - 40) \hat{i} - λ 44 \hat{j} \Rightarrow λ (a - 40) = - 20, - 44 λ = - 11 \Rightarrow λ = \frac{1}{4} \Rightarrow a - 40 = - 80 \Rightarrow a = - 40$

Page No 22.75:

Question 6:

If G is the intersection of diagonals of a parallelogram ABCD and O is any point, then $O \vec{A} + O \vec{B} + O \vec{C} + O \vec{D} =$
(a) $2 \vec{O G}$

(b) $4 \vec{O G}$

(c) $5 \vec{O G}$

(d) $3 \vec{O G}$

Answer:

(b) $4 \vec{O G}$

Let us consider the point O as origin.
G is the mid point of AC.

$∴ \vec{O G} = \frac{\vec{O A} + \vec{O C}}{2} 2 \vec{O G} = \vec{O A} + \vec{O C} . . . . . (1)$

Also, G is the mid point BD

$∴ \vec{O G} = \frac{\vec{O B} + \vec{O D}}{2} 2 \vec{O G} = \vec{O B} + \vec{O D} . . . . . (2)$

On adding (1) and (2) we get,

$2 \vec{O G} + 2 \vec{O G} = \vec{O A} + \vec{O B} + \vec{O C} + \vec{O D} 4 \vec{O G} = \vec{O A} + \vec{O B} + \vec{O C} + \vec{O D} ∴ \vec{O A} + \vec{O B} + \vec{O C} + \vec{O D} = 4 \vec{O G}$

Page No 22.75:

Question 7:

The vector cos α cos β $\hat{i}$ + cos α sin β $\hat{j}$ + sin α $\hat{k}$ is a
(a) null vector
(b) unit vector
(c) constant vector
(d) none of these

Answer:

(b) unit vector
Given: The vector $\cos α \cos β \hat{i} + \cos α \sin β \hat{j} + \sin α \hat{k} .$
Then,
$|\cos α \cos β \hat{i} + \cos α \sin β \hat{j} + \sin α \hat{k}| = \sqrt{\cos^{2} α \cos^{2} β + \cos^{2} α \sin^{2} β + \sin^{2} α} = \sqrt{\cos^{2} α + \sin^{2} α} = 1$
Hence, the given vector is a unit vector.

Page No 22.75:

Question 8:

In a regular hexagon ABCDEF, A $\vec{B}$ = a, B $\vec{C}$ = $\vec{b} and \vec{C D} = \vec{c}$ . Then, $\vec{A E}$ =
(a) $\vec{a} + \vec{b} + \vec{c}$

(b) $2 \vec{a} + \vec{b} + \vec{c}$

(c) $\vec{b} + \vec{c}$

(d) $\vec{a} + 2 \vec{b} + 2 \vec{c}$

Answer:

Option(c)
Given a regular hexagon $A B C D E F$ such that $\vec{A B} = \vec{a}, \vec{B C} = \vec{b}$ and $\vec{C D} = \vec{c}$ . Then,
In $△ A B C$ , we have
$\vec{A C} = \vec{a} + \vec{b} .$
In $△ A C D$ , we have
$\vec{A C} + \vec{C D} = \vec{A D} . \Rightarrow \vec{A D} = \vec{A C} + \vec{c} . \Rightarrow \vec{A D} = \vec{a} + \vec{b} + \vec{c} .$
Again, in $△ A D E$ , we have
$\vec{A E} = \vec{A D} + \vec{D E} . \Rightarrow \vec{A E} = \vec{a} + \vec{b} + \vec{c} - \vec{a} . \Rightarrow \vec{A E} = \vec{b} + \vec{c} .$
Hence option (c).

Page No 22.75:

Question 9:

The vector equation of the plane passing through $\vec{a}, \vec{b}, \vec{c}, is \vec{r} = α \vec{a} + β \vec{b} + γ \vec{c},$ provided that
(a) α + β + γ = 0
(b) α + β + γ =1
(c) α + β = γ
(d) α² + β² + γ² = 1

Answer:

(b) α + β + γ =1
Given: A plane passing through $\vec{a}, \vec{b}, \vec{c}$ .

⇒ Lines $\vec{a} - \vec{b}$ and $\vec{c} - \vec{a}$ lie on the plane.

The parmetric equation of the plane can be written as:

$\begin{array}{l} \vec{r} = \vec{a} + λ_{1} (\vec{a} - \vec{b}) + λ_{2} (\vec{c} - \vec{a}) \\ \vec{r} = \vec{a} (1 + λ_{1} - λ_{2}) - λ_{1} \vec{b} + λ_{2} \vec{c} \\ Given that \vec{r} = α \vec{a} + β \vec{b} + γ \vec{c} \\ ∴ α + β + γ = 1 + λ_{1} - λ_{2} - λ_{1} + λ_{2} \\ α + β + γ = 1 \end{array}$

Page No 22.75:

Question 10:

If O and O' are circumcentre and orthocentre of ∆ ABC, then $\vec{O A} + \vec{O B} + \vec{O C}$ equals
(a) 2 $\vec{O O}$ '
(b) $O \vec{O'}$
(c) $\vec{O' O}$
(d) $2 \vec{O' O}$

Answer:

Option (b).
Given: $O$ be the circumcentre and $O'$ be the orthocentre of $△ A B C$ . Let $G$ be the centroid of the triangle.
We know that $O, G$ and $H$ are collinear and by geometry $\vec{O' G} = 2 \vec{O G} .$ This yields,
$\vec{O' O} = \vec{O' G} + \vec{G O} = 2 \vec{G O} + \vec{G O} = 3 \vec{G O} .$
In other words $\vec{O O'} = 3 \vec{O G} .$
Since, $\vec{O G} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$ .

∴ $\vec{O O'} = 3 \times \frac{\vec{a} + \vec{b} + \vec{c}}{3} = \vec{a} + \vec{b} + \vec{c} = \vec{O A} + \vec{O B} + \vec{O C} .$

Page No 22.75:

Question 11:

If $\vec{a}$ , $\vec{b}$ , $\vec{c}$ and $\vec{d}$ are the position vectors of points A, B, C, D such that no three of them are collinear and $\vec{a} + \vec{c} = \vec{b} + \vec{d},$ then ABCD is a
(a) rhombus
(b) rectangle
(c) square
(d) parallelogram

Answer:

Given:
$\vec{a} + \vec{c} = \vec{b} + \vec{d} \Rightarrow \vec{c} - \vec{d} = \vec{b} - \vec{a} \Rightarrow \vec{A B} = \vec{D C} A n d \vec{a} + \vec{c} = \vec{b} + \vec{d} \Rightarrow \vec{c} - \vec{b} = \vec{d} - \vec{a} \Rightarrow \vec{A D} = \vec{B C} A l s o, \sin c e \vec{a} + \vec{c} = \vec{b} + \vec{d} \Rightarrow \frac{1}{2} (\vec{a} + \vec{c)} = \frac{1}{2} (\vec{b} + \vec{d}) s o, p o s i t i o n v e c t o r o f m i d p o i n t o f B D = p o s i t i o n v e c t o r o f m i d p o i n t o f A C . h e n c e d i a g o n a l s b i s e c t e a c h o t h e r . t h e g i v e n A B C D i s a p a r a l l e \log r a m .$
Option (d).

Page No 22.75:

Question 12:

Let G be the centroid of ∆ ABC. If $\vec{A B} = \vec{a,} \vec{A C} = \vec{b,}$ then the bisector $\vec{A G},$ in terms of $\vec{a} and \vec{b}$ is
(a) $\frac{2}{3} (\vec{a} + \vec{b})$

(b) $\frac{1}{6} (\vec{a} + \vec{b})$

(c) $\frac{1}{3} (\vec{a} + \vec{b})$

(d) $\frac{1}{2} (\vec{a} + \vec{b})$

Answer:

(c) $\frac{1}{3} (\vec{a} + \vec{b})$

Taking $A$ as origin. Then, position vector of $A, B$ and $C$ are $\vec{0}, \vec{a}$ and $\vec{b}$ respectively. Then,
Centroid $G$ has position vector $\frac{\vec{0} + \vec{a} + \vec{b}}{3} = \frac{\vec{a} + \vec{b}}{3}$
Therefore, $A G = \frac{\vec{a} + \vec{b}}{3} - \vec{0} = \frac{\vec{a} + \vec{b}}{3}$

Page No 22.76:

Question 13:

If ABCDEF is a regular hexagon, then $\vec{A D} + \vec{E B} + \vec{F C}$ equals
(a) $2 \vec{A B}$
(b) $\vec{0}$
(c) $3 \vec{A B}$
(d) $4 \vec{A B}$

Answer:

(d) $4 \vec{A B}$

$\begin{array}{l} \vec{AD} = 2 \vec{BC} \\ \vec{EB} = 2 \vec{FA} \\ \vec{FC} = 2 \vec{AB} \end{array}$

$\begin{array}{cll} \vec{A D} + \vec{E B} & = & 2 (\vec{B C} + \vec{F A}) \\ = & 2 (\vec{A O} + \vec{F A}) (∵ \vec{B C} = \vec{A O}) \end{array}$

In triangle AOF,

$\begin{array}{l} \vec{F A} + \vec{A O} + \vec{F O} = 0 \\ ∴ \vec{F A} + \vec{A O} = - \vec{F O} \\ ∴ \vec{A D} + \vec{E B} = - 2 \vec{F O} \end{array}$

And $\vec{A B} = - \vec{F O}$

$\begin{array}{l} ∴ \vec{A D} + \vec{E B} = 2 \vec{A B} \\ ∴ \vec{A D} + \vec{E B} + \vec{F C} = 2 \vec{A B} + 2 \vec{A B} = 4 \vec{A B} \end{array}$

Page No 22.76:

Question 14:

The position vectors of the points A, B, C are $2 \hat{i} + \hat{j} - \hat{k}, 3 \hat{i} - 2 \hat{j} + \hat{k} and \hat{i} + 4 \hat{j} - 3 \hat{k}$ respectively. These points
(a) form an isosceles triangle
(b) form a right triangle
(c) are collinear
(d) form a scalene triangle

Answer:

(a) form an isosceles triangle
Given: Position vectors of $A, B, C$ are $2 \hat{i} + \hat{j} - \hat{k}, 3 \hat{i} - 2 \hat{j} + \hat{k}$ and $\hat{i} + 4 \hat{j} - 3 \hat{k}$ . Then,
$\vec{A B} = (3 \hat{i} - 2 \hat{j} + \hat{k}) - (2 \hat{i} + \hat{j} - \hat{k}) = \hat{i} - 3 \hat{j} + 2 \hat{k} \vec{B C} = (\hat{i} + 4 \hat{j} - 3 \hat{k}) - (3 \hat{i} - 2 \hat{j} + \hat{k}) = - 2 \hat{i} + 6 \hat{j} - 4 \hat{k} \vec{C A} = (2 \hat{i} + \hat{j} - \hat{k}) - (\hat{i} + 4 \hat{j} - 3 \hat{k}) = \hat{i} - 3 \hat{j} + 2 \hat{k}$

$Now, \vec{|A B|} = \sqrt{1^{2} + {(- 3)}^{2} + 2^{2}} = \sqrt{1 + 9 + 4} = \sqrt{14} |\vec{C A}| = \sqrt{1^{2} + {(- 3)}^{2} + 2^{2}} = \sqrt{1 + 9 + 4} = \sqrt{14} |\vec{B C}| = \sqrt{{(- 2)}^{2} + 6^{2} + (- 4^{2})} = \sqrt{4 + 36 + 16} = \sqrt{56} ∴ |A B| = |\vec{C A}|$
Hence, the triangle is isosceles as two of its sides are equal.

