Rd Sharma XII Vol 2 2020 for Class 12 Science Maths Chapter 5 - Scalar Or Dot Product

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Rd Sharma XII Vol 2 2020 Solutions for Class 12 Science Maths Chapter 5 Scalar Or Dot Product are provided here with simple step-by-step explanations. These solutions for Scalar Or Dot Product are extremely popular among Class 12 Science students for Maths Scalar Or Dot Product 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 5 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 12.50:

Question 12:

If $\vec{a}, \vec{b}, \vec{c}$ are three mutually perpendicular vectors then $|\vec{a} + \vec{b} + \vec{c}|$ = _______________.

Answer:

Given:
$\vec{a}, \vec{b}, \vec{c}$ are three mutually perpendicular vectors
then, $\vec{a} \cdot \vec{b} = 0, \vec{a} \cdot \vec{c} = 0 and \vec{c} \cdot \vec{b} = 0$

${|\vec{a} + \vec{b} + \vec{c}|}^{2} = \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} + \vec{c} \cdot \vec{c} + 2 \vec{a} \cdot \vec{b} + 2 \vec{b} \cdot \vec{c} + 2 \vec{c} \cdot \vec{a} = {|\vec{a}|}^{2} + {|\vec{b}|}^{2} + {|\vec{c}|}^{2} + 2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = {|\vec{a}|}^{2} + {|\vec{b}|}^{2} + {|\vec{c}|}^{2} + 2 (0 + 0 + 0) (∵ \vec{a} \cdot \vec{b} = 0, \vec{a} \cdot \vec{c} = 0 and \vec{c} \cdot \vec{b} = 0) = {|\vec{a}|}^{2} + {|\vec{b}|}^{2} + {|\vec{c}|}^{2} \Rightarrow |\vec{a} + \vec{b} + \vec{c}| = \sqrt{{|\vec{a}|}^{2} + {|\vec{b}|}^{2} + {|\vec{c}|}^{2}}$

Hence, $|\vec{a} + \vec{b} + \vec{c}| = \sqrt{{|\vec{a}|}^{2} + {|\vec{b}|}^{2} + {|\vec{c}|}^{2}}$ .

Page No 12.50:

Question 13:

If $\vec{a}, \vec{b}$ are non-zero vectors such that $\vec{a} . \vec{b} = - |\vec{a}| |\vec{b}|$ ,then the angle between $\vec{a}$ and $\vec{b}$ is _________________.

Answer:

Given:
$\vec{a} . \vec{b} = - |\vec{a}| |\vec{b}|$

$\vec{a} . \vec{b} = - |\vec{a}| |\vec{b}| \Rightarrow |\vec{a}| |\vec{b}| \cos θ = - |\vec{a}| |\vec{b}| \Rightarrow \cos θ = - 1 \Rightarrow θ = π$

Hence, the angle between $\vec{a} and \vec{b}$ is $π$ .

Page No 12.50:

Question 14:

For any vector $\vec{r}, {(\vec{r} . \hat{i})}^{2} + {(\vec{r} . \hat{j})}^{2} + {(\vec{r} . \hat{k})}^{2}$ = ________________ .

Answer:

$Let \vec{r} = r_{1} \hat{i} + r_{2} \hat{j} + r_{3} \hat{k} {(\vec{r} \cdot \hat{i})}^{2} + {(\vec{r} \cdot \hat{j})}^{2} + {(\vec{r} \cdot \hat{k})}^{2} = {(r_{1})}^{2} + {(r_{2})}^{2} + {(r_{3})}^{2} = {(\sqrt{{r_{1}}^{2} + {r_{2}}^{2} + {r_{3}}^{2}})}^{2} = {|\vec{r}|}^{2}$

Hence, ${(\vec{r} \cdot \hat{i})}^{2} + {(\vec{r} \cdot \hat{j})}^{2} + {(\vec{r} \cdot \hat{k})}^{2} = {|\vec{r}|}^{2}$ .

Page No 12.50:

Question 15:

If $\vec{a}, \vec{b}$ are non-zero vectors of same magnitude such that the angle between $\vec{a} and \vec{b} is \frac{2 π}{3} and \vec{a} . \vec{b} = - 8, then |\vec{a}| =______________.$

Answer:

Given:
$|\vec{a}| = |\vec{b}|$ ...(1)
The angle between the vectors $\vec{a} and \vec{b} is \frac{2 π}{3}$ ...(2)

$\vec{a} \cdot \vec{b} = - 8$ ....(3)

$\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos θ \Rightarrow - 8 = {|\vec{a}|}^{2} \cos \frac{2 π}{3} (From (1), (2) and (3)) \Rightarrow - 8 = {|\vec{a}|}^{2} (- \frac{1}{2}) \Rightarrow {|\vec{a}|}^{2} = 16 \Rightarrow |\vec{a}| = 4$

Hence, $|\vec{a}| = 4$ .

Page No 12.50:

Question 16:

For any non-zero vectors $\vec{r},$ the expression $(\vec{r} . \hat{i}) \hat{i} + (\vec{r} . \hat{j}) \hat{j} + (\vec{r} . \hat{k}) \hat{k}$ equals ___________.

Answer:

$Let \vec{r} = r_{1} \hat{i} + r_{2} \hat{j} + r_{3} \hat{k} (\vec{r} \cdot \hat{i}) \hat{i} + (\vec{r} \cdot \hat{j}) \hat{j} + (\vec{r} \cdot \hat{k}) \hat{k} = (r_{1}) \hat{i} + (r_{2}) \hat{j} + (r_{3}) \hat{k} = r_{1} \hat{i} + r_{2} \hat{j} + r_{3} \hat{k} = \vec{r}$

Hence, the expression $(\vec{r} . \hat{i}) \hat{i} + (\vec{r} . \hat{j}) \hat{j} + (\vec{r} . \hat{k}) \hat{k}$ equals $\vec{r}$ .

Page No 12.50:

Question 17:

The vectors $\vec{a} = 3 \hat{i} - 2 \hat{j} + 2 \hat{k} and \vec{b} = - \hat{i} - 2 \hat{k}$ are the adjacent sides of a parallelogram. The acture angle between its diagonals is _____________.

Answer:

Given:
vectors $\vec{a} = 3 \hat{i} - 2 \hat{j} + 2 \hat{k} and \vec{b} = - \hat{i} - 2 \hat{k}$ are the adjacent sides of a parallelogram

$Let θ be the acute angle between the diagonals . The diagonals of the parallelogram are \vec{a} + \vec{b} = (3 \hat{i} - 2 \hat{j} + 2 \hat{k}) + (- \hat{i} - 2 \hat{k}) = 2 \hat{i} - 2 \hat{j} . . . (1) and \vec{a} - \vec{b} = (3 \hat{i} - 2 \hat{j} + 2 \hat{k}) - (- \hat{i} - 2 \hat{k}) = 4 \hat{i} - 2 \hat{j} + 4 \hat{k} . . . (2) The angle between the diagonals is : \cos θ = \frac{(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b})}{|\vec{a} + \vec{b}| |\vec{a} - \vec{b}|} = \frac{(2) (4) + (- 2) (- 2) + (0) (4)}{\sqrt{{(2)}^{2} + {(- 2)}^{2}} \sqrt{{(4)}^{2} + {(- 2)}^{2} + {(4)}^{2}}} = \frac{8 + 4}{\sqrt{4 + 4} \sqrt{16 + 4 + 16}} (From (1) and (2)) = \frac{12}{\sqrt{8} \sqrt{36}} = \frac{12}{6 \times 2 \sqrt{2}} = \frac{1}{\sqrt{2}} \Rightarrow θ = \frac{π}{4}$

Hence, the acute angle between its diagonals is $\frac{π}{4}$ .

Page No 23.29:

Question 1:

Find $\vec{a} \cdot \vec{b}$ when
(i) $\vec{a} = \hat{i} - 2 \hat{j} + \hat{k} and \vec{b} = 4 \hat{i} - 4 \hat{j} + 7 \hat{k}$

(ii) $\vec{a} = \hat{j} + 2 \hat{k} and \vec{b} = 2 \hat{i} + \hat{k}$

(iii) $\vec{a} = \hat{j} - \hat{k} and \vec{b} = 2 \hat{i} + 3 \hat{j} - 2 \hat{k}$

Answer:

$(i) We have \vec{a} = \overset{\land}{i} - 2 \overset{\land}{j} + \overset{\land}{k} and \vec{b} = 4 \overset{\land}{i} - \overset{\land}{j} +7 \overset{\land}{k} \vec{a} . \vec{b} = (\overset{\land}{i} - 2 \overset{\land}{j} + \overset{\land}{k}) . (4 \overset{\land}{i} - \overset{\land}{j} +7 \overset{\land}{k}) = (1) (4) + (- 2) (- 4) + (1) (7) = 4 + 8 + 7 = 19$

$(ii) We have \vec{a} = \hat{j} + 2 \hat{k} = 0 \overset{\land}{i} + \overset{\land}{j} + \overset{\land}{2 k} and \vec{b} =2 \overset{\land}{i} + \overset{\land}{k} = 2 \overset{\land}{i} +0 \overset{\land}{j} + \overset{\land}{k} \vec{a} . \vec{b} = (0 \overset{\land}{i} + \overset{\land}{j} + \overset{\land}{2 k}) . (2 \overset{\land}{i} +0 \overset{\land}{j} + \overset{\land}{k}) = (0) (2) + (1) (0) + (2) (1) = 0 + 0 + 2 = 2$

$(iii) We have \vec{a} = \overset{\land}{j} - \overset{\land}{k} = 0 \overset{\land}{i} + \overset{\land}{j} - \overset{\land}{k} and \vec{b} = 2 \overset{\land}{i} + \overset{\land}{3 j} -2 \overset{\land}{k} \vec{a} . \vec{b} = (0 \overset{\land}{i} + \overset{\land}{j} - \overset{\land}{k}) . (2 \overset{\land}{i} + \overset{\land}{3 j} -2 \overset{\land}{k}) = (0) (2) + (1) (3) + (- 1) (- 2) = 3 + 2 = 5$

Page No 23.30:

Question 2:

For what value of λ are the vectors $\vec{a} and \vec{b}$ perpendicular to each other if
(i) $\vec{a} = λ \hat{i} + 2 \hat{j} + \hat{k} and \vec{b} = 4 \hat{i} - 9 \hat{j} + 2 \hat{k}$

(ii) $\vec{a} = λ \hat{i} + 2 \hat{j} + \hat{k} and \vec{b} = 5 \hat{i} - 9 \hat{j} + 2 \hat{k}$

(iii) $\vec{a} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k} and \vec{b} = 3 \hat{i} + 2 \hat{j} - λ \hat{k}$

(iv) $\vec{a} = λ \hat{i} + 3 \hat{j} + 2 \hat{k} and \vec{b} = \hat{i} - \hat{j} + 3 \hat{k}$

Answer:

$(i) If the vectors \vec{a} and \vec{b} are perpendicular to each other, then \vec{a} . \vec{b} = 0 \Rightarrow (λ \overset{\land}{i} + 2 \overset{\land}{j} + \overset{\land}{k}) . (4 \overset{\land}{i} - 9 \overset{\land}{j} + 2 \overset{\land}{k}) = 0 \Rightarrow 4 λ - 18 + 2 = 0 \Rightarrow 4 λ - 16 = 0 \Rightarrow 4 λ = 16 \Rightarrow λ = 4$

$(ii) If the vectors \vec{a} and \vec{b} are perpendicular to each other, then \vec{a} . \vec{b} = 0 \Rightarrow (λ \overset{\land}{i} + 2 \overset{\land}{j} + \overset{\land}{k}) . (5 \overset{\land}{i} - 9 \overset{\land}{j} + 2 \overset{\land}{k}) = 0 \Rightarrow 5 λ - 18 + 2 = 0 \Rightarrow 5 λ - 16 = 0 \Rightarrow 5 λ = 16 \Rightarrow λ = \frac{16}{5}$