Page No 22.76:

Question 15:

If three points A, B and C have position vectors $\hat{i} + x \hat{j} + 3 \hat{k}, 3 \hat{i} + 4 \hat{j} + 7 \hat{k} and y \hat{i} - 2 \hat{j} - 5 \hat{k}$ respectively are collinear, then (x, y) =
(a) (2, −3)
(b) (−2, 3)
(c) (−2, −3)
(d) (2, 3)

Answer:

(a) (2, −3)
Given position vectors of $A, B$ and $C$ are $\hat{i} + x \hat{j} + 3 \hat{k}, 3 \hat{i} + 4 \hat{j} + 7 \hat{k}$ and $y \hat{i} - 2 \hat{j} - 5 \hat{k} .$ Then,
$\vec{A B} = 3 \hat{i} + 4 \hat{j} + 7 \hat{k} - \hat{i} - x \hat{j} - 3 \hat{k} = 2 \hat{i} + (4 - x) \hat{j} + 4 \hat{k} \vec{B C} = y \hat{i} - 2 \hat{j} - 5 \hat{k} - 3 \hat{i} - 4 \hat{j} - 7 \hat{k} = (y - 3) \hat{i} - 6 \hat{j} - 12 \hat{k}$
Since, the given vectors are collinear.
$∴ \vec{A B} = λ \vec{B C} \Rightarrow 2 \hat{i} + (4 - x) \hat{j} + 4 \hat{k} = λ (y - 3) \hat{i} - 6 λ \hat{j} - 12 λ \hat{k} \Rightarrow 2 = λ (y - 3), (4 - x) = - 6 λ, 4 = - 12 λ \Rightarrow 2 = λ (y - 3), (4 - x) = - 6 λ, λ = - \frac{1}{3} \Rightarrow 2 = - \frac{1}{3} (y - 3), (4 - x) = - 6 \times (- \frac{1}{3}) \Rightarrow - 6 = y - 3, 4 - x = 2 \Rightarrow y = - 3, x = 2$

Page No 22.76:

Question 16:

ABCD is a parallelogram with AC and BD as diagonals. Then, $\vec{A C} - \vec{B D} =$
(a) $4 \vec{A B}$
(b) $3 \vec{A B}$
(c) $2 \vec{A B}$
(d) $\vec{A B}$

Answer:

(c) $2 \vec{A B}$
Given: $A B C D$ , a parallelogram with diagonals $A C$ and $B D$ . Then,
$\vec{A C} = \vec{A B} + \vec{B C} \vec{A D} = \vec{A B} + \vec{B D} \Rightarrow \vec{B D} = \vec{A D} - \vec{A B}$

∴ $\vec{A C} - \vec{B D} = \vec{A B} + \vec{B C} - \vec{A D} + \vec{A B} = 2 \vec{A B}$ [∵ $\vec{A D} = \vec{B C}$ ]

Page No 22.76:

Question 17:

If OACB is a parallelogram with $\vec{O C} = \vec{a} and \vec{A B} = \vec{b},$ then $\vec{O A} =$
(a) $(\vec{a} + \vec{b})$

(b) $(\vec{a} - \vec{b})$

(c) $\frac{1}{2} (\vec{b} - \vec{a})$

(d) $\frac{1}{2} (\vec{a} - \vec{b})$

Answer:

(d) $\frac{1}{2} (\vec{a} - \vec{b})$
Given a parallelogram $O A C B$ such that $\vec{O C} = \vec{a}, \vec{A B} = \vec{b}$ . Then,
$\vec{O B} + \vec{B C} = \vec{O C} \Rightarrow \vec{O B} = \vec{O C} - \vec{B C}$
$\Rightarrow \vec{O B} = \vec{O C} - \vec{O A}$ [∵ $\vec{B C} = \vec{O A}$ ]
$\Rightarrow \vec{O B} = \vec{a} - \vec{O A} . . . (1)$

Therefore,
$\vec{O A} + \vec{A B} = \vec{O B} \Rightarrow \vec{O A} + \vec{b} = \vec{a} - \vec{O A} [Using (1)] \Rightarrow 2 \vec{O A} = \vec{a} - \vec{b} \Rightarrow \vec{O A} = \frac{1}{2} (\vec{a} - \vec{b})$

Page No 22.76:

Question 18:

If $\vec{a} and \vec{b}$ are two collinear vectors, then which of the following are incorrect?
(a) $\vec{b} = λ \vec{a}$ for some scalar λ
(b) $\vec{a} = \pm \vec{b}$
(c) the respective components of $\vec{a} and \vec{b}$ are proportional
(d) both the vectors $\vec{a} and \vec{b}$ have the same direction but different magnitudes

Answer:

(d) both the vectors $\vec{a} and \vec{b}$ have the same direction but different magnitudes
If $\vec{a}$ and $\vec{b}$ are collinear vectors, then they are paprallel. Therefore, we have
$\vec{b} = λ \vec{a}$ , for some scalar $λ$ .
If $λ = \pm 1$ $\Rightarrow \vec{a} = \pm \vec{b} .$
If $b = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$ and $\vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$ . Then,
$\vec{b} = λ \vec{a} . \Rightarrow b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k} = λ (a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}) . \Rightarrow b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k} = (λ a_{1}) \hat{i} + (λ a_{2}) \hat{j} + (λ a_{3}) \hat{k} . \Rightarrow b_{1} = λ a_{1}, b_{2} = λ a_{2}, b_{3} = λ a_{3} . \Rightarrow \frac{b_{1}}{a_{1}} = \frac{b_{2}}{a_{2}} = \frac{b_{3}}{a_{3}} = λ .$

Thus, the respective components of $\vec{a}$ and $\vec{b}$ can have different directions. Hence, the statement given in (d) is incorrect.

Page No 22.76:

Question 19:

In Figure, which of the following is not true?

(a) $\vec{A B} + \vec{B C} + \vec{C A} = \vec{0}$

(b) $\vec{A B} + \vec{B C} - \vec{A C} = \vec{0}$

(c) $\vec{A B} + \vec{B C} - \vec{C A} = \vec{0}$

(d) $\vec{A B} - \vec{C B} + \vec{C A} = \vec{0}$
Figure

Answer:

(c) $\vec{A B} + \vec{B C} - \vec{C A} = \vec{0}$

We have, LHS = $\vec{A B} + \vec{B C} - \vec{C A} = \vec{A C} - \vec{C A}$ [∵ $\vec{A B} + \vec{B C} = \vec{A C}$ ]
$= - \vec{C A} - \vec{C A} = - 2 \vec{C A}$
So, LHS $\neq$ RHS
Hence, It is not true.

Page No 22.76:

Question 20:

The position vector of the point which divides the join of points with position vectors $\vec{a} + \vec{b} and 2 \vec{a} - \vec{b} in the ration 1 : 2 is$
$(a) \frac{1}{3} (3 \vec{a} + 2 \vec{b}) (b) \vec{a} (c) \frac{5}{3} \vec{a} - \frac{1}{3} \vec{b} (d) \frac{1}{3} (4 \vec{a} + \vec{b})$

Answer:

Given:
X and Y are two points with position vectors $\vec{a} + \vec{b} and 2 \vec{a} - \vec{b}$ , respectively.
Z divides the line segment XY in the ratio 1 : 2

We know,
The position vector of R which divided P and Q internally in the ratio m : n is given by $\vec{O R} = \frac{m \vec{b} + n \vec{a}}{m + n}$ , where position vector of P and Q are $\vec{a} and \vec{b}$ , respectively.

Thus,
$\vec{O Z} = \frac{1 (2 \vec{a} - \vec{b}) + 2 (\vec{a} + \vec{b})}{1 + 2} \Rightarrow \vec{O Z} = \frac{2 \vec{a} - \vec{b} + 2 \vec{a} + 2 \vec{b}}{3} \Rightarrow \vec{O Z} = \frac{4 \vec{a} + \vec{b}}{3}$

Hence, the correct option is (d).

Page No 22.77:

Question 21:

The vector with initial point P (2, -3, 5) and terminal point Q (3, -4, 7) is
$(a) \hat{i} - \hat{j} + 2 \hat{k} (b) 5 \hat{i} - 7 \hat{j} + 12 \hat{k} (c) - \hat{i} + \hat{j} - 2 \hat{k} (d) none of these$

Answer:

Given:
vector with initial point P (2, −3, 5) and terminal point Q (3, −4, 7)

The required vector joining P and Q is the vector $\vec{P Q}$ , given by

$\vec{P Q} = (3 - 2) \hat{i} + (- 4 - (- 3)) \hat{j} + (7 - 5) \hat{k} \Rightarrow \vec{P Q} = \hat{i} - \hat{j} + 2 \hat{k}$

Hence, the correct option is (a).

Page No 22.77:

Question 22:

The vectors $2 \hat{j} + \hat{k} and 3 \hat{i} - \hat{j} + 4 \hat{k}$ represent the two sides AB and AC respectively of a ∆ABC. The length of the median through A is
$(a) \frac{\sqrt{34}}{2} (b) \sqrt{48} (c) \sqrt{18} (d) \sqrt{\frac{43}{2}}$

Answer:

Given:
The vectors $2 \hat{j} + \hat{k} and 3 \hat{i} - \hat{j} + 4 \hat{k}$ represent the two sides AB and AC respectively.

Let AD be the median of the triangle.

$in ∆ A B C, using triangle law of vector addition, \vec{B C} = \vec{A C} - \vec{A B} \Rightarrow \vec{B C} = (3 \hat{i} - \hat{j} + 4 \hat{k}) - (2 \hat{j} + \hat{k}) \Rightarrow \vec{B C} = 3 \hat{i} - \hat{j} + 4 \hat{k} - 2 \hat{j} - \hat{k} \Rightarrow \vec{B C} = 3 \hat{i} - 3 \hat{j} + 3 \hat{k} . . . (1) \Rightarrow \vec{B D} = \frac{1}{2} \vec{B C} (∵ A D is the median) \Rightarrow \vec{B D} = \frac{1}{2} (3 \hat{i} - 3 \hat{j} + 3 \hat{k}) Now, in ∆ A B D, using triangle law of vector addition, \vec{A D} = \vec{A B} + \vec{B D} \Rightarrow \vec{A D} = (2 \hat{j} + \hat{k}) + \frac{1}{2} (3 \hat{i} - 3 \hat{j} + 3 \hat{k}) \Rightarrow \vec{A D} = 2 \hat{j} + \hat{k} + \frac{3}{2} \hat{i} - \frac{3}{2} \hat{j} + \frac{3}{2} \hat{k} \Rightarrow \vec{A D} = \frac{3}{2} \hat{i} + \frac{1}{2} \hat{j} + \frac{5}{2} \hat{k} . . . (2) Thus, |\vec{A D}| = \sqrt{{(\frac{3}{2})}^{2} + {(\frac{1}{2})}^{2} + {(\frac{5}{2})}^{2}} = \sqrt{\frac{9}{4} + \frac{1}{4} + \frac{25}{4}} = \sqrt{\frac{35}{4}} = \frac{\sqrt{35}}{2}$

Hence, the correct option is (a).

Page No 22.77:

Question 23:

$If |\vec{a}| = 3 and - 1 \leq λ \leq 2, then |λ \vec{a}|$ lies in the interval
(a) [0, 6]
(b) [-3, 6]
(c) [3,6]
(d) [1, 2]

Answer:

Given: $|\vec{a}| = 3 and - 1 \leq λ \leq 2$

$|λ \vec{a}| = λ |\vec{a}| since, - 1 \leq λ \leq 2 \Rightarrow - 1 |\vec{a}| \leq λ |\vec{a}| \leq 2 |\vec{a}| \Rightarrow - 1 \times 3 \leq λ |\vec{a}| \leq 2 \times 3 (∵ |\vec{a}| = 3) \Rightarrow - 3 \leq |λ \vec{a}| \leq 6 \Rightarrow |λ \vec{a}| \in [- 3, 6]$

Hence, the correct option is (b).

Page No 22.77:

Question 24:

The value of λ for which the vectors $3 \hat{i} - 6 \hat{j} + \hat{k} and 2 \hat{i} - 4 \hat{j} + λ \hat{k}$ are parallel is
$(a) \frac{2}{3} (b) \frac{3}{2} (c) \frac{5}{2} (d) \frac{2}{5}$

Answer:

Given: the vectors $3 \hat{i} - 6 \hat{j} + \hat{k} and 2 \hat{i} - 4 \hat{j} + λ \hat{k}$ are parallel

$The vectors 3 \hat{i} - 6 \hat{j} + \hat{k} and 2 \hat{i} - 4 \hat{j} + λ \hat{k} are parallel \Rightarrow \frac{3}{2} = \frac{- 6}{- 4} = \frac{1}{λ} \Rightarrow \frac{3}{2} = \frac{1}{λ} \Rightarrow λ = \frac{2}{3}$

Hence, the correct option is (a).

Page No 22.77:

Question 25:

The position vector of the point which divides the join of points $2 \vec{a} - 3 \vec{b} and \vec{a} + \vec{b}$ in the ratio 3:1 is
$(a) \frac{1}{2} (8 \vec{a} - 2 \vec{b}) (b) \frac{1}{4} (7 \vec{a} - 8 \vec{b}) (c) \frac{3}{4} \vec{a} (d) \frac{5}{4} \vec{a}$

Answer:

Given:
X and Y are two points with position vectors $2 \vec{a} - 3 \vec{b} and \vec{a} + \vec{b}$ , respectively.
Z divides the line segment XY in the ratio 3 : 1

We know,
The position vector of R which divided P and Q internally in the ratio m : n is given by $\vec{O R} = \frac{m \vec{b} + n \vec{a}}{m + n}$ , where position vector of P and Q are $\vec{a} and \vec{b}$ , respectively.

Thus,
$\vec{O Z} = \frac{3 (\vec{a} + \vec{b}) + 1 (2 \vec{a} - 3 \vec{b})}{3 + 1} \Rightarrow \vec{O Z} = \frac{3 \vec{a} + 3 \vec{b} + 2 \vec{a} - 3 \vec{b}}{4} \Rightarrow \vec{O Z} = \frac{5 \vec{a}}{4}$

Hence, the correct option is (d).

Page No 22.77:

Question 1:

If $\vec{a}, \vec{b}, \vec{c}$ represent the sides of a triangle taken in order, then $\vec{a} + \vec{b} + \vec{c}$ = _______________.

Answer:

$Let A B C be a triangle such tha t \vec{A B} = \vec{c}, \vec{B C} = \vec{a} and \vec{C A} = \vec{b} . \vec{a} + \vec{b} + \vec{c} = \vec{B C} + \vec{C A} + \vec{A B} = \vec{B B} = \vec{0}$

Hence, $\vec{a} + \vec{b} + \vec{c}$ = $\vec{0} .$

Page No 22.77:

Question 2:

In a parallelogram ABCD, if $\vec{A B} = \vec{a} and \vec{B C} = \vec{b}, then \vec{A C} =_____________and \vec{B D} =_______________.$

Answer:

Given:
ABCD is a parallelogram
$\vec{A B} = \vec{a} and \vec{B C} = \vec{b}$

$in ∆ A B C, using triangle law of vector addition, \vec{A C} = \vec{A B} + \vec{B C} \Rightarrow \vec{A C} = \vec{a} + \vec{b} Now, in ∆ A B D, \vec{A D} = \vec{B C} = \vec{b} using triangle law of vector addition, \vec{A D} = \vec{A B} + \vec{B D} \vec{B D} = \vec{A D} - \vec{A B} \Rightarrow \vec{B D} = \vec{b} - \vec{a}$

Hence, $\vec{A C} = \vec{a} + \vec{b} and \vec{B D} = \vec{b} - \vec{a} .$

Page No 22.77:

Question 3:

If $\vec{a}, \vec{b}, \vec{c}$ are the position vectors of vertices of a triangle having its centroid at the origin, then $\vec{a} + \vec{b} + \vec{c}$ = _______________.

Answer:

Given:
$\vec{a}, \vec{b}, \vec{c}$ are the position vectors
centroid is at the origin

Let O be the centroid of the triangle.

$\vec{O O} = \frac{\vec{O A} + \vec{O B} + \vec{O C}}{3} \Rightarrow \vec{0} = \frac{\vec{O A} + \vec{O B} + \vec{O C}}{3} \Rightarrow \vec{0} = \frac{\vec{a} + \vec{b} + \vec{c}}{3} \Rightarrow \vec{0} = \vec{a} + \vec{b} + \vec{c}$

Hence, $\vec{a} + \vec{b} + \vec{c}$ = $\vec{0}$ .

Page No 22.77:

Question 4:

If $\vec{a}, \vec{b}, \vec{c}$ are the position vectors of the vertices of an equilateral triangle whose circumcentre is at the origin, then $\vec{a} + \vec{b} + \vec{c}$ = _____________.

Answer:

Given:
$\vec{a}, \vec{b}, \vec{c}$ are the position vectors of the vertices of an equilateral triangle
circumcentre is at the origin

Since, the triangle is equilateral
Therefore, circumcentre = centroid

Let O be the centroid of the triangle.

$\vec{O O} = \frac{\vec{O A} + \vec{O B} + \vec{O C}}{3} \Rightarrow \vec{0} = \frac{\vec{O A} + \vec{O B} + \vec{O C}}{3} \Rightarrow \vec{0} = \frac{\vec{a} + \vec{b} + \vec{c}}{3} \Rightarrow \vec{0} = \vec{a} + \vec{b} + \vec{c}$

Hence, $\vec{a} + \vec{b} + \vec{c}$ = $\vec{0}$ .

Page No 22.77:

Question 5:

If the vectors $\vec{a} = \hat{i} - 2 \hat{j} + 3 \hat{k} a n d \vec{b} = 3 \hat{i} - 6 \hat{j} + m \hat{k}$ are collinear, then m = ________________.

Answer:

Given: $\vec{a} = \hat{i} - 2 \hat{j} + 3 \hat{k} and \vec{b} = 3 \hat{i} - 6 \hat{j} + m \hat{k}$ are collinear

$The vectors \vec{a} = \hat{i} - 2 \hat{j} + 3 \hat{k} and \vec{b} = 3 \hat{i} - 6 \hat{j} + m \hat{k} are collinear \Rightarrow \frac{1}{3} = \frac{- 2}{- 6} = \frac{3}{m} \Rightarrow \frac{1}{3} = \frac{3}{m} \Rightarrow m = 9$

Hence, m = 9.

Page No 22.77:

Question 6:

Vectors of magnitude 21 units in the direction of the vector $3 \hat{i} - 6 \hat{j} + 2 \hat{k}$ are _____________.

Answer:

Given: $\vec{a} = 3 \hat{i} - 6 \hat{j} + 2 \hat{k}$

The unit vector in the direction of the given vector $\vec{a}$ is
$\hat{a} = \frac{\vec{a}}{\vec{|a|}} = \frac{3 \hat{i} - 6 \hat{j} + 2 \hat{k}}{\sqrt{{(3)}^{2} + {(- 6)}^{2} + {(2)}^{2}}} = \frac{3 \hat{i} - 6 \hat{j} + 2 \hat{k}}{\sqrt{9 + 36 + 4}} = \frac{3 \hat{i} - 6 \hat{j} + 2 \hat{k}}{\sqrt{49}} = \frac{1}{7} (3 \hat{i} - 6 \hat{j} + 2 \hat{k})$

Therefore, the vector having magnitude 21 units in the direction of $\vec{a}$ is
$\pm 21 \hat{a} = \pm 21 (\frac{1}{7} (3 \hat{i} - 6 \hat{j} + 2 \hat{k})) = \pm 3 (3 \hat{i} - 6 \hat{j} + 2 \hat{k}) = \pm (9 \hat{i} - 18 \hat{j} + 6 \hat{k})$

Hence, vectors of magnitude 21 units in the direction of the vector $3 \hat{i} - 6 \hat{j} + 2 \hat{k}$ are $\pm (9 \hat{i} - 18 \hat{j} + 6 \hat{k}) .$

Page No 22.77:

Question 7:

If D is the mid-point of side BC of $∆ A B C and \vec{A B} + \vec{A C} = k \vec{A D}$ , then k = ______________.

Answer:

Given:
D is the mid-point of side BC
$\vec{A B} + \vec{A C} = k \vec{A D}$

$in ∆ A B C, using triangle law of vector addition, \vec{B C} = \vec{A C} - \vec{A B} . . . (1) \Rightarrow \vec{B D} = \frac{1}{2} \vec{B C} (∵ D is the mid - point of B C) \Rightarrow \vec{B D} = \frac{1}{2} \vec{A C} - \frac{1}{2} \vec{A B} . . . (2) Now, in ∆ A B D, using triangle law of vector addition, \vec{A D} = \vec{A B} + \vec{B D} \Rightarrow \vec{A D} = \vec{A B} + \frac{1}{2} \vec{A C} - \frac{1}{2} \vec{A B} (From (2)) \Rightarrow \vec{A D} = \frac{1}{2} \vec{A B} + \frac{1}{2} \vec{A C} \Rightarrow 2 \vec{A D} = \vec{A B} + \vec{A C}$

Hence, k = $2 .$

Page No 22.77:

Question 8:

The cousines of the angles made by the vector $\hat{i} - 2 \hat{j} + 2 \hat{k}$ with the coordinate axes are: ______________.

Answer:

Given: $\vec{a} = \hat{i} - 2 \hat{j} + 2 \hat{k}$

Let l, m and n are the direction cosines of the given vector, then
$l = \frac{a_{1}}{\vec{|a|}} = \frac{1}{\sqrt{{(1)}^{2} + {(- 2)}^{2} + {(2)}^{2}}} = \frac{1}{\sqrt{1 + 4 + 4}} = \frac{1}{\sqrt{9}} = \frac{1}{3} m = \frac{a_{2}}{\vec{|a|}} = \frac{- 2}{\sqrt{{(1)}^{2} + {(- 2)}^{2} + {(2)}^{2}}} = \frac{- 2}{\sqrt{1 + 4 + 4}} = \frac{- 2}{\sqrt{9}} = - \frac{2}{3} n = \frac{a_{3}}{\vec{|a|}} = \frac{2}{\sqrt{{(1)}^{2} + {(- 2)}^{2} + {(2)}^{2}}} = \frac{2}{\sqrt{1 + 4 + 4}} = \frac{2}{\sqrt{9}} = \frac{2}{3} Thus, the direction cosines are (\frac{1}{3}, - \frac{2}{3}, \frac{2}{3}) .$

Hence, the cosines of the angles made by the vector $\hat{i} - 2 \hat{j} + 2 \hat{k}$ with the coordinate axes are $(\frac{1}{3}, - \frac{2}{3}, \frac{2}{3}) .$

Page No 22.77:

Question 9:

A vector of magnitude 6, making angle $\frac{π}{4} with x - axis, \frac{π}{3}$ with y-axis and an acute angle with z-axis is ____________.

Answer:

Given: A vector of magnitude 6, making angle $\frac{π}{4} with x - axis, \frac{π}{3}$ with y-axis and an acute angle with z-axis

Let a vector $\vec{a}$ makes an angle $\frac{π}{4} with x - axis, \frac{π}{3}$ with y-axis and an acute angle with z-axis and is of magnitude 6.

$Therefore, l = \cos \frac{π}{4} = \frac{1}{\sqrt{2}} m = \cos \frac{π}{3} = \frac{1}{2} n = \cos θ We know, l^{2} + m^{2} + n^{2} = 1 \Rightarrow {(\frac{1}{\sqrt{2}})}^{2} + {(\frac{1}{2})}^{2} + {(\cos θ)}^{2} = 1 \Rightarrow \frac{1}{2} + \frac{1}{4} + {(\cos θ)}^{2} = 1 \Rightarrow {(\cos θ)}^{2} = 1 - \frac{1}{2} - \frac{1}{4} \Rightarrow {(\cos θ)}^{2} = \frac{4 - 2 - 1}{4} \Rightarrow {(\cos θ)}^{2} = \frac{1}{4} \Rightarrow \cos θ = \frac{1}{2} (∵ θ is acute) \Rightarrow n = \frac{1}{2} T h u s, \vec{a} = |\vec{a}| (l \hat{i} + m \hat{j} + n \hat{k}) = 6 (\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{2} \hat{j} + \frac{1}{2} \hat{k}) = 3 \sqrt{2} \hat{i} + 3 \hat{j} + 3 \hat{k}$

Hence, a vector of magnitude 6, making angle $\frac{π}{4} with x - axis, \frac{π}{3}$ with y−axis and an acute angle with z−axis is $3 \sqrt{2} \hat{i} + 3 \hat{j} + 3 \hat{k} .$

Page No 22.78:

Question 10:

If the vectors $\vec{a} = 2 \hat{i} - (y + z) \hat{j} + 5 \hat{k} and \vec{b} = (x + y) \hat{i} + 3 \hat{j} + (z + x) \hat{k}$ are equal x+y+z= _________________.

Answer:

Given: $\vec{a} = 2 \hat{i} - (y + z) \hat{j} + 5 \hat{k} and \vec{b} = (x + y) \hat{i} + 3 \hat{j} + (z + x) \hat{k}$ are equal

Two vectors are equal if and only if their corresponding components are equal.
Thus,

$2 = x + y, - (y + z) = 3, 5 = z + x Now, x + y = 2 . . (1) - y - z = 3 . . (2) z + x = 5 . . (3) Subtracting (1) from (3), we get z - y = 3 . . (4) Adding (2) and (4), we get - 2 y = 6 \Rightarrow y = - 3 Substituting the value of y in (1) and (4), we get x - 3 = 2 \Rightarrow x = 5 and z + 3 = 3 \Rightarrow z = 0 Thus, x + y + z = 5 - 3 + 0 = 2$

Hence, x + y + z = 2.