$(iii) If the vectors \vec{a} and \vec{b} are perpendicular to each other, then \vec{a} . \vec{b} = 0 \Rightarrow (2 \overset{\land}{i} + 3 \overset{\land}{j} + 4 \overset{\land}{k}) . (3 \overset{\land}{i} + 2 \overset{\land}{j} - λ \overset{\land}{k}) = 0 \Rightarrow 6 + 6 - 4 λ = 0 \Rightarrow 12 - 4 λ = 0 \Rightarrow 4 λ = 12 \Rightarrow λ = 3$

$(iv) If the vectors \vec{a} and \vec{b} are perpendicular to each other, then \vec{a} . \vec{b} = 0 \Rightarrow (λ \overset{\land}{i} + 3 \overset{\land}{j} + 2 \overset{\land}{k}) . (\overset{\land}{i} - 1 \overset{\land}{j} + 3 \overset{\land}{k}) = 0 \Rightarrow λ - 3 + 6 = 0 \Rightarrow λ + 3 = 0 \Rightarrow λ = - 3$

Page No 23.30:

Question 3:

If $\vec{a} and \vec{b}$ are two vectors such that $|\vec{a}| = 4, |\vec{b}| = 3 and \vec{a} \cdot \vec{b} = 6$ , find the angle between $\vec{a} and \vec{b} .$

Answer:

$Let θ be the angle between \vec{a} and \vec{b} . Given that \vec{a} . \vec{b} = 6 \Rightarrow |\vec{a}| |\vec{b}| \cos θ = 6 \Rightarrow (4) (3) \cos θ = 6 \Rightarrow 12 \cos θ = 6 \Rightarrow \cos θ = \frac{6}{12} = \frac{1}{2} \Rightarrow θ = \cos^{- 1} (\frac{1}{2}) = \frac{π}{3}$

Page No 23.30:

Question 4:

$If \vec{a} = \hat{i} - \hat{j} and \vec{b} = - \hat{j} + 2 \hat{k}, find (\vec{a} - 2 \vec{b}) \cdot (\vec{a} + \vec{b}) .$

Answer:

$We have \vec{a} = \overset{\land}{i} - \overset{\land}{j} and \vec{b} = - \overset{\land}{j} + 2 \overset{\land}{k} \vec{a} - 2 \vec{b} = (\overset{\land}{i} - \overset{\land}{j}) - 2 (- \overset{\land}{j} + 2 \overset{\land}{k}) = \overset{\land}{i} - \overset{\land}{j} + 2 \overset{\land}{j} - 4 \overset{\land}{k} = \overset{\land}{i} + \overset{\land}{j} - 4 \overset{\land}{k} \vec{a} + \vec{b} = \overset{\land}{i} - \overset{\land}{j} - \overset{\land}{j} + 2 \overset{\land}{k} = \overset{\land}{i} - 2 \overset{\land}{j} + 2 \overset{\land}{k} (\vec{a} - 2 \vec{b}) . (\vec{a} + \vec{b}) = (\overset{\land}{i} + \overset{\land}{j} - 4 \overset{\land}{k}) . (\overset{\land}{i} - 2 \overset{\land}{j} + 2 \overset{\land}{k}) = 1 - 2 - 8 = - 9$

Page No 23.30:

Question 5:

Find the angle between the vectors $\vec{a} and \vec{b}$ , where
(i) $\vec{a} = \hat{i} - \hat{j} and \vec{b} = \hat{j} + \hat{k}$

(ii) $\vec{a} = 3 \hat{i} - 2 \hat{j} - 6 \hat{k} and \vec{b} = 4 \hat{i} - \hat{j} + 8 \hat{k}$

(iii) $\vec{a} = 2 \hat{i} - \hat{j} + 2 \hat{k} and \vec{b} = 4 \hat{i} + 4 \hat{j} - 2 \hat{k}$

(iv) $\vec{a} = 2 \hat{i} - 3 \hat{j} + \hat{k} and \vec{b} = \hat{i} + \hat{j} - 2 \hat{k}$

(v) $\vec{a} = \hat{i} + 2 \hat{j} - \hat{k}, \vec{b} = \hat{i} - \hat{j} + \hat{k}$

Answer:

$(i) Let θ be the angle between \vec{a} and \vec{b} . |\vec{a}| = \sqrt{{(1)}^{2} + {(- 1)}^{2}} = \sqrt{2} |\vec{b}| = \sqrt{{(1)}^{2} + {(1)}^{2}} = \sqrt{2} \vec{a} . \vec{b} = 0 - 1 + 0 = - 1 \cos θ = \frac{\vec{a} . \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{- 1}{\sqrt{2} \sqrt{2}} = \frac{- 1}{2} \Rightarrow θ = \cos^{- 1} (\frac{- 1}{2}) = \frac{2 π}{3}$

$(i i) Let θ be the angle between \vec{a} and \vec{b} . |\vec{a}| = \sqrt{{(3)}^{2} + {(- 2)}^{2} + {(- 6)}^{2}} = \sqrt{49} = 7 |\vec{b}| = \sqrt{{(4)}^{2} + {(- 1)}^{2} + {(8)}^{2}} = \sqrt{81} = 9 \vec{a} . \vec{b} = 12 + 2 - 48 = -34 \cos θ = \frac{\vec{a} . \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{-34}{(7) (9)} = \frac{-34}{63} \Rightarrow θ = \cos^{- 1} (\frac{-34}{63})$

$(i i i) Let θ be the angle between \vec{a} and \vec{b} . |\vec{a}| = \sqrt{{(2)}^{2} + {(- 1)}^{2} + {(2)}^{2}} = \sqrt{9} = 3 |\vec{b}| = \sqrt{{(4)}^{2} + {(4)}^{2} + {(- 2)}^{2}} = \sqrt{36} = 6 \vec{a} . \vec{b} = 8 - 4 - 4 = 0 \cos θ = \frac{\vec{a} . \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{0}{(3) (6)} = 0 \Rightarrow θ = \cos^{- 1} (0) = \frac{π}{2}$

$(i v) Let θ be the angle between \vec{a} and \vec{b} . |\vec{a}| = \sqrt{{(2)}^{2} + {(- 3)}^{2} + {(1)}^{2}} = \sqrt{14} |\vec{b}| = \sqrt{{(1)}^{2} + {(1)}^{2} + {(- 2)}^{2}} = \sqrt{6} \vec{a} . \vec{b} = 2 - 3 - 2 = - 3 \cos θ = \frac{\vec{a} . \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{- 3}{\sqrt{14} \sqrt{6}} = \frac{- 3}{\sqrt{84}} \Rightarrow θ = \cos^{- 1} (\frac{- 3}{\sqrt{84}})$

$(v) Let θ be the angle between \vec{a} and \vec{b} . |\vec{a}| = \sqrt{{(1)}^{2} + {(2)}^{2} + {(- 1)}^{2}} = \sqrt{6} |\vec{b}| = \sqrt{{(1)}^{2} + {(- 1)}^{2} + {(1)}^{2}} = \sqrt{3} \vec{a} . \vec{b} = 1 - 2 - 1 = - 2 \cos θ = \frac{\vec{a} . \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{- 2}{\sqrt{6} \sqrt{3}} = \frac{- 2}{\sqrt{18}} = \frac{- \sqrt{2} \times \sqrt{2}}{\sqrt{2} \times \sqrt{9}} = \frac{- \sqrt{2}}{3} \Rightarrow θ = \cos^{- 1} (\frac{- \sqrt{2}}{3})$

Page No 23.30:

Question 6:

Find the angles which the vector $\vec{a} = \hat{i} - \hat{j} + \sqrt{2} \hat{k}$ makes with the coordinate axes.

Answer:

$Let θ_{1} be the angle between \vec{a} and x - axis. |\vec{a}| = \sqrt{{(1)}^{2} + {(- 1)}^{2} + {(\sqrt{2})}^{2}} = \sqrt{4} = 2 \vec{b} = \overset{\land}{i} (Because \overset{\land}{i} is the unit vector along x -axis) |\vec{b}| = \sqrt{{(1)}^{2}} = \sqrt{1} = 1 \vec{a} . \vec{b} = 1 + 0 + 0 = 1 \cos θ_{1} = \frac{\vec{a} . \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{1}{(2) (1)} = \frac{1}{2} \Rightarrow θ_{1} = \cos^{- 1} (\frac{1}{2}) = \frac{π}{3} Let θ_{2} be the angle between \vec{a} and y - axis. |\vec{a}| = \sqrt{{(1)}^{2} + {(- 1)}^{2} + {(\sqrt{2})}^{2}} = \sqrt{4} = 2 \vec{b} = \overset{\land}{j} (Because \overset{\land}{j} is the unit vector along y -axis) |\vec{b}| = \sqrt{{(1)}^{2}} = \sqrt{1} = 1 \vec{a} . \vec{b} = 0 - 1 + 0 = - 1 \cos θ_{2} = \frac{\vec{a} . \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{- 1}{(2) (1)} = \frac{- 1}{2} \Rightarrow θ_{2} = \cos^{- 1} (\frac{- 1}{2}) = \frac{2 π}{3} Let θ_{3} be the angle between \vec{a} and z - axis. |\vec{a}| = \sqrt{{(1)}^{2} + {(- 1)}^{2} + {(\sqrt{2})}^{2}} = \sqrt{4} = 2 \vec{b} = \overset{\land}{k} (Because \overset{\land}{k} is the unit vector along z -axis) |\vec{b}| = \sqrt{{(1)}^{2}} = \sqrt{1} = 1 \vec{a} . \vec{b} = 0 + 0 + \sqrt{2} = \sqrt{2} \cos θ = \frac{\vec{a} . \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{\sqrt{2}}{(2) (1)} = \frac{1}{\sqrt{2}} \Rightarrow θ = \cos^{- 1} (\frac{1}{\sqrt{2}}) = \frac{π}{4}$

Page No 23.30:

Question 7:

(i) Dot product of a vector with $\hat{i} + \hat{j} - 3 \hat{k}, \hat{i} + 3 \hat{j} - 2 \hat{k} and 2 \hat{i} + \hat{j} + 4 \hat{k}$ are 0, 5 and 8 respectively. Find the vector.
(ii) Dot products of a vector with vectors $\hat{i} - \hat{j} + \hat{k}, 2 \hat{i} + \hat{j} - 3 \hat{k} and \hat{i} + \hat{j} + \hat{k}$ are respectively 4, 0 and 2. Find the vector.

Answer:

$(i) Let a \overset{\land}{i} + b \overset{\land}{j} + c \overset{\land}{k} be the required vector. Given that (a \overset{\land}{i} + b \overset{\land}{j} + c \overset{\land}{k}) . (\overset{\land}{i} + \overset{\land}{j} - 3 \overset{\land}{k}) = 0 \Rightarrow a + b - 3 c = 0 . . . (1) (a \overset{\land}{i} + b \overset{\land}{j} + c \overset{\land}{k}) . (\overset{\land}{i} + 3 \overset{\land}{j} - 2 \overset{\land}{k}) = 5 \Rightarrow a + 3 b - 2 c = 5 . . . (2) (a \overset{\land}{i} + b \overset{\land}{j} + c \overset{\land}{k}) . (2 \overset{\land}{i} + \overset{\land}{j} + 4 \overset{\land}{k}) = 5 \Rightarrow 2 a + b + 4 c = 8 . . . (3) Solving (1), (2) and (3), we get a = 1, b = 2, c = 1 So, a \overset{\land}{i} + b \overset{\land}{j} + c \overset{\land}{k} = \overset{\land}{i} + 2 \overset{\land}{j} + \overset{\land}{k}$

$(i i) Let a \overset{\land}{i} + b \overset{\land}{j} + c \overset{\land}{k} be the required vector. Given that (a \overset{\land}{i} + b \overset{\land}{j} + c \overset{\land}{k}) . (\overset{\land}{i} - \overset{\land}{j} + \overset{\land}{k}) = 4 \Rightarrow a - b + c = 4 . . . (1) (a \overset{\land}{i} + b \overset{\land}{j} + c \overset{\land}{k}) . (2 \overset{\land}{i} + \overset{\land}{j} - 3 \overset{\land}{k}) = 0 \Rightarrow 2 a + b - 3 c = 0 . . . (2) (a \overset{\land}{i} + b \overset{\land}{j} + c \overset{\land}{k}) . (\overset{\land}{i} + \overset{\land}{j} + \overset{\land}{k}) = 2 \Rightarrow a + b + c = 2 . . . (3) Solving (1), (2) and (3), we get a = 2; b = - 1; c = 1 So, a \overset{\land}{i} + b \overset{\land}{j} + c \overset{\land}{k} = 2 \overset{\land}{i} - \overset{\land}{j} + \overset{\land}{k}$