Page No 22.78:

Question 11:

If $\vec{a} a n d \vec{b}$ are the position vectors of A and B respectively, then the position vector of a point C on AB produced such that $\vec{A C} = 3 \vec{A B} is______________________.$

Answer:

Given:
$\vec{A C} = 3 \vec{A B}$

Let $\vec{c}$ is the position vectors of C.

$\vec{A C} = 3 \vec{A B} \Rightarrow \vec{c} - \vec{a} = 3 (\vec{b} - \vec{a}) \Rightarrow \vec{c} - \vec{a} = 3 \vec{b} - 3 \vec{a} \Rightarrow \vec{c} = 3 \vec{b} - 3 \vec{a} + \vec{a} \Rightarrow \vec{c} = 3 \vec{b} - 2 \vec{a}$

Hence, If $\vec{a} and \vec{b}$ are the position vectors of A and B respectively, then the position vector of a point C on AB produced such that $\vec{A C} = 3 \vec{A B} is$ $3 \vec{b} - 2 \vec{a} .$

Page No 22.78:

Question 12:

In a regular hexagon ABCDEF, $\vec{A C} + \vec{A F} - \vec{A B} =_________________________.$

Answer:

Given:
ABCDEF is a regular hexagon

$\vec{A C} + \vec{A F} - \vec{A B} = \vec{A C} + \vec{C D} - \vec{A B} (∵ \vec{A F} = \vec{C D}) = (\vec{A C} + \vec{C D}) - \vec{A B} = (\vec{A D}) - \vec{A B} = (\vec{A B} + \vec{B D}) - \vec{A B} = \vec{A B} + \vec{B D} - \vec{A B} = \vec{B D}$

Hence, $\vec{A C} + \vec{A F} - \vec{A B} = \vec{B D} .$

Page No 22.78:

Question 13:

If O, A, B, C and D are five points, such that $3 \vec{O D} + \vec{D A} + \vec{D B} + \vec{D C} = k (\vec{O A} + \vec{O B} + \vec{O C})$ then k = ________________.

Answer:

Given: $3 \vec{O D} + \vec{D A} + \vec{D B} + \vec{D C} = k (\vec{O A} + \vec{O B} + \vec{O C})$

$3 \vec{O D} + \vec{D A} + \vec{D B} + \vec{D C} = k (\vec{O A} + \vec{O B} + \vec{O C}) \Rightarrow \vec{O D} + \vec{D A} + \vec{O D} + \vec{D B} + \vec{O D} + \vec{D C} = k (\vec{O A} + \vec{O B} + \vec{O C}) \Rightarrow (\vec{O D} + \vec{D A}) + (\vec{O D} + \vec{D B}) + (\vec{O D} + \vec{D C}) = k (\vec{O A} + \vec{O B} + \vec{O C}) \Rightarrow (\vec{O A}) + (\vec{O B}) + (\vec{O C}) = k (\vec{O A} + \vec{O B} + \vec{O C}) \Rightarrow (\vec{O A} + \vec{O B} + \vec{O C}) = k (\vec{O A} + \vec{O B} + \vec{O C}) \Rightarrow 1 = k$

Hence, k = 1.

Page No 22.78:

Question 14:

A, B, C, D, E are five coplanar points such that $\vec{D A} + \vec{D B} + \vec{D C} + \vec{A E} + \vec{B E} + \vec{C E} = k \vec{D E,}$ then k = ___________________.

Answer:

Given: $\vec{D A} + \vec{D B} + \vec{D C} + \vec{A E} + \vec{B E} + \vec{C E} = k \vec{D E}$

$\vec{D A} + \vec{D B} + \vec{D C} + \vec{A E} + \vec{B E} + \vec{C E} = k \vec{D E} \Rightarrow (\vec{D A} + \vec{A E}) + (\vec{D B} + \vec{B E}) + (\vec{D C} + \vec{C E}) = k \vec{D E} \Rightarrow (\vec{D E}) + (\vec{D E}) + (\vec{D E}) = k \vec{D E} \Rightarrow 3 (\vec{D E}) = k (\vec{D E}) \Rightarrow 3 = k$

Hence, k = 3.

Page No 22.78:

Question 15:

The vectors $\vec{a} and \vec{b}$ are non-collinear. If vectors $(x - 2) \vec{a} + \vec{b} and (2 x + 1) \vec{a} - \vec{b}$ are collinear, then x = _________________.

Answer:

Given:
The vectors $\vec{a} and \vec{b}$ are non-collinear.
Vectors $(x - 2) \vec{a} + \vec{b} and (2 x + 1) \vec{a} - \vec{b}$ are collinear.

Let $\vec{p} = (x - 2) \vec{a} + \vec{b} and \vec{q} = (2 x + 1) \vec{a} - \vec{b}$ are collinear
Then, $\vec{p} = λ \vec{q}, where λ is some scalar .$

$(x - 2) \vec{a} + \vec{b} = λ [(2 x + 1) \vec{a} - \vec{b}] \Rightarrow (x - 2) \vec{a} + \vec{b} = λ (2 x + 1) \vec{a} + (- λ) \vec{b} \Rightarrow x - 2 = λ (2 x + 1) and 1 = (- λ) \Rightarrow x - 2 = λ (2 x + 1) and - 1 = λ \Rightarrow x - 2 = - 1 (2 x + 1) \Rightarrow x - 2 = - 2 x - 1 \Rightarrow x + 2 x = - 1 + 2 \Rightarrow 3 x = 1 \Rightarrow x = \frac{1}{3}$

Hence, x = $\frac{1}{3}$ .

Page No 22.78:

Question 16:

If A, B, C, D, E are five points in a plane such that $\vec{A B} + \vec{A E} + \vec{B C} + \vec{D C} + \vec{E D} = k \vec{A C},$ then the value of k is _______________.

Answer:

Given: $\vec{A B} + \vec{A E} + \vec{B C} + \vec{D C} + \vec{E D} = k \vec{A C}$

$\vec{A B} + \vec{A E} + \vec{B C} + \vec{D C} + \vec{E D} = k \vec{A C} \Rightarrow (\vec{A B} + \vec{B C}) + \vec{D C} + (\vec{A E} + \vec{E D}) = k \vec{A C} \Rightarrow (\vec{A C}) + \vec{D C} + (\vec{A D}) = k \vec{A C} \Rightarrow (\vec{A C}) + (\vec{A D} + \vec{D C}) = k \vec{A C} \Rightarrow (\vec{A C}) + (\vec{A C}) = k \vec{A C} \Rightarrow 2 (\vec{A C}) = k \vec{A C} \Rightarrow 2 = k$

Hence, the value of k is 2.

Page No 22.78:

Question 17:

Let P be the point of intersection of the diagonal of a parallelogram ABCD and O is any point. If $\vec{O A} + \vec{O B} + \vec{O C} + \vec{O D} = λ \vec{O P}, Then λ =__________________.$

Answer:

Given:
P is the point of intersection of the diagonal of a parallelogram ABCD
$\vec{O A} + \vec{O B} + \vec{O C} + \vec{O D} = λ \vec{O P}$

$We know, \vec{O A} + \vec{A P} = \vec{O P} \vec{O B} + \vec{B P} = \vec{O P} \vec{O C} + \vec{C P} = \vec{O P} \vec{O D} + \vec{D P} = \vec{O P} Adding all, we get \vec{O A} + \vec{O B} + \vec{O C} + \vec{O D} + \vec{A P} + \vec{B P} + \vec{C P} + \vec{D P} = 4 \vec{O P} \Rightarrow \vec{O A} + \vec{O B} + \vec{O C} + \vec{O D} - \vec{C P} + \vec{B P} + \vec{C P} + \vec{D P} = 4 \vec{O P} (∵ \vec{A P} = \vec{P C} \Rightarrow \vec{A P} = - \vec{C P}) \Rightarrow \vec{O A} + \vec{O B} + \vec{O C} + \vec{O D} - \vec{D P} + \vec{D P} = 4 \vec{O P} (∵ \vec{B P} = \vec{P D} \Rightarrow \vec{B P} = - \vec{D P}) \Rightarrow \vec{O A} + \vec{O B} + \vec{O C} + \vec{O D} = 4 \vec{O P} \Rightarrow λ \vec{O P} = 4 \vec{O P} (given) \Rightarrow λ = 4$

Hence, $λ = 4 .$

Page No 22.78:

Question 18:

A, B, C, D, E are five coplanar points such that $\vec{D A} + \vec{D B} + \vec{D C} + \vec{A E} + \vec{B E} + \vec{C E} = λ \vec{D E}, Then λ =___________________.$

Answer:

Given: $\vec{D A} + \vec{D B} + \vec{D C} + \vec{A E} + \vec{B E} + \vec{C E} = λ \vec{D E}$

$\vec{D A} + \vec{D B} + \vec{D C} + \vec{A E} + \vec{B E} + \vec{C E} = λ \vec{D E} \Rightarrow (\vec{D A} + \vec{A E}) + (\vec{D B} + \vec{B E}) + (\vec{D C} + \vec{C E}) = λ \vec{D E} \Rightarrow (\vec{D E}) + (\vec{D E}) + (\vec{D E}) = λ \vec{D E} \Rightarrow 3 (\vec{D E}) = λ (\vec{D E}) \Rightarrow 3 = λ$

Hence, $λ$ = 3.

Page No 22.78:

Question 19:

If D, E, F are mid-points of the sides BC, CA and AB respectively of $∆ A B C, then \vec{A D} + \vec{B E} + \vec{C F} =______________________.$

Answer:

Given:
D, E, F are mid-points of the sides BC, CA and AB respectively

Let $\vec{a}, \vec{b} and \vec{c}$ are the position vectors of A, B and C, respectively.

$Now, \vec{B D} = \frac{1}{2} \vec{B C} = \frac{1}{2} (\vec{c} - \vec{b}) (∵ D is the mid - point of B C) Thus, in ∆ A B D, using triangle law of vector addition, \vec{A D} = \vec{A B} + \vec{B D} \Rightarrow \vec{A D} = \vec{b} - \vec{a} + \frac{1}{2} \vec{c} - \frac{1}{2} \vec{b} \Rightarrow \vec{A D} = \frac{1}{2} \vec{b} + \frac{1}{2} \vec{c} - \vec{a} . . . (1) Similarly, \vec{C E} = \frac{1}{2} \vec{C A} = \frac{1}{2} (\vec{a} - \vec{c}) (∵ E is the mid - point of A C) Thus, in ∆ B C E, using triangle law of vector addition, \vec{B E} = \vec{B C} + \vec{C E} \Rightarrow \vec{B E} = \vec{c} - \vec{b} + \frac{1}{2} \vec{a} - \frac{1}{2} \vec{c} \Rightarrow \vec{B E} = \frac{1}{2} \vec{c} + \frac{1}{2} \vec{a} - \vec{b} . . . (2) And \vec{A F} = \frac{1}{2} \vec{A B} = \frac{1}{2} (\vec{b} - \vec{a}) (∵ F is the mid - point of A B) Thus, in ∆ A C F, using triangle law of vector addition, \vec{C F} = \vec{C A} + \vec{A F} \Rightarrow \vec{C F} = \vec{a} - \vec{c} + \frac{1}{2} \vec{b} - \frac{1}{2} \vec{a} \Rightarrow \vec{C F} = \frac{1}{2} \vec{b} + \frac{1}{2} \vec{a} - \vec{c} . . . (3) Adding (1), (2) and (3), we get \vec{A D} + \vec{B E} + \vec{C F} = \frac{1}{2} \vec{b} + \frac{1}{2} \vec{c} - \vec{a} + \frac{1}{2} \vec{c} + \frac{1}{2} \vec{a} - \vec{b} + \frac{1}{2} \vec{b} + \frac{1}{2} \vec{a} - \vec{c} = \vec{0}$

Hence, $\vec{A D} + \vec{B E} + \vec{C F} = \vec{0} .$

Page No 22.78:

Question 20:

The algebraic sum of the vectors directed from the vertices to the mid-points of the opposite side is equal to ________________.