Page No 23.30:

Question 8:

If $\hat{a} and \hat{b}$ are unit vectors inclined at an angle θ, prove that
(i) $\cos \frac{θ}{2} = \frac{1}{2} |\hat{a} + \hat{b}|$

(ii) $\tan \frac{θ}{2} = \frac{|\hat{a} - \hat{b}|}{|\hat{a} + \hat{b}|}$

Answer:

$Given that \overset{⏞}{a} and \overset{⏞}{b} are unit vectors. So, |\hat{a}| =1, |\hat{b}| =1 We have {|\hat{a} + \hat{b}|}^{2} = {|\hat{a}|}^{2} + {|\hat{b}|}^{2} + 2 \hat{a} . \hat{b} = 1 + 1 + 2 |\hat{a}| |\hat{b}| \cos θ = 2 + 2 \cos θ \Rightarrow \cos θ = \frac{{|\hat{a} + \hat{b}|}^{2} - 2}{2} . . . (1) {|\hat{a} - b|}^{2} = {|\hat{a}|}^{2} + {|\hat{b}|}^{2} - 2 \hat{a} . \hat{b} = 1 + 1 - 2 |\hat{a}| |\hat{b}| \cos θ = 2 - 2 \cos θ \Rightarrow \cos θ = \frac{2 - {|\hat{a} - \hat{b}|}^{2}}{2} . . . (2) (i) Now, \cos \frac{θ}{2} = \sqrt{\frac{1 + \cos θ}{2}} = \sqrt{\frac{1 + \frac{{|\hat{a} + \hat{b}|}^{2} - 2}{2}}{2}} [From (1)] = \sqrt{\frac{2 + {|\hat{a} + \hat{b}|}^{2} - 2}{4}} = \sqrt{\frac{{|\hat{a} + \hat{b}|}^{2}}{4}} = \frac{1}{2} |\hat{a} + \hat{b}| (ii) \sin \frac{θ}{2} = \sqrt{\frac{1 - \cos θ}{2}} = \sqrt{\frac{1 - \frac{2 - {|\hat{a} - \hat{b}|}^{2}}{2}}{2}} [From (2)] = \sqrt{\frac{2 + {|\hat{a} - \hat{b}|}^{2} - 2}{4}} = \sqrt{\frac{{|\hat{a} - \hat{b}|}^{2}}{4}} = \frac{1}{2} |\hat{a} - \hat{b}| Now, \tan \frac{θ}{2} = \frac{\sin \frac{θ}{2}}{\cos \frac{θ}{2}} = \frac{\frac{1}{2} |\hat{a} - \hat{b}|}{\frac{1}{2} |\hat{a} + \hat{b}|} = \frac{|\hat{a} - \hat{b}|}{|\hat{a} + \hat{b}|}$

Page No 23.30:

Question 9:

If the sum of two unit vectors is a unit vector prove that the magnitude of their difference is $\sqrt{3}$ .

Answer:

$Given that \overset{⏞}{a}, \overset{⏞}{b} and |\hat{a} + \hat{b}| are unit vectors. So, |\hat{a}| =1, |\hat{b}| =1 and ||\hat{a} + \hat{b}|| =1 We have {|\hat{a} + \hat{b}|}^{2} + {|\hat{a} - \hat{b}|}^{2} = 2 ({|\hat{a}|}^{2} + {|\hat{b}|}^{2}) \Rightarrow 1 + {|\hat{a} - \hat{b}|}^{2} = 2 (1 + 1) \Rightarrow 1 + {|\hat{a} - \hat{b}|}^{2} = 4 \Rightarrow {|\hat{a} - \hat{b}|}^{2} = 3 \Rightarrow |\hat{a} - \hat{b}| = \sqrt{3}$

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Question 10:

If $\vec{a,} \vec{b,} \vec{c}$ are three mutually perpendicular unit vectors, then prove that $|\vec{a} + \vec{b} + \vec{c}| = \sqrt{3}$ .

Answer:

$Given that \vec{a}, \vec{b} and \vec{c} are unit vectors. So, |\vec{a}| =1, |\vec{b}| =1 and |\vec{c}| =1 Since they are mutually perpendicular, \vec{a} . \vec{b} = \vec{b} . \vec{c} = \vec{c} . \vec{a} = 0 Now, {|\vec{a} + \vec{b} + \vec{c}|}^{2} = {|\vec{a}|}^{2} + {|\vec{b}|}^{2} + {|\vec{c}|}^{2} + 2 \vec{a} . \vec{b} + 2 \vec{b} . \vec{c} + 2 \vec{c} . \vec{a} = 1 + 1 + 1 + 0 + 0 + 0 = 3 ∴ |\vec{a} + \vec{b} + \vec{c}| = \sqrt{3}$

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Question 11:

If $|\vec{a} + \vec{b}| = 60, |\vec{a} - \vec{b}| = 40 and |\vec{b}| = 46, find |\vec{a}|$

Answer:

$We know that {|\vec{a} + \vec{b}|}^{2} + {|\vec{a} - \vec{b}|}^{2} = 2 ({|\vec{a}|}^{2} + {|\vec{b}|}^{2}) \Rightarrow 60^{2} + 40^{2} = 2 ({|\vec{a}|}^{2} + 46^{2}) (Given) \Rightarrow 3600 + 1600 = 2 {|\vec{a}|}^{2} + 4232 \Rightarrow 968 = 2 {|\vec{a}|}^{2} \Rightarrow {|\vec{a}|}^{2} = 484 \Rightarrow |\vec{a}| = 22$

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Question 12:

Show that the vector $\hat{i} + \hat{j} + \hat{k}$ is equally inclined to the coordinate axes.

Answer:

$Let θ_{1} be the angle between \vec{a} and x - axis. |\vec{a}| = \sqrt{{(1)}^{2} + {(1)}^{2} + {(1)}^{2}} = \sqrt{3} \vec{b} = \overset{\land}{i} (Because \overset{\land}{i} is the unit vector along x -axis) |\vec{b}| = \sqrt{{(1)}^{2}} = \sqrt{1} = 1 \vec{a} . \vec{b} = 1 + 0 + 0 = 1 \cos θ_{1} = \frac{\vec{a} . \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{1}{(\sqrt{3}) (1)} = \frac{1}{\sqrt{3}} \Rightarrow θ_{1} = \cos^{- 1} (\frac{1}{\sqrt{3}}) . . . (1) Let θ_{2} be the angle between \vec{a} and y - axis. |\vec{a}| = \sqrt{{(1)}^{2} + {(1)}^{2} + {(1)}^{2}} = \sqrt{3} \vec{b} = \overset{\land}{j} (Because \overset{\land}{j} is the unit vector along y -axis) |\vec{b}| = \sqrt{{(1)}^{2}} = \sqrt{1} = 1 \vec{a} . \vec{b} = 0 + 1 + 0 = 1 \cos θ_{2} = \frac{\vec{a} . \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{1}{(\sqrt{3}) (1)} = \frac{1}{\sqrt{3}} \Rightarrow θ_{2} = \cos^{- 1} (\frac{1}{\sqrt{3}}) . . . (2) Let θ_{3} be the angle between \vec{a} and z - axis. |\vec{a}| = \sqrt{{(1)}^{2} + {(1)}^{2} + {(1)}^{2}} = \sqrt{3} \vec{b} = \overset{\land}{k} (Because \overset{\land}{k} is the unit vector along z -axis) |\vec{b}| = \sqrt{{(1)}^{2}} = \sqrt{1} = 1 \vec{a} . \vec{b} = 0 + 0 + 1 = 1 \cos θ = \frac{\vec{a} . \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{1}{(\sqrt{3}) (1)} = \frac{1}{\sqrt{3}} \Rightarrow θ = \cos^{- 1} (\frac{1}{\sqrt{3}}) . . . (3) From (1), (2) and (3), the given vector is equally inclined to the coordinate axes.$

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Question 13:

Show that the vectors $\vec{a} = \frac{1}{7} (2 \hat{i} + 3 \hat{j} + 6 \hat{k}), \vec{b} = \frac{1}{7} (3 \hat{i} - 6 \hat{j} + 2 \hat{k}), \vec{c} = \frac{1}{7} (6 \hat{i} + 2 \hat{j} - 3 \hat{k})$ are mutually perpendicular unit vectors.

Answer:

$We have |\vec{a}| = \frac{1}{7} \sqrt{2^{2} + 3^{2} + 6^{2}} = \frac{1}{7} \sqrt{49} = \frac{7}{7} = 1 |\vec{b}| = \frac{1}{7} \sqrt{3^{2} + {(- 6)}^{2} + 2^{2}} = \frac{1}{7} \sqrt{49} = \frac{7}{7} = 1 |\vec{c}| = \frac{1}{7} \sqrt{6^{2} + 2^{2} + {(- 3)}^{2}} = \frac{1}{7} \sqrt{49} = \frac{7}{7} = 1 And \vec{a} . \vec{b} = \frac{1}{7} (2 \hat{i} + 3 \hat{j} + 6 \hat{k}) . \frac{1}{7} (3 \hat{i} - 6 \hat{j} + 2 \hat{k}) = \frac{1}{49} (6 - 18 + 12) = 0 \vec{b} . \vec{c} = \frac{1}{7} (3 \hat{i} - 6 \hat{j} + 2 \hat{k}) . \frac{1}{7} (6 \hat{i} + 2 \hat{j} - 3 \hat{k}) = \frac{1}{49} (18 - 12 - 6) = 0 \vec{c} . \vec{a} = \frac{1}{7} (6 \hat{i} + 2 \hat{j} - 3 \hat{k}) . \frac{1}{7} (2 \hat{i} + 3 \hat{j} + 6 \hat{k}) = \frac{1}{49} (12 + 6 - 18) = 0 So, |\vec{a}| = |\vec{b}| = |\vec{c}| = 1 and \vec{a} . \vec{b} = \vec{b} . \vec{c} = \vec{c} . \vec{a} = 0 So, the given vectors are mutually perpendicular unit vectors.$

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Question 14:

For any two vectors $\vec{a} and \vec{b}$ , show that $(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 0 \Leftrightarrow |\vec{a}| = |\vec{b}|$ .

Answer:

$We have (\vec{a} + \vec{b}) . (\vec{a} - \vec{b}) = 0 \Rightarrow {|\vec{a}|}^{2} - {|\vec{b}|}^{2} = 0 \Rightarrow {|\vec{a}|}^{2} = {|\vec{b}|}^{2} \Rightarrow |\vec{a}| = |\vec{b}|$

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Question 15:

If $\vec{a} = 2 \hat{i} - \hat{j} + \hat{k}$ , $\vec{b} = \hat{i} + \hat{j} - 2 \hat{k}$ and $\vec{c} = \hat{i} + 3 \hat{j} - \hat{k}$ , find λ such that $\vec{a}$ is perpendicular to $λ \vec{b} + \vec{c}$ . [NCERT EXEMPLAR]

Answer:

The given vectors are $\vec{a} = 2 \hat{i} - \hat{j} + \hat{k}$ , $\vec{b} = \hat{i} + \hat{j} - 2 \hat{k}$ and $\vec{c} = \hat{i} + 3 \hat{j} - \hat{k}$ .

Now,

$λ \vec{b} + \vec{c} = λ (\hat{i} + \hat{j} - 2 \hat{k}) + (\hat{i} + 3 \hat{j} - \hat{k}) = (λ + 1) \hat{i} + (λ + 3) \hat{j} - (2 λ + 1) \hat{k}$

It is given that

$\vec{a} ⊥ (λ \vec{b} + \vec{c}) \Rightarrow \vec{a} . (λ \vec{b} + \vec{c}) = 0 \Rightarrow (2 \hat{i} - \hat{j} + \hat{k}) . [(λ + 1) \hat{i} + (λ + 3) \hat{j} - (2 λ + 1) \hat{k}] = 0 \Rightarrow 2 (λ + 1) - (λ + 3) - (2 λ + 1) = 0 \Rightarrow 2 λ + 2 - λ - 3 - 2 λ - 1 = 0 \Rightarrow λ = - 2$

Thus, the value of λ is −2.

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Question 16:

If $\vec{p} = 5 \hat{i} + λ \hat{j} - 3 \hat{k} and \vec{q} = \hat{i} + 3 \hat{j} - 5 \hat{k},$ then find the value of λ, so that $\vec{p} + \vec{q}$ and $\vec{p} - \vec{q}$ are perpendicular vectors.

Answer:

$Given that \vec{p} = 5 \hat{i} + λ \hat{j} - 3 \hat{k} a n d \vec{q} = \hat{i} + 3 \hat{j} - 5 \hat{k} \vec{p} + \vec{q} = (5 \hat{i} + λ \hat{j} - 3 \hat{k}) + (\hat{i} + 3 \hat{j} - 5 \hat{k}) = 6 \hat{i} + (λ + 3) \hat{j} - 8 \hat{k} \vec{p} - \vec{q} = (5 \hat{i} + λ \hat{j} - 3 \hat{k}) - (\hat{i} + 3 \hat{j} - 5 \hat{k}) = 4 \hat{i} + (λ - 3) \hat{j} + 2 \hat{k} Given that \vec{p} + \vec{q} is orthogonal to \vec{p} - \vec{q} . \Rightarrow (\vec{p} + \vec{q}) . (\vec{p} - \vec{q}) = 0 \Rightarrow [6 \hat{i} + (λ + 3) \hat{j} - 8 \hat{k}] . [4 \hat{i} + (λ - 3) \hat{j} + 2 \hat{k}] = 0 \Rightarrow 24 + λ^{2} - 9 - 16 = 0 \Rightarrow λ^{2} = 1 ∴ λ = \pm 1$

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Question 17:

If $\vec{α} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k} and \vec{β} = 2 \hat{i} + \hat{j} - 4 \hat{k},$ then express $\vec{β}$ in the form of $\vec{β} = \vec{β_{1}} + \vec{β_{2}},$ where $\vec{β_{1}}$ is parallel to $\vec{α} and \vec{β_{2}}$ is perpendicular to $\vec{α}$ .

Answer:

$Given that \vec{α} = 3 \hat{i} + 4 \hat{j} +5 \hat{k} and \vec{β} =2 \hat{i} + \hat{j} - 4 \hat{k} Also, \vec{β} = \vec{β_{1}} + \vec{β_{2}} \Rightarrow \vec{β_{2}} = \vec{β} - {\vec{β}}_{1} . . . (1) Since {\vec{β}}_{1} is parallel to \vec{α}, \vec{β_{1}} = t \vec{α} \Rightarrow \vec{β_{1}} = t (3 \hat{i} + 4 \hat{j} +5 \hat{k}) = 3 t \hat{i} + 4 t \hat{j} +5 t \hat{k} ...(2) Substituting the values of \vec{β_{1}} and \vec{α} in (1), we get \vec{β_{2}} = 2 \hat{i} + \hat{j} - 4 \hat{k} - (3 t \hat{i} + 4 t \hat{j} +5 t \hat{k}) = (2 - 3 t) \hat{i} + (1 - 4 t) \hat{j} + (- 4 - 5 t) \hat{k} . . . (3) Since \vec{β_{2}} is perpendicular to \vec{α}, \vec{β_{2}} . \vec{α} = 0 \Rightarrow [(2 - 3 t) \hat{i} + (1 - 4 t) \hat{j} + (- 4 - 5 t) \hat{k}] . (3 \hat{i} + 4 \hat{j} +5 \hat{k}) = 0 \Rightarrow 3 (2 - 3 t) + 4 (1 - 4 t) + 5 (- 4 - 5 t) = 0 \Rightarrow 6 - 9 t + 4 - 16 t - 20 - 25 t = 0 \Rightarrow - 50 t = 10 \Rightarrow t = \frac{- 1}{5} From (2) and (3), we get \vec{β_{1}} = \frac{- 1}{5} (3 \hat{i} + 4 \hat{j} +5 \hat{k}) \vec{β_{2}} = \frac{13}{5} \hat{i} + \frac{9}{5} \hat{j} - 3 \hat{k} = \frac{1}{5} (13 \hat{i} + 9 \hat{j} - 15 \hat{k})$

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Question 18:

If either $\vec{a} = \vec{0} or \vec{b} = \vec{0}$ , then $\vec{a} \cdot \vec{b} = 0 .$ But the converse need not be true. Justify your answer with an example.

Answer:

$Let us assume that either |\vec{a}| =0 or |\vec{b}| = 0 Then, \vec{a} . \vec{b} = |\vec{a}| |\vec{b}| \cos θ = 0 (θ is the angle between \vec{a} and \vec{b}) Now, let us assume that \vec{a} . \vec{b} = 0 \Rightarrow |\vec{a}| |\vec{b}| \cos θ = 0 But here we cannot say that either |\vec{a}| =0 or |\vec{b}| = 0 . (Because even cos θ can be zero) For example, let \vec{a} = 2 \hat{i} + \hat{j} + 3 \hat{k} and \vec{b} = - 3 \hat{i} + 2 \hat{k} Here, |\vec{a}| = \sqrt{4 + 1 + 9} = \sqrt{14} ≠0 |\vec{b}| = \sqrt{9 + 4} = \sqrt{13} ≠0 But \vec{a} . \vec{b} = (2 \hat{i} + \hat{j} + 3 \hat{k}) . (- 3 \hat{i} + 2 \hat{k}) = - 6 + 0 + 6 = 0$

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Question 19:

Show that the vectors $\vec{a} = 3 \hat{i} - 2 \hat{j} + \hat{k}, \vec{b} = \hat{i} - 3 \hat{j} + 5 \hat{k}, \vec{c} = 2 \hat{i} + \hat{j} - 4 \hat{k}$ form a right-angled triangle.

Answer:

$Let ABC be the given triangle and \vec{A C} = \vec{b} = \hat{i} - 3 \hat{j} + 5 \hat{k} \vec{C B} = \vec{a} = 3 \hat{i} - 2 \hat{j} + \hat{k} \vec{A B} = \vec{c} = 2 \hat{i} + \hat{j} - 4 \hat{k} \vec{a} . \vec{b} = 3 + 6 + 5 = 14 \vec{b} . \vec{c} = 2 - 3 - 20 = - 21 \vec{c} . \vec{a} = 6 - 2 - 4 = 0 So, \vec{A B} is perpendicular to \vec{C B} . Thus, ∆ ABC is a right-angled triangle.$

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Question 20:

If $\vec{a} = 2 \hat{i} + 2 \hat{j} + 3 \hat{k}, \vec{b} = - \hat{i} + 2 \hat{j} + \hat{k} and \vec{c} = 3 \hat{i} + \hat{j}$ are such that $\vec{a} + λ \vec{b}$ is perpendicular to $\vec{c}$ , then find the value of λ.

Answer:

$We have \vec{a} = 2 \hat{i} + 2 \hat{j} + 3 \hat{k} \vec{b} = - \hat{i} + 2 \hat{j} + \hat{k} and \vec{c} = 3 \hat{i} + \hat{j} ∴ \vec{a} + λ \vec{b} = 2 \hat{i} + 2 \hat{j} + 3 \hat{k} + λ (- \hat{i} + 2 \hat{j} + \hat{k}) = (2 - λ) \hat{i} + (2 + 2 λ) \hat{j} + (3 + λ) \hat{k} Given that \vec{a} + λ \vec{b} is perpendicular to \vec{c} . \Rightarrow (\vec{a} + λ \vec{b}) . \vec{c} = 0 \Rightarrow [(2 - λ) \hat{i} + (2 + 2 λ) \hat{j} + (3 + λ) \hat{k}] . (3 \hat{i} + \hat{j} + 0 \hat{k}) = 0 \Rightarrow 3 (2 - λ) + 1 (2 + 2 λ) + 0 = 0 \Rightarrow 6 - 3 λ + 2 + 2 λ = 0 ∴ λ = 8$

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Question 21:

Find the angles of a triangle whose vertices are A (0, −1, −2), B (3, 1, 4) and C (5, 7, 1).

Answer:

$Given that \vec{O A} = 0 \hat{i} - 1 \hat{j} - 2 \hat{k}; \vec{O B} = 3 \hat{i} + 1 \hat{j} + 4 \hat{k}; \vec{O C} = 5 \hat{i} + 7 \hat{j} + 1 \hat{k} \vec{A B} = \vec{O B} - \vec{O A} = 3 \hat{i} + 2 \hat{j} + 6 \hat{k} \Rightarrow |\vec{A B}| = \sqrt{9 + 4 + 36} = 7 \vec{B A} = \vec{O A} - \vec{O B} = - 3 \hat{i} - 2 \hat{j} - 6 \hat{k} \Rightarrow |\vec{B A}| = \sqrt{9 + 4 + 36} = 7 \vec{B C} = \vec{O C} - \vec{O B} = 2 \hat{i} + 6 \hat{j} - 3 \hat{k} \Rightarrow |\vec{B C}| = \sqrt{4 + 36 + 9} = 7 \vec{C B} = \vec{O B} - \vec{O C} = - 2 \hat{i} - 6 \hat{j} + 3 \hat{k} \Rightarrow |\vec{C B}| = \sqrt{4 + 36 + 9} = 7 \vec{C A} = \vec{O A} - \vec{O C} = - 5 \hat{i} - 8 \hat{j} - 3 \hat{k} \Rightarrow |\vec{C A}| = \sqrt{25 + 64 + 9} = \sqrt{98} = 7 \sqrt{2} \vec{A C} = \vec{O C} - \vec{O A} = 5 \hat{i} + 8 \hat{j} + 3 \hat{k} \Rightarrow |\vec{A C}| = \sqrt{25 + 64 + 9} = \sqrt{98} = 7 \sqrt{2} \cos A = \frac{\vec{A B} . \vec{A C}}{|\vec{A B}| |\vec{A C}|} = \frac{15 + 16 + 18}{(7) (7 \sqrt{2})} = \frac{49}{49 \sqrt{2}} = \frac{1}{\sqrt{2}} \Rightarrow A = \cos^{- 1} (\frac{1}{\sqrt{2}}) = \frac{π}{4} \cos B = \frac{\vec{B A} . \vec{B C}}{|\vec{B A}| |\vec{B C}|} = \frac{- 6 - 12 + 18}{(7) (7)} = \frac{0}{49} = 0 \Rightarrow B = \cos^{- 1} (0) = \frac{π}{2} \cos C = \frac{\vec{C B} . \vec{C A}}{|\vec{C B}| |\vec{C A}|} = \frac{10 + 48 - 9}{(7) (7 \sqrt{2})} = \frac{49}{49 \sqrt{2}} = \frac{1}{\sqrt{2}} \Rightarrow C = \cos^{- 1} (\frac{1}{\sqrt{2}}) = \frac{π}{4}$

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Question 22:

Find the magnitude of two vectors $\vec{a} and \vec{b}$ that are of the same magnitude, are inclined at 60° and whose scalar product is 1/2.

Answer:

$Given that the angle between \vec{a} and \vec{b} {is 30}^{0} . Also, |\vec{a}| = |\vec{b}|; \vec{a} . \vec{b} = \frac{1}{2} We know that \vec{a} . \vec{b} = |\vec{a}| |\vec{b}| \cos θ \Rightarrow \frac{1}{2} = |\vec{a}| |\vec{a}| \cos 60 \Rightarrow \frac{1}{2} = {|\vec{a}|}^{2} (\frac{1}{2}) \Rightarrow {|\vec{a}|}^{2} = 1 \Rightarrow |\vec{a}| = 1 ∴ |\vec{a}| = |\vec{b}| = 1$

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Question 23:

Show that the points whose position vectors are $\vec{a} = 4 \hat{i} - 3 \hat{j} + \hat{k}, \vec{b} = 2 \hat{i} - 4 \hat{j} + 5 \hat{k}, \vec{c} = \hat{i} - \hat{j}$ form a right triangle.