Answer:

Let D, E, F are mid-points of the sides BC, CA and AB respectively.
To find: $\vec{A D} + \vec{B E} + \vec{C F}$

Let $\vec{a}, \vec{b} and \vec{c}$ are the position vectors of A, B and C, respectively.

$Now, \vec{B D} = \frac{1}{2} \vec{B C} = \frac{1}{2} (\vec{c} - \vec{b}) (∵ D is the mid - point of B C) Thus, in ∆ A B D, using triangle law of vector addition, \vec{A D} = \vec{A B} + \vec{B D} \Rightarrow \vec{A D} = \vec{b} - \vec{a} + \frac{1}{2} \vec{c} - \frac{1}{2} \vec{b} \Rightarrow \vec{A D} = \frac{1}{2} \vec{b} + \frac{1}{2} \vec{c} - \vec{a} . . . (1) Similarly, \vec{C E} = \frac{1}{2} \vec{C A} = \frac{1}{2} (\vec{a} - \vec{c}) (∵ E is the mid - point of A C) Thus, in ∆ B C E, using triangle law of vector addition, \vec{B E} = \vec{B C} + \vec{C E} \Rightarrow \vec{B E} = \vec{c} - \vec{b} + \frac{1}{2} \vec{a} - \frac{1}{2} \vec{c} \Rightarrow \vec{B E} = \frac{1}{2} \vec{c} + \frac{1}{2} \vec{a} - \vec{b} . . . (2) And \vec{A F} = \frac{1}{2} \vec{A B} = \frac{1}{2} (\vec{b} - \vec{a}) (∵ F is the mid - point of A B) Thus, in ∆ A C F, using triangle law of vector addition, \vec{C F} = \vec{C A} + \vec{A F} \Rightarrow \vec{C F} = \vec{a} - \vec{c} + \frac{1}{2} \vec{b} - \frac{1}{2} \vec{a} \Rightarrow \vec{C F} = \frac{1}{2} \vec{b} + \frac{1}{2} \vec{a} - \vec{c} . . . (3) Adding (1), (2) and (3), we get \vec{A D} + \vec{B E} + \vec{C F} = \frac{1}{2} \vec{b} + \frac{1}{2} \vec{c} - \vec{a} + \frac{1}{2} \vec{c} + \frac{1}{2} \vec{a} - \vec{b} + \frac{1}{2} \vec{b} + \frac{1}{2} \vec{a} - \vec{c} = \vec{0}$

Hence, the algebraic sum of the vectors directed from the vertices to the mid-points of the opposite side is equal to $\vec{0} .$

Page No 22.78:

Question 21:

The values of k for which $|k \vec{a}| < |\vec{a}| and k \vec{a} + \frac{1}{2} \vec{a}$ is parallel to $\vec{a}$ holds true, are _______________.

Answer:

Given:
$|k \vec{a}| < |\vec{a}| and k \vec{a} + \frac{1}{2} \vec{a}$ is parallel to $\vec{a}$

$|k \vec{a}| < |\vec{a}| \Rightarrow |k| |\vec{a}| < |\vec{a}| \Rightarrow |k| < 1 \Rightarrow - 1 < k < 1 . . . (1) Also, k \vec{a} + \frac{1}{2} \vec{a} is parallel to \vec{a} But at k = - \frac{1}{2}, k \vec{a} + \frac{1}{2} \vec{a} = - \frac{1}{2} \vec{a} + \frac{1}{2} \vec{a} = \vec{0} Thus, k \neq - \frac{1}{2} . . . (2)$

Hence, the values of k for which $|k \vec{a}| < |\vec{a}| and k \vec{a} + \frac{1}{2} \vec{a}$ is parallel to $\vec{a}$ holds true, are $(- 1, 1) ~ \{- \frac{1}{2}\}$ .

Page No 22.78:

Question 22:

The vector $\vec{a} + \vec{b}$ bisects the angle between the non-collinear vectors $\vec{a} and \vec{b}$ , if ___________.

Answer:

Given: Vector $\vec{a} + \vec{b}$ bisects the angle between the non-collinear vectors $\vec{a} and \vec{b}$ .

Non-collinear vectors $\vec{a} and \vec{b}$ forms a parallelogram.
Vector $\vec{a} + \vec{b}$ is the diagonal of the parallelogram.
And the diagonal bisects the angle if it is a rhombus.
Thus, $|\vec{a}| = |\vec{b}|$ .

Hence, the vector $\vec{a} + \vec{b}$ bisects the angle between the non-collinear vectors $\vec{a} and \vec{b}$ , if $|\vec{a}| = |\vec{b}| .$

Page No 22.78:

Question 1:

Define "zero vector".

Answer:

A vector whose initial and terminal point are coincident is called a zero vector or null vector. The null vector is denoted by $\vec{0}$ . The magnitude of null vectors is zero.

Page No 22.78:

Question 2:

Define unit vector.

Answer:

A vector whose modulus is unity is called a unit vector.The unit vector in the direction of a vector $\vec{a}$ is denoted by $\hat{a} .$
Thus, $|\hat{a}| = 1$

Page No 22.79:

Question 3:

Define position vector of a point.

Answer:

A point $O$ is fixed as origin in space (or plane) and $P$ is any point, then $\vec{O P}$ is called a position vector of $P$ with respect to $O .$

Page No 22.79:

Question 4:

Write $\vec{P Q} + \vec{R P} + \vec{Q R}$ in the simplified form.

Answer:

We have, $\vec{P Q} + \vec{R P} + \vec{Q R}$ $= \vec{P Q} + \vec{Q R} + \vec{R P}$
$= \vec{P R} + \vec{R P}$ [∴ $\vec{P Q} + \vec{Q R} = \vec{P R}$ ]
$= \vec{0}$

Page No 22.79:

Question 5:

If $\vec{a}$ and $\vec{b}$ are two non-collinear vectors such that $x \vec{a} + y \vec{b} = \vec{0},$ then write the values of x and y.

Answer:

We have, $x \vec{a} + y \vec{b} = \vec{0}$
$\Rightarrow x = 0 and y = 0$ [ $∵$ $\vec{a}$ and $\vec{b}$ are non-collinear vectors]

Page No 22.79:

Question 6:

If $\vec{a}$ and $\vec{b}$ represent two adjacent sides of a parallelogram, then write vectors representing its diagonals.

Answer:

Let $\vec{a}$ and $\vec{b}$ represents two adjacent sides of a parallelogram $A B C D$ .
∴ $A B = D C and A D = B C$

$\Rightarrow \vec{D C} = \vec{A B} = \vec{a}$ and $\vec{A D} = \vec{B C} = \vec{b}$
In $△ A B C$ ,
$\vec{A B} + \vec{B C} = \vec{A C} \Rightarrow \vec{a} + \vec{b} = \vec{A C}$
In $△ A B D$ ,
$\vec{A D} + \vec{D B} = \vec{A B} \Rightarrow \vec{b} + \vec{D B} = \vec{a} \Rightarrow \vec{D B} = \vec{a} - \vec{b}$

Page No 22.79:

Question 7:

If $\vec{a}$ , $\vec{b}$ , $\vec{c}$ represent the sides of a triangle taken in order, then write the value of $\vec{a} + \vec{b} + \vec{c} .$

Answer:

Let $A B C$ be a triangle such that $\vec{B C} = \vec{a}, \vec{C A} = \vec{b}$ and $\vec{A B} = \vec{c} .$ Then,
$\vec{a} + \vec{b} + \vec{c} = \vec{B C} + \vec{C A} + \vec{A B}$
$= \vec{B A} + \vec{A B} = \vec{0}$ [∵ $\vec{B C} + \vec{C A} = \vec{B A}$ ]

Page No 22.79:

Question 8:

If $\vec{a}$ , $\vec{b}$ , $\vec{c}$ are position vectors of the vertices A, B and C respectively, of a triangle ABC, write the value of $\vec{A B} + \vec{B C} + \vec{C A} .$

Answer:

Given: $\vec{a}, \vec{b}$ and $\vec{c}$ are the position vectors of $A, B$ and $C$ respectively. Then,
$\vec{A B} = \vec{b} - \vec{a} \vec{B C} = \vec{c} - \vec{b} \vec{C A} = \vec{a} - \vec{c}$
Consider,
$\vec{A B} + \vec{B C} + \vec{C A} = \vec{b} - \vec{a} + \vec{c} - \vec{b} + \vec{a} - \vec{c} = \vec{0}$

Page No 22.79:

Question 9:

If $\vec{a}$ , $\vec{b}$ , $\vec{c}$ are position vectors of the points A, B and C respectively, write the value of $\vec{A B} + \vec{B C} + \vec{A C} .$

Answer:

Given: $\vec{a}, \vec{b}, \vec{c}$ are the position vectors of $A, B, C$ respectively. Then,
$\vec{A B} = \vec{b} - \vec{a} \vec{B C} = \vec{c} - \vec{b} \vec{A C} = \vec{c} - \vec{a}$
Therefore,
$\vec{A B} + \vec{B C} + \vec{A C} = \vec{b} - \vec{a} + \vec{c} - \vec{b} + \vec{c} - \vec{a} = 2 (\vec{c} - \vec{a})$

Page No 22.79:

Question 10:

If $\vec{a}$ , $\vec{b}$ , $\vec{c}$ are the position vectors of the vertices of a triangle, then write the position vector of its centroid.

Answer:

Let $A B C$ be a triangle and $D, E$ and $F$ are the midpoints of the sides $B C, C A$ and $A B$ respectively. Also, Let $\vec{a}, \vec{b}, \vec{c}$ are the position vectors of $A, B, C$ respectively. Then the position vectors of $D, E, F$ are $(\frac{\vec{b} + \vec{c}}{2}), (\frac{\vec{c} + \vec{a}}{2}), (\frac{\vec{a} + \vec{b}}{2})$ respectively.
The position vector of a point divides $A D$ in the ratio of 2 ; is $\frac{1 . \vec{a} + 2 \frac{\vec{b} + \vec{c}}{2}}{2} = \frac{\vec{a} + \vec{b} + \vec{c}}{3} .$
Similarly, Position vectors of the points divides $B E, C F$ in the ratio of $2 : 1$ are equal to $\frac{\vec{a} + \vec{b} + \vec{c}}{3}$ .
Thus, the point dividing $A D$ in the ratio 2 : 1 also divides $B E, C F$ in the same ratio.
Hence, the medians of a triangle are concurrent and the position vector of the centroid is $\frac{\vec{a} + \vec{b} + \vec{c}}{3}$ .

Page No 22.79:

Question 11:

If G denotes the centroid of ∆ABC, then write the value of $\vec{G A} + \vec{G B} + \vec{G C} .$

Answer:

Let $\vec{a}, \vec{b}, \vec{c}$ be the position vectors of the vertices $A, B, C$ respectively. Then, the position vector of the centroid $G$ is $\frac{\vec{a} + \vec{b} + \vec{c}}{3}$
Thus,
$\vec{G A} + \vec{G B} + \vec{G C} = \vec{a} - (\frac{\vec{a} + \vec{b} + \vec{c}}{3}) + \vec{b} - (\frac{\vec{a} + \vec{b} + \vec{c}}{3}) + \vec{c} - (\frac{\vec{a} + \vec{b} + \vec{c}}{3}) = (\vec{a} + \vec{b} + \vec{c}) - 3 (\frac{\vec{a} + \vec{b} + \vec{c}}{3}) = (\vec{a} + \vec{b} + \vec{c}) - (\vec{a} + \vec{b} + \vec{c}) = \vec{0}$

Page No 22.79:

Question 12:

If $\vec{a}$ and $\vec{b}$ denote the position vectors of points A and B respectively and C is a point on AB such that 3AC = 2AB, then write the position vector of C.

Answer:

Given: $\vec{a}$ and $\vec{b}$ are the position vectors of points $A$ and $B$ respectively and $C$ is a point on $A B$ such that $3 A C = 2 A B .$
Let $\vec{c}$ is the position vector of C
Now,
$\vec{A B} = \vec{b} - \vec{a}$
$\vec{A C} = \vec{c} - \vec{a}$
Consider,
$3 A C = 2 A B \Rightarrow 3 (\vec{c} - \vec{a}) = 2 (\vec{b} - \vec{a}) \Rightarrow 3 \vec{c} - 3 \vec{a} = 2 \vec{b} - 2 \vec{a} \Rightarrow 3 \vec{c} = 2 \vec{b} + \vec{a} \Rightarrow \vec{c} = \frac{1}{3} (2 \vec{b} + \vec{a}) \Rightarrow \vec{c} = \frac{1}{3} (\vec{a} + 2 \vec{b})$

Hence, the position vector of $C$ is $\frac{1}{3} (\vec{a} + 2 \vec{b})$

Page No 22.79:

Question 13:

If D is the mid-point of side BC of a triangle ABC such that $\vec{A B} + \vec{A C} = λ \vec{A D},$ write the value of λ.

Answer:

Given: $D$ is the midpoint of the side $B C$ of a triangle $A B C$ such that $\vec{A B} + \vec{B C} = λ \vec{A D} .$
Let $\vec{a}, \vec{b}, \vec{c}$ are the position vectors of AB, BC and CA.
Now, the position vector of $D$ is $\frac{\vec{b} + \vec{c}}{2}$ . Then,
$\vec{A B} = \vec{b} - \vec{a} \vec{A C} = \vec{c} - \vec{a} \vec{A D} = \frac{\vec{b} + \vec{c}}{2} - \vec{a}$
Now, we have,
$\vec{A B} + \vec{A C} = λ \vec{A D} \Rightarrow \vec{b} - \vec{a} + \vec{c} - \vec{a} = λ (\frac{\vec{b} + \vec{c}}{2} - \vec{a}) \Rightarrow \vec{b} + \vec{c} - 2 \vec{a} = λ (\frac{\vec{b} + \vec{c} - 2 \vec{a}}{2}) \Rightarrow λ = 2$

Page No 22.79:

Question 14:

If D, E, F are the mid-points of the sides BC, CA and AB respectively of a triangle ABC, write the value of $\vec{A D} + \vec{B E} + \vec{C F} .$

Answer:

Given: $D, E, F$ are the midpoints of the sides $B C, C A, A B$ respectively. Then, the position vectors of the midpoints $D, E, F$ are given by $\frac{\vec{b} + \vec{c}}{2}, \frac{\vec{c} + \vec{a}}{2}, \frac{\vec{a} + \vec{b}}{2}$
$Now, \vec{A D} + \vec{B E} + \vec{C F} = (\frac{\vec{b} + \vec{c}}{2}) - \vec{a} + (\frac{\vec{c} + \vec{a}}{2}) - \vec{b} + (\frac{\vec{a} + \vec{b}}{2}) - \vec{c} = 2 (\frac{\vec{a} + \vec{b} + \vec{c}}{2}) - (\vec{a} + \vec{b} + \vec{c}) = (\vec{a} + \vec{b} + \vec{c}) - (\vec{a} + \vec{b} + \vec{c}) = \vec{0}$

Page No 22.79:

Question 15:

If $\vec{a}$ is a non-zero vector of modulus a and m is a non-zero scalar such that m $\vec{a}$ is a unit vector, write the value of m.

Answer:

Given $\vec{a}$ a non zero vector with modulus $a$ . Also, $m \vec{a}$ is the unit vector. Therefore,
$|m \vec{a}| = 1 \Rightarrow |m| |\vec{a}| = 1 \Rightarrow |m| a = 1 \Rightarrow |m| = \frac{1}{a} \Rightarrow m = \pm \frac{1}{a}$

Page No 22.79:

Question 16:

If $\vec{a}$ , $\vec{b}$ , $\vec{c}$ are the position vectors of the vertices of an equilateral triangle whose orthocentre is at the origin, then write the value of $\vec{a} + \vec{b} + \vec{c} .$

Answer:

Let, ABC be a given equilateral triangle and its vertices are A( $\overset{⇀}{a}$ ), B( $\overset{⇀}{b}$ ) and C( $\overset{⇀}{c}$ ).
Also, O( $\overset{⇀}{0}$ ) be the orthocentre of triangle ABC.
We know that centroid and orthocentre of equilateral triangle coincide at one point.
$Orthocentre of △ A B C = \overset{⇀}{0} \Rightarrow Centroid △ A B C = \overset{⇀}{0} \Rightarrow \frac{\overset{⇀}{a} + \overset{⇀}{b} + \overset{⇀}{c}}{3} = \overset{⇀}{0} ∴ \overset{⇀}{a} + \overset{⇀}{b} + \overset{⇀}{c} = \overset{⇀}{0}$

Page No 22.79:

Question 17:

Write a unit vector making equal acute angles with the coordinates axes.

Answer:

Suppose $\vec{r}$ makes an angle $α$ with each of the axis $Ο Χ$ , $Ο Υ$ and $Ο Ζ$ .
Then, its direction cosines are $l = \cos α, m = \cos α, n = \cos α$ .
Now,
$Ι^{2} + m^{2} + n^{2} = 1 \Rightarrow l^{2} + l^{2} + l^{2} = 1 [∵ l = m = n] \Rightarrow 3 l^{2} = 1 \Rightarrow l^{2} = \frac{1}{3} \Rightarrow l = \pm \frac{1}{\sqrt{3}}$
Since, the angle is acute Hence, we take only positive value
Therefore, unit vector is $(\frac{1}{\sqrt{3}} \hat{i} + \frac{1}{\sqrt{3}} \hat{j} + \frac{1}{\sqrt{3}} \hat{k}) .$

Page No 22.79:

Question 18:

If a vector makes angles α, β, γ with OX, OY and OZ respectively, then write the value of sin² α + sin² β + sin² γ.

Answer:

Suppose, a vector $\vec{O P}$ makes an angle $α, β, γ$ with $O X, O Y, O Z$ respectively. Then, direction cosines of the vector are given by
$l = \cos α, m = \cos β and n = \cos γ$
Consider,
$\sin^{2} α + \sin^{2} β + \sin^{2} γ = 1 - \cos^{2} α + 1 - \cos^{2} β + 1 - \cos^{2} γ = 3 - (\cos^{2} α + \cos^{2} β + \cos^{2} γ) = 3 - (l^{2} + m^{2} + n^{2})$
$= 3 - 1$ [∵ $l^{2} + m^{2} + n^{2} = 1$ ]
= 2

Page No 22.79:

Question 19:

Write a vector of magnitude 12 units which makes 45° angle with X-axis, 60° angle with Y-axis and an obtuse angle with Z-axis.

Answer:

Suppose a vector $\vec{r}$ makes an angle 45 $°$ with $Ο Χ$ , 60 $°$ with $Ο Υ$ and having magnitude 12 units.
$l = \cos 45 ° = \frac{1}{\sqrt{2}} and m = \cos 60 ° = \frac{1}{2}$
$Now, l^{2} + m^{2} + n^{2} = 1 \Rightarrow \frac{1}{2} + \frac{1}{4} + n^{2} = 1 \Rightarrow n^{2} = \frac{1}{4} \Rightarrow n = - \frac{1}{2} [∵ The angle with the z - axis is obtuse]$
Therefore,
$\vec{r} = |\vec{r}| (l \hat{i} + m \hat{j} + n \hat{k}) = 12 (\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{2} \hat{j} - \frac{1}{2} \hat{k}) = 6 (\sqrt{2} \hat{i} + \hat{j} - \hat{k})$

Page No 22.79:

Question 20:

Write the length (magnitude) of a vector whose projections on the coordinate axes are 12, 3 and 4 units.

Answer:

Given: Projection on the coordinate axes are $12, 3, 4$ units. Therefore,
Length of vector $= \sqrt{12^{2} + 3^{2} + 4^{2}}$
= $\sqrt{169}$
= 13

Page No 22.79:

Question 21:

Write the position vector of a point dividing the line segment joining points A and B with position vectors $\vec{a}$ and $\vec{b}$ externally in the ratio 1 : 4, where $\vec{a} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k} and \vec{b} = - \hat{i} + \hat{j} + \hat{k} .$

Answer:

The position vectors of $A$ and $B$ are
$\vec{a} = 2 \hat{i} + 3 j + 4 \hat{k} \vec{b} = - \hat{i} + \hat{j} + \hat{k}$
Let $C$ divides $A B$ in the ratio such that AB : CB = 1 : 4
Position vector of $C$ = $\frac{1 (- \hat{i} + \hat{j} + \hat{k}) - 4 (2 \hat{i} + 3 \hat{j} + 4 \hat{k})}{1 - 4}$
= $\frac{- \hat{i} + \hat{j} + \hat{k} - 8 \hat{i} - 12 \hat{j} - 16 \hat{k}}{- 3}$
= $\frac{- 9 \hat{i} - 11 \hat{j} - 15 \hat{k}}{- 3}$
= $3 \hat{i} + \frac{11 \hat{j}}{3} + 5 \hat{k}$

Page No 22.79:

Question 22:

Write the direction cosines of the vector $\vec{r} = 6 \hat{i} - 2 \hat{j} + 3 \hat{k} .$

Answer:

Given: $\vec{r} = 6 \hat{i} - 2 \hat{j} + 3 \hat{k}$
Then, direction cosines of $\overset{⏜}{r}$ are $\frac{6}{\sqrt{6^{2} + {(- 2)}^{2} + 3^{2}}}, \frac{- 2}{\sqrt{6^{2} + {(- 2)}^{2} + 3^{2}}}, \frac{3}{\sqrt{6^{2} + {(- 2)}^{2} + 3^{2}}}$ or, $\frac{6}{7}, \frac{- 2}{7}, \frac{3}{7}$

Page No 22.79:

Question 23:

If $\vec{a} = \hat{i} + \hat{j}, \vec{b} = \hat{j} + \hat{k} and \vec{c} = \hat{k} + \hat{i},$ write unit vectors parallel to $\vec{a} + \vec{b} - 2 \vec{c} .$

Answer:

Given: $\vec{a} = \hat{i} + \hat{j}, \vec{b} = \hat{j} + \hat{k}, \vec{c} = \hat{k} + \hat{i}$
Now, $\vec{a} + \vec{b} - 2 \vec{c} = \hat{i} + \hat{j} + \hat{j} + \hat{k} - 2 \hat{k} - 2 \hat{i}$
$= - \hat{i} + 2 \hat{j} - \hat{k}$
Unit vector parallel to $\vec{a} + \vec{b} - 2 \vec{c} = \frac{- \hat{i} + 2 \hat{j} - \hat{k}}{\sqrt{{(- 1)}^{2} + 2^{2} + {(- 1)}^{2}}}$
$= \frac{- \hat{i} + 2 \hat{j} - \hat{k}}{\sqrt{6}}$

Page No 22.80:

Question 24:

If $\vec{a} = \hat{i} + 2 \hat{j}, \vec{b} = \hat{j} + 2 \hat{k},$ write a unit vector along the vector $3 \vec{a} - 2 \vec{b} .$

Answer:

Given: $\vec{a} = \hat{i} + 2 \hat{j}, \vec{b} = \hat{j} + 2 \hat{k}$
Therefore,
$3 \vec{a} - 2 \vec{b} = 3 \hat{i} + 6 \hat{j} - 2 \hat{j} - 4 \hat{k} = 3 \hat{i} + 4 \hat{j} - 4 \hat{k}$

Hence, Unit vector along $3 \vec{a} - 2 \vec{b} = \frac{3 \hat{i} + 4 \hat{j} - 4 \hat{k}}{\sqrt{3^{2} + 4^{2} + {(- 4)}^{2}}} = \frac{3 \hat{i} + 4 \hat{j} - 4 \hat{k}}{\sqrt{9 + 16 + 16}} = \frac{1}{\sqrt{41}} (3 \hat{i} + 4 \hat{j} - 4 \hat{k})$

Page No 22.80:

Question 25:

Write the position vector of a point dividing the line segment joining points having position vectors $\hat{i} + \hat{j} - 2 \hat{k} and 2 \hat{i} - \hat{j} + 3 \hat{k}$ externally in the ratio 2:3.

Answer:

Let $A$ and $B$ be the points with position vectors $\vec{a} = \hat{i} + \hat{j} - 2 \hat{k}, \vec{b} = 2 \hat{i} - \hat{j} + 3 \hat{k}$ respectively.
Let $C$ divide $A B$ externally in the ratio 2 : 3 such that $A C : C B = 2 : 3$
∴ Position vector of $C$ = $\frac{2 (2 \hat{i} - \hat{j} + 3 \hat{k}) - 3 (\hat{i} + \hat{j} - 2 \hat{k})}{2 - 3}$
   = $\frac{4 \hat{i} - 2 \hat{j} + 6 \hat{k} - 3 \hat{i} - 3 \hat{j} + 6 \hat{k}}{- 1}$
   = $\frac{\hat{i} - 5 \hat{j} + 12 \hat{k}}{- 1}$
   = $- \hat{i} + 5 \hat{j} - 12 \hat{k}$

Page No 22.80:

Question 26:

If $\vec{a} = \hat{i} + \hat{j}, \vec{b} = \hat{j} + \hat{k}, \vec{c} = \hat{k} + \hat{i},$ find the unit vector in the direction of $\vec{a} + \vec{b} + \vec{c}$ .