Answer:

$Given that \vec{a} = \vec{O A} = 4 \hat{i} - 3 \hat{j} + \hat{k}; \vec{b} = \vec{O B} = 2 \hat{i} - 4 \hat{j} + 5 \hat{k}; \vec{c} = \vec{O C} = \hat{i} - \hat{j} + 0 \hat{k} \vec{A B} = \vec{O B} - \vec{O A} = - 2 \hat{i} - \hat{j} + 4 \hat{k} \vec{B C} = \vec{O C} - \vec{O B} = - \hat{i} + 3 \hat{j} - 5 \hat{k} \vec{C A} = \vec{O A} - \vec{O C} = 3 \hat{i} - 2 \hat{j} + \hat{k} \vec{A B} . \vec{B C} = 2 - 3 - 20 = - 21 \neq 0 \vec{B C} . \vec{C A} = - 3 - 6 - 5 = - 14 \neq 0 \vec{A B} . \vec{C A} = - 6 + 2 + 4 = 0 So, \vec{A B} is perpendicular to \vec{C A} . So, ∆ ABC is a right-angled triangle.$

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Question 24:

If the vertices A, B and C of ∆ABC have position vectors (1, 2, 3), (−1, 0, 0) and (0, 1, 2), respectively, what is the magnitude of ∠ABC?

Answer:

$Given that \vec{O A} = \hat{i} + 2 \hat{j} + 3 \hat{k}; \vec{O B} = - 1 \hat{i} + 0 \hat{j} + 0 \hat{k}; \vec{O C} = 0 \hat{i} + 1 \hat{j} + 2 \hat{k} \vec{A B} = \vec{O B} - \vec{O A} = - 2 \hat{i} - 2 \hat{j} - 3 \hat{k} \Rightarrow |\vec{A B}| = \sqrt{4 + 4 + 9} = \sqrt{17} \vec{B C} = \vec{O C} - \vec{O B} = \hat{i} + \hat{j} + 2 \hat{k} \Rightarrow |\vec{B C}| = \sqrt{1 + 1 + 4} = \sqrt{6} \vec{C A} = \vec{O A} - \vec{O C} = \hat{i} + \hat{j} + \hat{k} \Rightarrow |\vec{C A}| = \sqrt{1 + 1 + 1} = \sqrt{3} \cos ∠ A B C = \frac{|\vec{A B} . \vec{B C}|}{|\vec{A B}| |\vec{B C}|} = \frac{|- 2 - 2 - 6|}{(\sqrt{17}) (\sqrt{6})} = \frac{10}{\sqrt{102}} \Rightarrow ∠ A B C = \cos^{- 1} (\frac{10}{\sqrt{102}})$

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Question 25:

If A, B and C have position vectors (0, 1, 1), (3, 1, 5) and (0, 3, 3) respectively, show that ∆ ABC is right-angled at C.

Answer:

$Given that \vec{O A} = 0 \hat{i} + \hat{j} + \hat{k}; \vec{O B} = 3 \hat{i} + \hat{j} + 5 \hat{k}; \vec{O C} = 0 \hat{i} + 3 \hat{j} + 3 \hat{k} \vec{B C} = \vec{O C} - \vec{O B} = - 3 \hat{i} + 2 \hat{j} - 2 \hat{k} \vec{C A} = \vec{O A} - \vec{O C} = 0 \hat{i} - 2 \hat{j} - 2 \hat{k} Now, \vec{B C} . \vec{C A} = 0 - 4 + 4 = 0 So, \vec{B C} is perpendicular to \vec{C A} . So, ∆ ABC is right-angled at C .$

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Question 26:

Find the projection of $\vec{b} + \vec{c} on \vec{a}$ , where $\vec{a} = 2 \hat{i} - 2 \hat{j} + \hat{k}, \vec{b} = \hat{i} + 2 \hat{j} - 2 \hat{k} and \vec{c} = 2 \hat{i} - \hat{j} + 4 \hat{k} .$

Answer:

$Given that \vec{a} = 2 \hat{i} - 2 \hat{j} + \hat{k} \vec{b} = \hat{i} + 2 \hat{j} - 2 \hat{k} and \vec{c} = 2 \hat{i} - \hat{j} + 4 \hat{k} ∴ \vec{b} + \vec{c} = \hat{i} + 2 \hat{j} - 2 \hat{k} + 2 \hat{i} - \hat{j} + 4 \hat{k} = 3 \hat{i} + \hat{j} + 2 \hat{k} Projection of \vec{b} + \vec{c} on \vec{a} is \frac{(\vec{b} + \vec{c}) . \vec{a}}{|\vec{a}|} = \frac{(3 \hat{i} + \hat{j} + 2 \hat{k}) . (2 \hat{i} - 2 \hat{j} + \hat{k})}{\sqrt{4 + 4 + 1}} = \frac{6 - 2 + 2}{3} = 2$

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Question 27:

If $\vec{a} = 5 \hat{i} - \hat{j} - 3 \hat{k} and \vec{b} = \hat{i} + 3 \hat{j} - 5 \hat{k},$ then show that the vectors $\vec{a} + \vec{b} and \vec{a} - \vec{b}$ are orthogonal.

Answer:

$Given that \vec{a} = 5 \hat{i} - \hat{j} - 3 \hat{k}; \vec{b} = \hat{i} + 3 \hat{j} - 5 \hat{k} ∴ \vec{a} + \vec{b} = 5 \hat{i} - \hat{j} - 3 \hat{k} + \hat{i} + 3 \hat{j} - 5 \hat{k} = 6 \hat{i} + 2 \hat{j} - 8 \hat{k} And \vec{a} - \vec{b} = 5 \hat{i} - \hat{j} - 3 \hat{k} - (\hat{i} + 3 \hat{j} - 5 \hat{k}) = 4 \hat{i} - 4 \hat{j} + 2 \hat{k} Now, (\vec{a} + \vec{b}) . (\vec{a} - \vec{b}) = (6 \hat{i} + 2 \hat{j} - 8 \hat{k}) . (4 \hat{i} - 4 \hat{j} + 2 \hat{k}) = 24 - 8 - 16 = 0 So, \vec{a} + \vec{b} is orthogonal to \vec{a} - \vec{b} .$

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Question 28:

A unit vector $\vec{a}$ makes angles $\frac{π}{4} and \frac{π}{3}$ with $\hat{i}$ and $\hat{j}$ respectively and an acute angle θ with $\hat{k}$ . Find the angle θ and components of $\vec{a}$ .

Answer:

$Let \vec{a} = a_{1} i \hat{} + a_{2} \hat{j} {+a}_{3} \hat{k}, where a_{1}, a_{2} and a_{3} are components of \vec{a} . \Rightarrow {a_{1}}^{2} + {a_{2}}^{2} + {a_{3}}^{2} = 1 (Because \vec{a} is a unit vector) . . . (1) Now, \vec{a} . \hat{i} = a_{1} \Rightarrow |\vec{a}| |\hat{i}| \cos \frac{π}{4} = a_{1} (Because the angle between \vec{a} and \hat{i} is \frac{π}{4}) \Rightarrow (1) (1) \frac{1}{\sqrt{2}} = a_{1} (Because \vec{a} and \hat{i} are unit vectors) \Rightarrow a_{1} = \frac{1}{\sqrt{2}} Again, \vec{a} . \hat{j} = a_{2} \Rightarrow |\vec{a}| |\hat{i}| \cos \frac{π}{3} = a_{2} (Because the angle between \vec{a} and \hat{i} is \frac{π}{3}) \Rightarrow (1) (1) \frac{1}{2} = a_{2} (Because \vec{a} and \hat{i} are unit vectors) \Rightarrow a_{2} = \frac{1}{2} Now from (1), {(\frac{1}{\sqrt{2}})}^{2} + {(\frac{1}{2})}^{2} + {a_{3}}^{2} = 1 \Rightarrow \frac{1}{2} + \frac{1}{4} + {a_{3}}^{2} = 1 \Rightarrow \frac{3}{4} + {a_{3}}^{2} = 1 \Rightarrow {a_{3}}^{2} = \frac{1}{4} \Rightarrow a_{3} = \frac{1}{2} Now, \vec{a} . \hat{k} = a_{3} \Rightarrow |\vec{a}| |\hat{k}| \cos θ = \frac{1}{2} (Because the angle between \vec{a} and \hat{k} is θ) \Rightarrow (1) (1) \cos θ = \frac{1}{2} (Because \vec{a} and \hat{i} are unit vectors) \Rightarrow θ = \cos^{- 1} (\frac{1}{2}) = \frac{π}{3} And \vec{a} = \frac{1}{\sqrt{2}} i \hat{} + \frac{1}{2} \hat{j} + \frac{1}{2} \hat{k}$

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Question 29:

If two vectors $\vec{a} and \vec{b}$ are such that $|\vec{a}| = 2, |\vec{b}| = 1 and \vec{a} \cdot \vec{b} = 1,$ then find the value of $(3 \vec{a} - 5 \vec{b}) \cdot (2 \vec{a} + 7 \vec{b}) .$

Answer:

$Given that |\vec{a}| = 2, |\vec{b}| = 1 and \vec{a} . \vec{b} = 1 . . . (1) Now, (3 \vec{a} - 5 \vec{b}) . (2 \vec{a} + 7 \vec{b}) = 6 {|\vec{a}|}^{2} + 21 \vec{a} . \vec{b} - 10 \vec{b} . \vec{a} - 35 {|\vec{b}|}^{2} = 6 {|\vec{a}|}^{2} + 21 \vec{a} . \vec{b} - 10 \vec{a} . \vec{b} - 35 {|\vec{b}|}^{2} (We know that \vec{a} . \vec{b} = \vec{b} . \vec{a}) = 6 {|\vec{a}|}^{2} + 11 \vec{a} . \vec{b} - 35 {|\vec{b}|}^{2} = 6 {(2)}^{2} + 11 (1) - 35 {(1)}^{2} [From (1)] = 24 + 11 - 35 = 0$

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Question 30:

If $\vec{a}$ is a unit vector, then find $|\vec{x}|$ in each of the following.
(i) $(\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = 8$

(ii) $(\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = 12$

Answer:

$Given that \vec{a} is a unit vector. \Rightarrow |\vec{a}| = 1 . . . (1) (i) (\vec{x} - \vec{a}) . (\vec{x} + \vec{a}) = 8 \Rightarrow {|\vec{x}|}^{2} - {|\vec{a}|}^{2} = 8 \Rightarrow {|\vec{x}|}^{2} - 1^{2} = 8 [From (1)] \Rightarrow {|\vec{x}|}^{2} = 9 \Rightarrow |\vec{x}| = 3 (ii) (\vec{x} - \vec{a}) . (\vec{x} + \vec{a}) = 12 \Rightarrow {|\vec{x}|}^{2} - {|\vec{a}|}^{2} = 12 \Rightarrow {|\vec{x}|}^{2} - 1^{2} = 12 [From (1)] \Rightarrow {|\vec{x}|}^{2} = 13 \Rightarrow |\vec{x}| = \sqrt{13}$

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Question 31:

Find $|\vec{a}| and |\vec{b}|$ if
(i) $(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 12 and |\vec{a}| = 2 |\vec{b}|$

(ii) $(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 8 and |\vec{a}| = 8 |\vec{b}|$

(iii) $(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 3 and |\vec{a}| = 2 |\vec{b}|$

Answer:

$(i) Given that |\vec{a}| = 2 |\vec{b}| . . . (1) And (\vec{a} + \vec{b}) . (\vec{a} - \vec{b}) = 12 \Rightarrow {|\vec{a}|}^{2} - {|\vec{b}|}^{2} = 12 \Rightarrow {(2 |\vec{b}|)}^{2} - {|\vec{b}|}^{2} = 12 [From (1)] \Rightarrow 4 {|\vec{b}|}^{2} - {|\vec{b}|}^{2} = 12 \Rightarrow 3 {|\vec{b}|}^{2} = 12 \Rightarrow {|\vec{b}|}^{2} = 4 \Rightarrow |\vec{b}| = 2 |\vec{a}| = 2 |\vec{b}| = 2 (2) = 4 ∴ |\vec{a}| = 4 and |\vec{b}| = 2$