Answer:

Let $\vec{a} = \hat{i} + \hat{j}, \vec{b} = \hat{j} + \hat{k}, \vec{c} = \hat{k} + \hat{i}$
Then, $\vec{a} + \vec{b} + \vec{c} = \hat{i} + \hat{j} + \hat{j} + \hat{k} + \hat{k} + \hat{i} = 2 (\hat{i} + \hat{j} + \hat{k})$
∴ $|\vec{a} + \vec{b} + \vec{c}| = \sqrt{2^{2} + 2^{2} + 2^{2}} = \sqrt{12} = 2 \sqrt{3}$
Therefore, unit vector in the direction of $\vec{a} + \vec{b} + \vec{c} = \frac{2 (\hat{i} + \hat{j} + \hat{k})}{2 \sqrt{3}} = \frac{1}{\sqrt{3}} (\hat{i} + \hat{j} + \hat{k})$

Page No 22.80:

Question 27:

$If \vec{a} = 3 \hat{i} - \hat{j} - 4 \hat{k}, \vec{b} = - 2 \hat{i} + 4 \hat{j} - 3 \hat{k} and \vec{c} = \hat{i} + 2 \hat{j} - \hat{k}, find |3 \vec{a} - 2 \vec{b} + 4 \vec{c}| .$

Answer:

Given $\vec{a} = 3 \hat{i} - \hat{j} - 4 \hat{k}, \vec{b} = - 2 \hat{i} + 4 \hat{j} - 3 \hat{k}, \vec{c} = \hat{i} + 2 \hat{j} - \hat{k}$
Now, $3 \vec{a} - 2 \vec{b} + 4 \vec{c} = 3 (3 \hat{i} - \hat{j} - 4 \hat{k}) - 2 (- 2 \hat{i} + 4 \hat{j} - 3 \hat{k}) + 4 (\hat{i} + 2 \hat{j} - \hat{k})$
= $9 \hat{i} - 3 \hat{j} - 12 \hat{k} + 4 \hat{i} - 8 \hat{j} + 6 \hat{k} + 4 \hat{i} + 8 \hat{j} - 4 \hat{k}$
= $17 \hat{i} - 3 \hat{j} - 10 \hat{k}$
$∴$ $|3 \vec{a} - 2 \vec{b} + 4 \vec{c}| = \sqrt{17^{2} + {(- 3)}^{2} + {(- 10)}^{2}} = \sqrt{289 + 9 + 100} = \sqrt{398}$

Page No 22.80:

Question 28:

A unit vector $\vec{r}$ makes angles $\frac{π}{3}$ and $\frac{π}{2}$ with $\hat{j} and \hat{k}$ respectively and an acute angle θ with $\hat{i}$ . Find θ.

Answer:

A unit vector makes an angle $\frac{π}{3}$ and $\frac{π}{2}$ with $\hat{j}$ and $\hat{k}$
Let $Ι, m, n$ be its direction cosines
∴ $l = \cos θ, m = \cos (\frac{π}{3}) = \frac{1}{2}, n = \cos (\frac{π}{2}) = 0$
Now
∴ $l^{2} + m^{2} + n^{2} = 1$
$\Rightarrow$ $l^{2} + \frac{1}{4} + 0 = 1$
$\Rightarrow$ $l^{2} = 1 - \frac{1}{4} = \frac{3}{4}$
$\Rightarrow$ $l = \pm \frac{\sqrt{3}}{2}$
∴ $\vec{r}$ makes an angle $30 °, 150 °$ with $\overset{⏜}{i}$
Since, angle $θ$ is acute.
∴ $θ = 30 °$

Page No 22.80:

Question 29:

Write a unit vector in the direction of $\vec{a} = 3 \hat{i} + 2 \hat{j} + 6 \hat{k} .$

Answer:

We have,
$\vec{a} = 3 \hat{i} - 2 \hat{j} + 6 \hat{k} |\vec{a}| = \sqrt{3^{2} + {(- 2)}^{2} + 6^{2}} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$

∴ Unit vector in the direction of $\vec{a}$ = $\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{1}{7} (3 \hat{i} - 2 \hat{j} + 6 \hat{k}) = \frac{3}{7} \hat{i} - \frac{2}{7} \hat{j} + \frac{6}{7} \hat{k}$

Page No 22.80:

Question 30:

If $\vec{a} = \hat{i} + 2 \hat{j} - 3 \hat{k} and \vec{b} = 2 \hat{i} + 4 \hat{j} + 9 \hat{k},$ find a unit vector parallel to $\vec{a} + \vec{b}$ .

Answer:

Given:
$\vec{a} = \hat{i} + 2 \hat{j} - 3 \hat{k}, \vec{b} = 2 \hat{i} + 4 \hat{j} + 9 \hat{k} Now, \vec{a} + \vec{b} = 3 \hat{i} + 6 \hat{j} + 6 \hat{k} |\vec{a} + \vec{b}| = \sqrt{3^{2} + 6^{2} + 6^{2}} = \sqrt{9 + 36 + 36} = \sqrt{81} = 9$

Unit vector parallel to $\vec{a} + \vec{b} = \frac{\vec{a} + \vec{b}}{|\vec{a} + \vec{b}|} = \frac{3 \hat{i} + 6 \hat{j} + 6 \hat{k}}{9} = \frac{1}{9} \times 3 (\hat{i} + 2 \hat{j} + 2 \hat{k}) = \frac{1}{3} (\hat{i} + 2 \hat{j} + 2 \hat{k})$

Page No 22.80:

Question 31:

Write a unit vector in the direction of $\vec{b} = 2 \hat{i} + \hat{j} + 2 \hat{k}$ .

Answer:

Given:
$\vec{b} = 2 \hat{i} + \hat{j} + 2 \hat{k} |\vec{b}| = \sqrt{2^{2} + 1^{2} + 2^{2}} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$
∴ Unit vector = $\frac{\vec{b}}{|\vec{b}|} = \frac{1}{3} (2 \hat{i} + \hat{j} + 2 \hat{k}) = \frac{2}{3} \hat{i} + \frac{1}{3} \overset{⏜}{j} + \frac{2}{3} \hat{k}$

Page No 22.80:

Question 32:

Find the position vector of the mid-point of the line segment AB, where A is the point (3, 4, −2) and B is the point (1, 2, 4).

Answer:

Given: A (3, 4, −2) and B(1, 2, 4)
Let C is the mid point of AB
∴ Position vector of C = $\frac{3 \hat{i} + 4 \hat{j} - 2 \hat{k} + \hat{i} + 2 \hat{j} + 4 \hat{k}}{2}$
$= \frac{4 \hat{i} + 6 \hat{j} + 2 \hat{k}}{2} = 2 \hat{i} + 3 \hat{j} + \hat{k}$

Page No 22.80:

Question 33:

Find a vector in the direction of $\vec{a} = 2 \hat{i} - \hat{j} + 2 \hat{k},$ which has magnitude of 6 units.

Answer:

Given:
$\vec{a} = 2 \hat{i} - \hat{j} + 2 \hat{k} |\vec{a}| = \sqrt{2^{2} + {(- 1)}^{2} + 2^{2}} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$
∴ Required Vector $= 6 \times \frac{\vec{a}}{|\vec{a}|} = 6 \times \frac{(2 \hat{i} - \hat{j} + 2 \hat{k})}{3} = 4 \hat{i} - 2 \hat{j} + 4 \hat{k}$

Page No 22.80:

Question 34:

What is the cosine of the angle which the vector $\sqrt{2} \hat{i} + \hat{j} + \hat{k}$ makes with y-axis?

Answer:

Given $\sqrt{2} \hat{i} + \hat{j} + \hat{k}$ .
Therefore , direction cosines are $\frac{\sqrt{2}}{\sqrt{{(\sqrt{2})}^{2} + 1^{2} + 1^{2}}}, \frac{1}{\sqrt{{(\sqrt{2})}^{2} + 1^{2} + 1^{2}}}, \frac{1}{\sqrt{{(\sqrt{2})}^{2} + 1^{2} + 1^{2}}}$ or $\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2}$

So, cosine angle with respect to y-axis is $\frac{1}{2}$

Page No 22.80:

Question 35:

Write two different vectors having same magnitude.

Answer:

Let $\vec{a} = 2 \hat{i} - \hat{j} + 3 \hat{k}$ and $\vec{b} = - 2 \hat{i} + \hat{j} - 3 \hat{k}$
It can be observed that
$|\vec{a}| = \sqrt{2^{2} + {(- 1)}^{2} + 3^{2}} = \sqrt{14} |\vec{b}| = \sqrt{{(- 2)}^{2} + 1^{2} + {(- 3)}^{2}} = \sqrt{14}$

Hence, $\vec{a}$ and $\vec{b}$ are two vectors having same magnitude.

Page No 22.80:

Question 36:

Write two different vectors having same direction.

Answer:

Let $\vec{p} = \hat{i} + \hat{2 j} + 3 \hat{k}$ and $\vec{q} = 2 \hat{i} + 4 \hat{j} + 6 \hat{k}$
Then, direction cosines of $\vec{p}$ are
$l = \frac{1}{\sqrt{1^{2} + 2^{2} + 3^{2}}} = \frac{2}{\sqrt{14}}, m = \frac{2}{\sqrt{1^{2} + 2^{2} + 3^{2}}} = \frac{2}{\sqrt{14}} and n = \frac{3}{\sqrt{1^{2} + 2^{2} + 3^{2}}} = \frac{3}{\sqrt{14}}$
Direction cosines of $\vec{q}$ are
$l = \frac{2}{\sqrt{2^{2} + 4^{2} + 6^{2}}} = \frac{2}{2 \sqrt{14}} = \frac{1}{\sqrt{14}}, m = \frac{4}{\sqrt{2^{2} + 4^{2} + 6^{2}}} = \frac{4}{2 \sqrt{14}} = \frac{2}{\sqrt{14}} and n = \frac{6}{\sqrt{2^{2} + 4^{2} + 6^{2}}} = \frac{6}{2 \sqrt{14}} = \frac{3}{\sqrt{14}}$

The direction cosines of two vectors are same. Hence the two diffrent vectors $\vec{p}, \vec{q}$ have same directions.

Page No 22.80:

Question 37:

Write a vector in the direction of vector $5 \hat{i} - \hat{j} + 2 \hat{k}$ which has magnitude of 8 unit.

Answer:

Given:
$\vec{a} = 5 \hat{i} - \hat{j} + 2 \hat{k} |\vec{a}| = \sqrt{5^{2} + {(- 1)}^{2} + 2^{^{2}}}$
$= \sqrt{25 + 1 + 4} = \sqrt{30}$
∴ Position Vector in the direction of vector $= 8 \times \frac{\vec{a}}{|\vec{a}|} = \frac{8}{\sqrt{30}} (5 \hat{i} - \hat{j} + 2 \hat{k})$

Page No 22.80:

Question 38:

Write the direction cosines of the vector $\hat{i} + 2 \hat{j} + 3 \hat{k}$ .

Answer:

Given: $\hat{i} + 2 \hat{j} + 3 \hat{k}$
Then, direction cosines are
$\frac{1}{\sqrt{1^{2} + 2^{2} + 3^{2}}}, \frac{2}{\sqrt{1^{2} + 2^{2} + 3^{2}}}, \frac{3}{\sqrt{1^{2} + 2^{2} + 3^{2}}}$ or, $\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}$ .

Page No 22.80:

Question 39:

Find a unit vector in the direction of $\vec{a} = 2 \hat{i} - 3 \hat{j} + 6 \hat{k}$ .

Answer:

Given:
$\vec{a} = 2 \hat{i} - 3 \hat{j} + 6 \hat{k} |\vec{a}| = \sqrt{2^{2} + {(- 3)}^{2} + 6^{2}} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$
Unit vector = $\frac{\vec{a}}{|\vec{a}|} = \frac{2 \hat{i} - 3 \hat{j} + 6 \hat{k}}{7} = \frac{2}{7} \hat{i} - \frac{3}{7} \hat{j} + \frac{6}{7} \hat{k}$

Page No 22.80:

Question 40:

For what value of 'a' the vectors $2 \hat{i} - 3 \hat{j} + 4 \hat{k} and a \hat{i} + 6 \hat{j} - 8 \hat{k}$ are collinear?