$(ii) Given that And |\vec{a}| = 8 |\vec{b}| . . . (1) (\vec{a} + \vec{b}) . (\vec{a} - \vec{b}) = 8 \Rightarrow {|\vec{a}|}^{2} - {|\vec{b}|}^{2} = 8 \Rightarrow {(8 |\vec{b}|)}^{2} - {|\vec{b}|}^{2} = 8 [From (1)] \Rightarrow 64 {|\vec{b}|}^{2} - {|\vec{b}|}^{2} = 8 \Rightarrow 63 {|\vec{b}|}^{2} = 8 \Rightarrow {|\vec{b}|}^{2} = \frac{8}{63} \Rightarrow |\vec{b}| = \sqrt{\frac{8}{63}} |\vec{a}| = 8 |\vec{b}| = 8 \sqrt{\frac{8}{63}} = \frac{8 \sqrt{8}}{\sqrt{63}} ∴ |\vec{a}| = \frac{8 \sqrt{8}}{\sqrt{63}} and |\vec{b}| = \sqrt{\frac{8}{63}}$

$(iii) Given that |\vec{a}| = 2 |\vec{b}| . . . (1) And (\vec{a} + \vec{b}) . (\vec{a} - \vec{b}) = 3 \Rightarrow {|\vec{a}|}^{2} - {|\vec{b}|}^{2} = 3 \Rightarrow {(2 |\vec{b}|)}^{2} - {|\vec{b}|}^{2} = 3 [From (1)] \Rightarrow 4 {|\vec{b}|}^{2} - {|\vec{b}|}^{2} = 3 \Rightarrow 3 {|\vec{b}|}^{2} = 3 \Rightarrow {|\vec{b}|}^{2} = 1 \Rightarrow |\vec{b}| = 1 |\vec{a}| = 2 |\vec{b}| = 2 (1) = 2 ∴ |\vec{a}| = 2 and |\vec{b}| = 1$

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Question 32:

Find $|\vec{a} - \vec{b}|$ if
(i) $|\vec{a}| = 2, |\vec{b}| = 5 and \vec{a} \cdot \vec{b} = 8$

(ii) $|\vec{a}| = 3, |\vec{b}| = 4 and \vec{a} \cdot \vec{b} = 1$

(iii) $|\vec{a}| = 2, |\vec{b}| = 3 and \vec{a} \cdot \vec{b} = 4$

Answer:

$(i) Given that |\vec{a}| = 2, |\vec{b}| = 5 and \vec{a} . \vec{b} = 8 . . . (1) We know that {|\vec{a} - \vec{b}|}^{2} = {|\vec{a}|}^{2} + {|\vec{b}|}^{2} - 2 \vec{a} . \vec{b} = 2^{2} + 5^{2} - 2 (8) [Using (1)] = 4 + 25 - 16 = 13 ∴ |\vec{a} - \vec{b}| = \sqrt{13}$

$(ii) Given that |\vec{a}| = 3, |\vec{b}| = 4 and \vec{a} . \vec{b} = 1 . . . (1) We know that {|\vec{a} - \vec{b}|}^{2} = {|\vec{a}|}^{2} + {|\vec{b}|}^{2} - 2 \vec{a} . \vec{b} = 3^{2} + 4^{2} - 2 (1) [Using (1)] = 9 + 16 - 2 = 23 ∴ |\vec{a} - \vec{b}| = \sqrt{23}$

$(iii) Given that |\vec{a}| = 2, |\vec{b}| = 3 and \vec{a} . \vec{b} = 4 . . . (1) We know that {|\vec{a} - \vec{b}|}^{2} = {|\vec{a}|}^{2} + {|\vec{b}|}^{2} - 2 \vec{a} . \vec{b} = 2^{2} + 3^{2} - 2 (4) [Using (1)] = 4 + 9 - 8 = 5 ∴ |\vec{a} - \vec{b}| = \sqrt{5}$

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Question 33:

Find the angle between two vectors $\vec{a} and \vec{b}$ if
(i) $|\vec{a}| = \sqrt{3}, |\vec{b}| = 2 and \vec{a} \cdot \vec{b} = \sqrt{6}$

(ii) $|\vec{a}| = 3, |\vec{b}| = 3 and \vec{a} \cdot \vec{b} = 1$

Answer:

$(i) Let θ be the angle between \vec{a} and \vec{b} . Given that |\vec{a}| = \sqrt{3}, |\vec{b}| = 2 and \vec{a} . \vec{b} = \sqrt{6} . . . (1) We know that \vec{a} . \vec{b} = |\vec{a}| |\vec{b}| \cos θ \Rightarrow \sqrt{6} = (\sqrt{3}) (2) \cos θ [Using (1)] \Rightarrow \cos θ = \frac{\sqrt{6}}{2 \sqrt{3}} = \frac{1}{\sqrt{2}} \Rightarrow θ = \cos^{- 1} (\frac{1}{\sqrt{2}}) = \frac{π}{4}$

$(ii) Let θ be the angle between \vec{a} and \vec{b} . Given that |\vec{a}| = 3, |\vec{b}| = 3 and \vec{a} . \vec{b} = 1 . . . (1) We know that \vec{a} . \vec{b} = |\vec{a}| |\vec{b}| \cos θ \Rightarrow 1 = (3) (3) \cos θ [Using (1)] \Rightarrow \cos θ = \frac{1}{(3) (3)} = \frac{1}{9} \Rightarrow θ = \cos^{- 1} (\frac{1}{9})$

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Question 34:

Express the vector $\vec{a} = 5 \hat{i} - 2 \hat{j} + 5 \hat{k}$ as the sum of two vectors such that one is parallel to the vector $\vec{b} = 3 \hat{i} + \hat{k}$ and other is perpendicular to $\vec{b}$ .

Answer:

$Given that \vec{a} = 5 \hat{i} - 2 \hat{j} + 5 \hat{k} and \vec{b} =3 \hat{i} + \hat{k} Let \vec{x} and \vec{y} be such that \vec{a} = \vec{x} + \vec{y} \Rightarrow \vec{y} = \vec{a} - \vec{x} . . . (1) Since \vec{x} is parallel to \vec{b}, \Rightarrow \vec{x} = t \vec{b} (t is constant) \Rightarrow \vec{x} = t (3 \hat{i} + \hat{k}) = 3 t \hat{i} +t \hat{k} Substituting the values of \vec{x} and \vec{a} in (1), we get \vec{y} = 5 \hat{i} - 2 \hat{j} + 5 \hat{k} - (3 t \hat{i} +t \hat{k}) = (5 - 3 t) \hat{i} - 2 \hat{j} + (5 - t) \hat{k} . . . (2) Since \vec{y} is perpendicular to \vec{b}, \vec{y} . \vec{b} = 0 \Rightarrow [(5 - 3 t) \hat{i} - 2 \hat{j} + (5 - t) \hat{k}] . (3 \hat{i} + \hat{k}) = 0 \Rightarrow 3 (5 - 3 t) + 0 + (5 - t) = 0 \Rightarrow 15 - 9 t + 5 - t = 0 \Rightarrow 20 - 10 t = 0 \Rightarrow t = 2 From (1) and (2), we get \vec{x} = 6 \hat{i} +2 \hat{k} \vec{y} = - \hat{i} - 2 \hat{j} + 3 \hat{k}$

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Question 35:

If $\vec{a} and \vec{b}$ are two vectors of the same magnitude inclined at an angle of 30°, such that $\vec{a} \cdot \vec{b} = 3, find |\vec{a}|, |\vec{b}| .$

Answer:

$Given that the angle between \vec{a} and \vec{b} {is 30}^{0} . Also, |\vec{a}| = |\vec{b}| and \vec{a} . \vec{b} = 3 We know that \vec{a} . \vec{b} = |\vec{a}| |\vec{b}| \cos θ \Rightarrow 3 = |\vec{a}| |\vec{a}| \cos 30 \Rightarrow 3 = {|\vec{a}|}^{2} (\frac{\sqrt{3}}{2}) \Rightarrow {|\vec{a}|}^{2} = \frac{6}{\sqrt{3}} = 2 \sqrt{3} \Rightarrow |\vec{a}| = \sqrt{2 \sqrt{3}} = |\vec{b}| ∴ |\vec{a}| = |\vec{b}| = \sqrt{2 \sqrt{3}}$

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Question 36:

Express $2 \hat{i} - \hat{j} + 3 \hat{k}$ as the sum of a vector parallel and a vector perpendicular to $2 \hat{i} + 4 \hat{j} - 2 \hat{k} .$

Answer:

$Let \vec{a} =2 \hat{i} - \hat{j} + 3 \hat{k} and \vec{b} = 2 \hat{i} + 4 \hat{j} - 2 \hat{k} and \vec{x} and \vec{y} be such that \vec{a} = \vec{x} + \vec{y} \Rightarrow \vec{y} = \vec{a} - \vec{x} ...(1) Since \vec{x} is parallel to \vec{b}, \vec{x} = t \vec{b} \Rightarrow \vec{x} = t (2 \hat{i} + 4 \hat{j} - 2 \hat{k}) = 2 t \hat{i} + 4 t \hat{j} - 2 t \hat{k} ...(2) Substituting the values of \vec{x} and \vec{a} in (1), \vec{y} = 2 \hat{i} - \hat{j} + 3 \hat{k} - (2 t \hat{i} + 4 t \hat{j} - 2 t \hat{k}) = (2 - 2 t) \hat{i} + (- 1 - 4 t) \hat{j} + (3 + 2 t) \hat{k} . . . (3) Since \vec{y} is perpendicular to \vec{b}, \vec{y} . \vec{b} = 0 \Rightarrow [(2 - 2 t) \hat{i} + (- 1 - 4 t) \hat{j} + (3 + 2 t) \hat{k}] . (2 \hat{i} + 4 \hat{j} - 2 \hat{k}) = 0 \Rightarrow 2 (2 - 2 t) + 4 (- 1 - 4 t) - 2 (3 + 2 t) = 0 \Rightarrow 4 - 4 t - 4 - 16 t - 6 - 4 t = 0 \Rightarrow - 24 t = 6 \Rightarrow t = (\frac{- 1}{4}) From (2) and (3), \vec{x} = 2 (\frac{- 1}{4}) \hat{i} + 4 (\frac{- 1}{4}) \hat{j} - 2 (\frac{- 1}{4}) \hat{k} = \frac{- 1}{2} \hat{i} - \hat{j} + \frac{1}{2} \hat{k} \vec{y} = [2 - 2 (\frac{- 1}{4})] \hat{i} + [- 1 - 4 (\frac{- 1}{4})] \hat{j} + [3 + 2 (\frac{- 1}{4})] \hat{k} = \frac{5}{2} \hat{i} + \frac{5}{2} \hat{k} = \frac{5}{2} (\hat{i} + \hat{k}) So, \vec{a} = \vec{x} + \vec{y} = (\frac{- 1}{2} \hat{i} - \hat{j} + \frac{1}{2} \hat{k}) + \frac{5}{2} (\hat{i} + \hat{k})$

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Question 37:

Decompose the vector $6 \hat{i} - 3 \hat{j} - 6 \hat{k}$ into vectors which are parallel and perpendicular to the vector $\hat{i} + \hat{j} + \hat{k} .$

Answer:

$Let \vec{a} =6 \hat{i} - 3 \hat{j} - 6 \hat{k} and \vec{b} = \hat{i} + \hat{j} + \hat{k} and \vec{x} and \vec{y} be such that \vec{a} = \vec{x} + \vec{y} \Rightarrow \vec{y} = \vec{a} - \vec{x} . . . (1) Since \vec{x} is parallel to \vec{b}, \vec{x} = t \vec{b} \Rightarrow \vec{x} = t (\hat{i} + \hat{j} + \hat{k}) = t \hat{i} + t \hat{j} + t \hat{k} ...(2) Substituting the values of \vec{x} and \vec{a} in (1), we get \vec{y} = 6 \hat{i} - 3 \hat{j} - 6 \hat{k} - (t \hat{i} + t \hat{j} + t \hat{k}) = (6 - t) \hat{i} + (- 3 - t) \hat{j} + (- 6 - t) \hat{k} . . . (3) Since \vec{y} is perpendicular to \vec{b}, \vec{y} . \vec{b} = 0 \Rightarrow [(6 - t) \hat{i} + (- 3 - t) \hat{j} + (- 6 - t) \hat{k}] . (\hat{i} + \hat{j} + \hat{k}) = 0 \Rightarrow 1 (6 - t) + 1 (- 3 - t) + 1 (- 6 - t) = 0 \Rightarrow - 3 - 3 t = 0 \Rightarrow t = - 1 From (2) and (3), we get \vec{x} = - \hat{i} - \hat{j} - \hat{k} \vec{y} = 7 \hat{i} - 2 \hat{j} - 5 \hat{k} So, \vec{a} = \vec{x} + \vec{y} = (- \hat{i} - \hat{j} - \hat{k}) + (7 \hat{i} - 2 \hat{j} - 5 \hat{k})$

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Question 38:

Let $\vec{a} = 5 \hat{i} - \hat{j} + 7 \hat{k} a n d \vec{b} = \hat{i} - \hat{j} + λ \hat{k} .$ Find λ such that $\vec{a} + \vec{b}$ is orthogonal to $\vec{a} - \vec{b}$ .

Answer:

$Given that \vec{a} = 5 \hat{i} - \hat{j} + 7 \hat{k}; \vec{b} = \hat{i} - \hat{j} + λ \hat{k} ∴ \vec{a} + \vec{b} = 5 \hat{i} - \hat{j} + 7 \hat{k} + \hat{i} - \hat{j} + λ \hat{k} = 6 \hat{i} - 2 \hat{j} + (7 + λ) \hat{k} and \vec{a} - \vec{b} = 5 \hat{i} - \hat{j} + 7 \hat{k} - (\hat{i} - \hat{j} + λ \hat{k}) = 4 \hat{i} + 0 \hat{j} + (7 - λ) \hat{k} Given that \vec{a} + \vec{b} is orthogonal to \vec{a} - \vec{b} . \Rightarrow (\vec{a} + \vec{b}) . (\vec{a} - \vec{b}) = 0 \Rightarrow [6 \hat{i} - 2 \hat{j} + (7 + λ) \hat{k}] . [4 \hat{i} + 0 \hat{j} + (7 - λ) \hat{k}] = 0 \Rightarrow 24 + 0 + 49 - λ^{2} = 0 \Rightarrow λ^{2} = 73 \Rightarrow λ = \sqrt{73}$

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Question 39:

If $\vec{a} \cdot \vec{a} = 0 and \vec{a} \cdot \vec{b} = 0,$ what can you conclude about the vector $\vec{b}$ ?

Answer:

$Given that \vec{a} . \vec{a} = 0 \Rightarrow {|\vec{a}|}^{2} = 0 \Rightarrow |\vec{a}| = 0 . . . (1) Also given that \vec{a} . \vec{b} = 0 \Rightarrow |\vec{a}| |\vec{b}| \cos θ = 0 (Where θ is the angle between \vec{a} and \vec{b}) \Rightarrow 0 |\vec{b}| \cos θ = 0 [From (1)] \Rightarrow 0 = 0 So, it means that for any vector \vec{b}, the given equation \vec{a} . \vec{b} = 0 is satisfied.$

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Question 40:

If $\vec{c}$ is perpendicular to both $\vec{a} and \vec{b}$ , then prove that it is perpendicular to both $\vec{a} + \vec{b} and \vec{a} - \vec{b}$ .

Answer:

$Given that \vec{c} is perpendicular to both \vec{a} and \vec{b} . \Rightarrow \vec{c} . \vec{a} = 0 and \vec{c} . \vec{b} = 0 . . . (1) Now, \vec{c} . (\vec{a} + \vec{b}) = \vec{c} . \vec{a} + \vec{c} . \vec{b} = 0 + 0 = 0 [From (1)] So, \vec{c} is perpendicular to \vec{a} + \vec{b} . Again, \vec{c} . (\vec{a} - \vec{b}) = \vec{c} . \vec{a} - \vec{c} . \vec{b} = 0 - 0 = 0 [From (1)] So, \vec{c} is perpendicular to \vec{a} - \vec{b} .$

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Question 41:

If $|\vec{a}| = a a n d |\vec{b}| = b,$ prove that ${(\frac{\vec{a}}{a^{2}} - \frac{\vec{b}}{b^{2}})}^{2} = {(\frac{\vec{a} - \vec{b}}{a b})}^{2} .$

Answer:

${(\frac{\vec{a}}{a^{2}} - \frac{\vec{b}}{b^{2}})}^{2} = {|\frac{\vec{a}}{a^{2}}|}^{2} + {|\frac{\vec{b}}{b^{2}}|}^{2} - \frac{2 \vec{a} . \vec{b}}{a^{2} b^{2}} = \frac{{|\vec{a}|}^{2}}{a^{4}} + \frac{{|\vec{b}|}^{2}}{b^{4}} - \frac{2 \vec{a} . \vec{b}}{a^{2} b^{2}} = \frac{a^{2}}{a^{4}} + \frac{b^{2}}{b^{4}} - \frac{2 \vec{a} . \vec{b}}{a^{2} b^{2}} (From the given information) = \frac{1}{a^{2}} + \frac{1}{b^{2}} - \frac{2 \vec{a} . \vec{b}}{a^{2} b^{2}} = \frac{b^{2} + a^{2} - 2 \vec{a} . \vec{b}}{a^{2} b^{2}} = \frac{a^{2} + b^{2} - 2 \vec{a} . \vec{b}}{a^{2} b^{2}} = \frac{{|\vec{a}|}^{2} + {|\vec{a}|}^{2} - 2 \vec{a} . \vec{b}}{a^{2} b^{2}} (From the given information) = \frac{{(\vec{a} - \vec{b})}^{2}}{a^{2} b^{2}} = {(\frac{\vec{a} - \vec{b}}{a b})}^{2}$

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Question 42:

If $\vec{a,} \vec{b,} \vec{c}$ are three non-coplanar vectors, such that $\vec{d} \cdot \vec{a} = \vec{d} \cdot \vec{b} = \vec{d} \cdot \vec{c} = 0,$ then show that $\vec{d}$ is the null vector.

Answer:

Given that: $\vec{d} \cdot \vec{a} = 0$
so, either $\vec{d}$ = 0 or $\vec{d} ⊥ \vec{a}$

similarly, $\vec{d} \cdot \vec{b} = 0$
so, $\vec{d}$ = 0 or $\vec{d} ⊥ \vec{b}$

Also, $\vec{d} \cdot \vec{c} = 0$
so, $\vec{d}$ = 0 or $\vec{d} ⊥ \vec{c}$

But $\vec{d}$ cannot be perpendicular to $\vec{a}, \vec{b}, \vec{c}$ as $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar.

so, $\vec{d}$ =0. $\vec{d}$ is a null vector.

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Question 43:

If a vector $\vec{a}$ is perpendicular to two non-collinear vectors $\vec{b} and \vec{c}, then show that \vec{a}$ is perpendicular to every vector in the plane of $\vec{b} and \vec{c} .$

Answer:

$Given that \vec{a} is perpendicular to \vec{b} and \vec{c} . \Rightarrow \vec{a} . \vec{b} = 0 and \vec{a} . \vec{c} =0 ... (1) Now, let \vec{r} be any vector in the plane of \vec{b} and \vec{c} . Then, \vec{r} is the linear combination of \vec{b} and \vec{c} . \vec{r} = x \vec{b} + y \vec{c}, for some x and y . Now, \vec{a} . \vec{r} = \vec{a} . (x \vec{b} + y \vec{c}) = x (\vec{a} . \vec{b}) + y (\vec{a} . \vec{c}) = x (0) + y (0) [From (1)] = 0 Thus, \vec{a} is perpendicular to \vec{r} . That is, \vec{a} is perpendicular to every vector in the plane of \vec{b} and \vec{c} .$

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Question 44:

If $\vec{a} + \vec{b} + \vec{c} = \vec{0},$ show that the angle θ between the vectors $\vec{b} and \vec{c}$ is given by

cos θ = $\frac{{|\vec{a}|}^{2} - {|\vec{b}|}^{2} - {|\vec{c}|}^{2}}{2 |\vec{b}| |\vec{c}|} .$

Answer:

$Given, \vec{a} + \vec{b} + \vec{c} = 0 \Rightarrow \vec{b} + \vec{c} = - \vec{a} \Rightarrow {|\vec{b} + \vec{c}|}^{2} = {|- \vec{a}|}^{2} \Rightarrow {|\vec{b}|}^{2} + {|\vec{c}|}^{2} + 2 \vec{b} . \vec{c} = {|\vec{a}|}^{2} \Rightarrow 2 \vec{b} . \vec{c} = {|\vec{a}|}^{2} - {|\vec{b}|}^{2} - {|\vec{c}|}^{2} \Rightarrow 2 |\vec{b}| |\vec{c}| \cos θ = {|\vec{a}|}^{2} - {|\vec{b}|}^{2} - {|\vec{c}|}^{2} ∴ \cos θ = \frac{{|\vec{a}|}^{2} - {|\vec{b}|}^{2} - {|\vec{c}|}^{2}}{2 |\vec{b}| |\vec{c}|}$

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Question 45:

Let $\vec{u,} \vec{v} and \vec{w}$ be vectors such $\vec{u} + \vec{v} + \vec{w} = \vec{0} .$ If $|\vec{u}| = 3, |\vec{v}| = 4 and |\vec{w}| = 5,$ then find $\vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u} .$

Answer:

$Given that \vec{u} + \vec{v} + \vec{w} = 0 \Rightarrow |\vec{u} + \vec{v} + \vec{w}| = 0 \Rightarrow {|\vec{u} + \vec{v} + \vec{w}|}^{2} = 0 \Rightarrow {|\vec{u}|}^{2} + {|\vec{v}|}^{2} + {|\vec{w}|}^{2} + 2 (\vec{u} . \vec{v} + \vec{v} . \vec{w} + \vec{w} . \vec{u}) = 0 \Rightarrow 3^{2} + 4^{2} + 5^{2} + 2 (\vec{u} . \vec{v} + \vec{v} . \vec{w} + \vec{w} . \vec{u}) = 0 (Given : |\vec{u}| = 3, |\vec{v}| = 4 and |\vec{w}| = 5) \Rightarrow 9 + 16 + 25 + 2 (\vec{u} . \vec{v} + \vec{v} . \vec{w} + \vec{w} . \vec{u}) = 0 \Rightarrow 50 + 2 (\vec{u} . \vec{v} + \vec{v} . \vec{w} + \vec{w} . \vec{u}) = 0 \Rightarrow 2 (\vec{u} . \vec{v} + \vec{v} . \vec{w} + \vec{w} . \vec{u}) = - 50 ∴ \vec{u} . \vec{v} + \vec{v} . \vec{w} + \vec{w} . \vec{u} = \frac{- 50}{2} = - 25$

Page No 23.32:

Question 46:

Let $\vec{a} = x^{2} \hat{i} + 2 \hat{j} - 2 \hat{k}, \vec{b} = \hat{i} - \hat{j} + \hat{k} and \vec{c} = x^{2} \hat{i} + 5 \hat{j} - 4 \hat{k}$ be three vectors. Find the values of x for which the angle between $\vec{a} and \vec{b}$ is acute and the angle between $\vec{b} and \vec{c}$ is obtuse.

Answer:

$We have \vec{a} = = x^{2} \overset{⏜}{i} + 2 \hat{j} - 2 \hat{k}, \vec{b} = \hat{i} - \hat{j} + \hat{k} and \vec{c} = x^{2} \hat{i} + 5 \hat{j} - 4 \hat{k} {Let θ}_{1} be the angle between \vec{a} and \vec{b} {and θ}_{2} be the angle between \vec{b} and \vec{c} . {Given that θ}_{1} {is acute and θ}_{2} is obtuse. \Rightarrow \cos θ_{1} > 0 and \cos θ_{2} < 0 \Rightarrow \frac{\vec{a} . \vec{b}}{|\vec{a}| . |\vec{b}|} > o and \frac{\vec{b} . \vec{c}}{|\vec{b}| . \vec{|c|}} < 0 \Rightarrow \frac{x^{2} - 4}{\sqrt{x^{4} + 4 + 4} \sqrt{1 + 1 + 1}} > 0 and \frac{x^{2} - 9}{\sqrt{1 + 1 + 1} \sqrt{x^{4} + 25 + 16}} < 0 \Rightarrow x^{2} - 4 > 0 and x^{2} - 9 < 0 \Rightarrow x \in (- \infty, - 2) \cup (2, \infty) and x \in (- 3, 3) \Rightarrow x \in (- 3, - 2) \cup (2, 3)$

Page No 23.32:

Question 47:

Find the values of x and y if the vectors $\vec{a} = 3 \hat{i} + x \hat{j} - \hat{k} and \vec{b} = 2 \hat{i} + \hat{j} + y \hat{k}$ are mutually perpendicular vectors of equal magnitude.