Answer:

Given: Two vectors , let $\vec{p} =$ $2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ and $\vec{q} =$ $a \hat{i} + 6 \hat{j} - 8 \hat{k}$
Since the given vectors are collinear, we have,
$\vec{p} = λ \vec{q}$
$\Rightarrow 2 \hat{i} - 3 \hat{j} + 4 \hat{k} = λ (a \hat{i} + 6 \hat{j} - 8 \hat{k}) \Rightarrow 2 \hat{i} - 3 \hat{j} + 4 \hat{k} = a λ \hat{i} + 6 λ \hat{j} - 8 λ \hat{k}$

$\Rightarrow λ a = 2, 6 λ = - 3 and - 8 λ = 4 \Rightarrow λ = - \frac{1}{2} and a = - 4$

Page No 22.80:

Question 41:

Write the direction cosines of the vectors $- 2 \hat{i} + \hat{j} - 5 \hat{k}$ .

Answer:

Given: $- 2 \hat{i} + \hat{j} - 5 \hat{k}$
Then, its direction cosines are:
$\frac{- 2}{\sqrt{{(- 2)}^{2} + 1^{2} + {(- 5)}^{2}}}, \frac{1}{\sqrt{{(- 2)}^{2} + 1^{2} + {(- 5)}^{2}}}, \frac{- 5}{\sqrt{{(- 2)}^{2} + 1^{2} + {(- 5)}^{2}}}$ or, $\frac{- 2}{\sqrt{30}}, \frac{1}{\sqrt{30}}, \frac{- 5}{\sqrt{30}}$

Page No 22.80:

Question 42:

Find the sum of the following vectors: $\vec{a} = \hat{i} - 2 \hat{j}, \vec{b} = 2 \hat{i} - 3 \hat{j}, \vec{c} = 2 \hat{i} + 3 \hat{k} .$

Answer:

Given $\vec{a} = \hat{i} - 2 \hat{j}, \vec{b} = 2 \hat{i} - 3 \hat{j}, \vec{c} = 2 \hat{i} + 3 \hat{k}$

So, Sum of the three vectors = $\vec{a} + \vec{b} + \vec{c} = \hat{i} - 2 \hat{j} + 2 \hat{i} - 3 \hat{j} + 2 \hat{i} + 3 \hat{k}$
= $5 \hat{i} - 5 \hat{j} + 3 \hat{k}$

Page No 22.80:

Question 43:

Find a unit vector in the direction of the vector $\vec{a} = 3 \hat{i} - 2 \hat{j} + 6 \hat{k}$ .

Answer:

Given: $\vec{a} = 3 \hat{i} - 2 \hat{j} + 6 \hat{k}$
Then,
$\vec{|a|} = \sqrt{3^{2} + {(- 2)}^{2} + 6^{2}} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$

∴ Unit vector = $\frac{\vec{a}}{|\vec{a}|} = \frac{3 \hat{i} - 2 \hat{j} + 6 \hat{k}}{7}$ $= \frac{3}{7} \hat{i} - \frac{2}{7} \hat{j} + \frac{6}{7} \hat{k}$

Page No 22.80:

Question 44:

If $\vec{a} = x \hat{i} + 2 \hat{j} - z \hat{k} and \vec{b} = 3 \hat{i} - y \hat{j} + \hat{k}$ are two equal vectors, then write the value of x + y + z.

Answer:

Given: $\vec{a} = x \hat{i} + 2 \hat{j} - z \hat{k}$ and $\vec{b} = 3 \hat{i} - y \hat{j} + \hat{k}$
Since the two vectors are equal. We have,
$x \hat{i} + 2 \hat{j} - z \hat{k} = 3 \hat{i} - y \hat{j} + \hat{k} \Rightarrow x = 3, y = - 2, z = - 1$
∴ $x + y + z = 3 - 2 - 1 = 0$

Page No 22.80:

Question 45:

Write a unit vector in the direction of the sum of the vectors $\vec{a} = 2 \hat{i} + 2 \hat{j} - 5 \hat{k}$ and $\vec{b} = 2 \hat{i} + \hat{j} - 7 \hat{k}$ . [CBSE 2014]

Answer:

We have, $\vec{a} = 2 \hat{i} + 2 \hat{j} - 5 \hat{k}$ and $\vec{b} = 2 \hat{i} + \hat{j} - 7 \hat{k}$ .

$∴ \vec{a} + \vec{b} = (2 \hat{i} + 2 \hat{j} - 5 \hat{k}) + (2 \hat{i} + \hat{j} - 7 \hat{k}) = 4 \hat{i} + 3 \hat{j} - 12 \hat{k}$

$\Rightarrow |\vec{a} + \vec{b}| = \sqrt{4^{2} + 3^{2} + {(- 12)}^{2}} = \sqrt{16 + 9 + 144} = \sqrt{169} = 13$

∴ Required unit vector = $\frac{\vec{a} + \vec{b}}{|\vec{a} + \vec{b}|} = \frac{4 \hat{i} + 3 \hat{j} - 12 \hat{k}}{13} = \frac{4}{13} \hat{i} + \frac{3}{13} \hat{j} - \frac{12}{13} \hat{k}$

Page No 22.80:

Question 46:

Find the value of 'p' for which the vectors $3 \hat{i} + 2 \hat{j} + 9 \hat{k}$ and $\hat{i} - 2 p \hat{j} + 3 \hat{k}$ are parallel. [CBSE 2014]

Answer:

Let $\vec{a} = 3 \hat{i} + 2 \hat{j} + 9 \hat{k}$ and $\vec{b} = \hat{i} - 2 p \hat{j} + 3 \hat{k}$ be the two given vectors.

If $\vec{a}$ and $\vec{b}$ are parallel, then

$\vec{b} = λ \vec{a}$ for some scalar λ

$∴ \hat{i} - 2 p \hat{j} + 3 \hat{k} = λ (3 \hat{i} + 2 \hat{j} + 9 \hat{k}) \Rightarrow \hat{i} - 2 p \hat{j} + 3 \hat{k} = 3 λ \hat{i} + 2 λ \hat{j} + 9 λ \hat{k} \Rightarrow 1 = 3 λ and - 2 p = 2 λ (Equating coefficients of \hat{i}, \hat{j}, \hat{k}) \Rightarrow p = - λ = - \frac{1}{3}$

Thus, the value of p is $- \frac{1}{3}$ .

Page No 22.80:

Question 47:

Find a vector $\vec{a}$ of magnitude $5 \sqrt{2}$ , making an angle of $\frac{π}{4}$ with x-axis, $\frac{π}{2}$ with y-axis and an acute angle θ with z-axis. [CBSE 2014]

Answer:

It is given that vector $\vec{a}$ makes an angle of $\frac{π}{4}$ with x-axis, $\frac{π}{2}$ with y-axis and an acute angle θ with z-axis.
$∴ l = \cos \frac{π}{4} = \frac{1}{\sqrt{2}}, m = \cos \frac{π}{2} = 0, n = \cos θ$
Now,

$l^{2} + m^{2} + n^{2} = 1 \Rightarrow \frac{1}{2} + 0 + \cos^{2} θ = 1 \Rightarrow \cos^{2} θ = 1 - \frac{1}{2} = \frac{1}{2} \Rightarrow \cos θ = \frac{1}{\sqrt{2}} (θ is acute)$

We know that

$\vec{a} = |\vec{a}| (l \hat{i} + m \hat{j} + n \hat{k}) \Rightarrow \vec{a} = 5 \sqrt{2} (\frac{1}{\sqrt{2}} \hat{i} + 0 \hat{j} + \frac{1}{\sqrt{2}} \hat{k}) \Rightarrow \vec{a} = 5 (\hat{i} + 0 \hat{j} + \hat{k})$

Page No 22.81:

Question 48:

Write a unit vector in the direction of $\vec{PQ}$ , where P and Q are the points (1, 3, 0) and (4, 5, 6) respectively. [CBSE 2014]

Answer:

P(1, 3, 0) and Q(4, 5, 6) are the given points.

$∴ \vec{PQ} = (4 \hat{i} + 5 \hat{j} + 6 \hat{k}) - (\hat{i} + 3 \hat{j} + 0 \hat{k}) = 3 \hat{i} + 2 \hat{j} + 6 \hat{k} \Rightarrow |\vec{PQ}| = \sqrt{3^{2} + 2^{2} + 6^{2}} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$

∴ Unit vector in the direction of $\vec{PQ}$ = $\frac{\vec{PQ}}{|\vec{PQ}|} = \frac{3 \hat{i} + 2 \hat{j} + 6 \hat{k}}{7} = \frac{1}{7} (3 \hat{i} + 2 \hat{j} + 6 \hat{k})$

Page No 22.81:

Question 49:

Find a vector in the direction of vector $2 \hat{i} - 3 \hat{j} + 6 \hat{k}$ which has magnitude 21 units. [CBSE 2014]

Answer:

Let $\vec{a} = 2 \hat{i} - 3 \hat{j} + 6 \hat{k}$ .

$∴ |\vec{a}| = \sqrt{2^{2} + {(- 3)}^{2} + 6^{2}} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$

Unit vector in the direction of $\vec{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{2 \hat{i} - 3 \hat{j} + 6 \hat{k}}{7}$

∴ Vector in the direction of vector $\vec{a}$ which has magnitude 21 units
$= 21 \times (\frac{2 \hat{i} - 3 \hat{j} + 6 \hat{k}}{7}) = 3 (2 \hat{i} - 3 \hat{j} + 6 \hat{k}) = 6 \hat{i} - 9 \hat{j} + 18 \hat{k}$

Page No 22.81:

Question 50:

If $|\vec{a}| = 4$ and $- 3 \leq λ \leq 2$ , then write the range of $|λ \vec{a}|$ .

Answer:

It is given that

$- 3 \leq λ \leq 2 \Rightarrow - 3 \times |\vec{a}| \leq λ |\vec{a}| \leq 2 \times |\vec{a}| \Rightarrow - 3 \times 4 \leq |λ \vec{a}| \leq 2 \times 4 (k |\vec{a}| = |k \vec{a}|, k is scalar) \Rightarrow - 12 \leq |λ \vec{a}| \leq 8$

Thus, the range of $|λ \vec{a}|$ is [−12, 8].

Page No 22.81:

Question 51:

In a triangle OAC, if B is the mid-point of side AC and $\vec{OA} = \vec{a}, \vec{OB} = \vec{b}$ , then what is $\vec{OC}$ ? [CBSE 2015]

Answer:

In ∆OAC, $\vec{OA} = \vec{a}$ and $\vec{OB} = \vec{b}$ .

It is given that B is the mid-point of AC.

∴ Position vector of B = $\frac{Position vector of A + Position vector of C}{2}$

$\Rightarrow \vec{OB} = \frac{\vec{OA} + \vec{OC}}{2} \Rightarrow \vec{b} = \frac{\vec{a} + \vec{OC}}{2} \Rightarrow \vec{a} + \vec{OC} = 2 \vec{b} \Rightarrow \vec{OC} = 2 \vec{b} - \vec{a}$

Page No 22.81:

Question 52:

Write the position vector of the point which divides the join of points with position vectors $3 \vec{a} - 2 \vec{b} and 2 \vec{a} + 3 \vec{b}$ in the ratio 2 : 1.

Answer:

Suppose R be the point which divides the line joining the points with position vectors $3 \vec{a} - 2 \vec{b} and 2 \vec{a} + 3 \vec{b}$ in the ratio 2 : 1
And, $\vec{OA} = 3 \vec{a} - 2 \vec{b} and \vec{OB} = 2 \vec{a} + 3 \vec{b}$
Here, m : n = 2 : 1
Therefore, position vector $\vec{OR}$ is as follows:
$\vec{OR} = \frac{m \vec{OB} + n \vec{OA}}{m + n} = \frac{2 (2 \vec{a} + 3 \vec{b}) + 1 (3 \vec{a} - 2 \vec{b})}{2 + 1} = \frac{7 \vec{a} + 4 \vec{b}}{3} = \frac{7}{3} \vec{a} + \frac{4}{3} \vec{b}$

View NCERT Solutions for all chapters of Class 12