Answer:

$We have \vec{a} = 3 \hat{i} + x \hat{j} - \hat{k} and \vec{b} = 2 \hat{i} + \hat{j} + y \hat{k} It is given that the vectors are perpendicular. \Rightarrow \vec{a} . \vec{b} = 0 \Rightarrow 6 + x - y = 0 \Rightarrow x - y = - 6 . . . (1) Also, it is given that |\vec{a}| = |\vec{b}| \Rightarrow \sqrt{9 + x^{2} + 1} = \sqrt{4 + 1 + y^{2}} \Rightarrow \sqrt{10 + x^{2}} = \sqrt{5 + y^{2}} \Rightarrow 10 + x^{2} = 5 + y^{2} \Rightarrow x^{2} - y^{2} = - 5 \Rightarrow (x + y) (x - y) = - 5 \Rightarrow - 6 (x + y) = - 5 [Using (1)] \Rightarrow x + y = \frac{5}{6} . . . (2) (1)+(2) gives 2 x = \frac{- 31}{6} \Rightarrow x = \frac{- 31}{12} From (1), \frac{- 31}{12} - y = - 6 \Rightarrow y = \frac{- 31}{12} + 6 = \frac{41}{12} ∴ x = \frac{- 31}{12} and y = \frac{41}{12}$

Page No 23.32:

Question 48:

If $\vec{a} and \vec{b}$ are two non-collinear unit vectors such that $|\vec{a} + \vec{b}| = \sqrt{3},$ find $(2 \vec{a} - 5 \vec{b}) \cdot (3 \vec{a} + \vec{b}) .$

Answer:

$We have |\vec{a} + \vec{b}| = \sqrt{3} Squaring both sides, we get {|\vec{a} + \vec{b}|}^{2} = 3 \Rightarrow {|\vec{a}|}^{2} + {|\vec{b}|}^{2} + 2 \vec{a} . \vec{b} = 3 \Rightarrow 1 + 1 + 2 \vec{a} . \vec{b} = 3 (Because \vec{a} and \vec{b} are unit vectors) \Rightarrow 2 + 2 \vec{a} . \vec{b} = 3 \Rightarrow 2 \vec{a} . \vec{b} = 1 \Rightarrow 2 \vec{a} . \vec{b} = 1 \Rightarrow \vec{a} . \vec{b} = \frac{1}{2} . . . (1) Now, (2 \vec{a} - 5 \vec{b}) . (3 \vec{a} + \vec{b}) = 6 {|\vec{a}|}^{2} + 2 \vec{a} . \vec{b} - 15 \vec{b} . \vec{a} - 5 {|\vec{b}|}^{2} = 6 {|\vec{a}|}^{2} + 2 \vec{a} . \vec{b} - 15 \vec{a} . \vec{b} - 5 {|\vec{b}|}^{2} (\vec{a} . \vec{b} = \vec{b} . \vec{a)} = 6 {|\vec{a}|}^{2} - 13 \vec{a} . \vec{b} - 5 {|\vec{b}|}^{2} = 6 (1) - 13 (\frac{1}{2}) - 5 (1) [From (1)] = 1 - \frac{13}{2} = \frac{- 11}{2}$

Page No 23.33:

Question 49:

If $\vec{a}$ , $\vec{b}$ are two vectors such that $|\vec{a} + \vec{b}| = |\vec{b}|$ , then prove that $\vec{a} + 2 \vec{b}$ is perpendicular to $\vec{a}$ .

Answer:

$Given that |\vec{a} + \vec{b}| = |\vec{b}| Squaring both sides, we get {|\vec{a} + \vec{b}|}^{2} = {|\vec{b}|}^{2} \Rightarrow {|\vec{a}|}^{2} + {|\vec{b}|}^{2} + 2 \vec{a} . \vec{b} = {|\vec{b}|}^{2} \Rightarrow {|\vec{a}|}^{2} + 2 \vec{a} . \vec{b} = 0 . . . (1) Now, (\vec{a} + 2 \vec{b}) . \vec{a} = \vec{a} . \vec{a} + 2 \vec{b} . \vec{a} = {|\vec{a}|}^{2} + 2 \vec{a} . \vec{b} = 0 [Using (1)] So, \vec{a} + 2 \vec{b} is perpendicular to \vec{a} .$

Page No 23.46:

Question 1:

In a triangle OAB, $∠$ AOB = 90º. If P and Q are points of trisection of AB, prove that ${OP}^{2} + {OQ}^{2} = \frac{5}{9} {AB}^{2}$ .

Answer:

In triangle OAB, $∠$ AOB = 90º. P and Q are points of trisection of AB.

Taking O as the origin, let the position vectors of A and B be $\vec{a}$ and $\vec{b}$ , respectively.

Since P and Q are the points of trisection of AB, so AP : PB = 1 : 2 and AQ : QB = 2 : 1.

Position vector of P, $\vec{OP} = \frac{2 \vec{a} + \vec{b}}{3}$ (Using section formula)
Position vector of Q, $\vec{OQ} = \frac{\vec{a} + 2 \vec{b}}{3}$

Also, $\vec{OA} ⊥ \vec{OB}$ .

$∴ \vec{a} . \vec{b} = 0$ .....(1)

Now,

${OP}^{2} + {OQ}^{2} = {|\vec{OP}|}^{2} + {|\vec{OQ}|}^{2} = (\frac{2 \vec{a} + \vec{b}}{3}) . (\frac{2 \vec{a} + \vec{b}}{3}) + (\frac{\vec{a} + 2 \vec{b}}{3}) . (\frac{\vec{a} + 2 \vec{b}}{3}) = \frac{4 {|\vec{a}|}^{2} + 4 \vec{a} . \vec{b} + {|\vec{b}|}^{2} + {|\vec{a}|}^{2} + 4 \vec{a} . \vec{b} + 4 {|\vec{b}|}^{2}}{9}$
$= \frac{5 {|\vec{a}|}^{2} + 5 {|\vec{b}|}^{2}}{9} [Using (1)] = \frac{5}{9} ({|\vec{a}|}^{2} + {|\vec{b}|}^{2}) = \frac{5}{9} {|\vec{AB}|}^{2} [Using Pythagoras Theorem] = \frac{5}{9} {AB}^{2}$

Page No 23.46:

Question 2:

Prove that: If the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Answer:

Let OACB be a quadrilateral such that diagonals OC and AB bisect each other at 90º.

Taking O as the origin, let the poisition vectors of A and B be $\vec{a}$ and $\vec{b}$ , respectively. Then,

$\vec{OA} = \vec{a}$ and $\vec{OB} = \vec{b}$
Position vector of mid-point of AB, $\vec{OE} = \frac{\vec{a} + \vec{b}}{2}$

∴ Position vector of C, $\vec{OC} = \vec{a} + \vec{b}$

By the triangle law of vector addition, we have

$\vec{OA} + \vec{AB} = \vec{OB} \Rightarrow \vec{AB} = \vec{OB} - \vec{OA} = \vec{b} - \vec{a}$

Since $\vec{AB} ⊥ \vec{OC}$ ,

$\Rightarrow \vec{AB} . \vec{OC} = 0 \Rightarrow (\vec{b} - \vec{a}) . (\vec{a} + \vec{b}) = 0 \Rightarrow {|\vec{b}|}^{2} - {|\vec{a}|}^{2} = 0 \Rightarrow {|\vec{a}|}^{2} = {|\vec{b}|}^{2} \Rightarrow |\vec{a}| = |\vec{b}| \Rightarrow OA = OB$

In a quadrilateral if diagonals bisects each other at right angle and adjacent sides are equal, then it is a rhombus.

Page No 23.46:

Question 3:

(Pythagoras's Theorem) Prove by vector method that in a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Answer:

Let ABC be a right triangle with $∠$ BAC = 90º. Taking A as the origin, let the position vectors of B and C be $\vec{b}$ and $\vec{c}$ , respectively. Then,

$\vec{AB} = \vec{b}$ and $\vec{AC} = \vec{c}$

Since $\vec{AB} ⊥ \vec{AC}$ $\Rightarrow \vec{b} . \vec{c} = 0$ .....(1)

Now,

${|\vec{AB}|}^{2} + {|\vec{AC}|}^{2} = {|\vec{b}|}^{2} + {|\vec{c}|}^{2}$ .....(2)

Also,

${|\vec{BC}|}^{2} = {|\vec{c} - \vec{b}|}^{2} = (\vec{c} - \vec{b}) . (\vec{c} - \vec{b}) = {|\vec{c}|}^{2} - 2 \vec{b} . \vec{c} + {|\vec{b}|}^{2} = {|\vec{c}|}^{2} + {|\vec{b}|}^{2} . . . . . (3) [Using (1)]$

From (2) and (3), we have

${|\vec{AB}|}^{2} + {|\vec{AC}|}^{2} = {|\vec{BC}|}^{2}$

Page No 23.46:

Question 4:

Prove by vector method that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

Answer:

Let ABCD be a parallelogram such that AC and BD are its two diagonals. Taking A as the origin, let the position vectors of B and D be $\vec{b}$ and $\vec{d}$ , respectively. Then,

$\vec{AB} = \vec{b}$ and $\vec{AD} = \vec{d}$

Using triangle law of vector addition, we have

$\vec{AD} + \vec{DB} = \vec{AB} \Rightarrow \vec{DB} = \vec{b} - \vec{d}$

In ∆ABC,

$\vec{AC} = \vec{AB} + \vec{BC} = \vec{AB} + \vec{AD} = \vec{b} + \vec{d}$

Now,

${|\vec{AB}|}^{2} + {|\vec{BC}|}^{2} + {|\vec{CD}|}^{2} + {|\vec{DA}|}^{2} = {|\vec{AB}|}^{2} + {|\vec{AD}|}^{2} + {|- \vec{AB}|}^{2} + {|- \vec{AD}|}^{2} = 2 {|\vec{AB}|}^{2} + 2 {|\vec{AD}|}^{2} = 2 {|\vec{b}|}^{2} + 2 {|\vec{d}|}^{2} . . . . . (1)$

Also,

${|\vec{DB}|}^{2} + {|\vec{AC}|}^{2} = {|\vec{b} - \vec{d}|}^{2} + {|\vec{b} + \vec{d}|}^{2} = (\vec{b} - \vec{d}) . (\vec{b} - \vec{d}) + (\vec{b} + \vec{d}) . (\vec{b} + \vec{d}) = {|\vec{b}|}^{2} - 2 \vec{b} . \vec{d} + {|\vec{d}|}^{2} + {|\vec{b}|}^{2} + 2 \vec{b} . \vec{d} + {|\vec{d}|}^{2} = 2 {|\vec{b}|}^{2} + 2 {|\vec{d}|}^{2} . . . . . (2)$

From (1) and (2), we have

${|\vec{AB}|}^{2} + {|\vec{BC}|}^{2} + {|\vec{CD}|}^{2} + {|\vec{DA}|}^{2} = {|\vec{DB}|}^{2} + {|\vec{AC}|}^{2}$

Page No 23.46:

Question 5:

Prove using vectors: The quadrilateral obtained by joining mid-points of adjacent sides of a rectangle is a rhombus.

Answer:

ABCD is a rectangle. Let P, Q, R and S be the mid-points of the sides AB, BC, CD and DA, respectively.

Now,

$\vec{PQ} = \vec{PB} + \vec{BQ} = \frac{1}{2} \vec{AB} + \frac{1}{2} \vec{BC} = \frac{1}{2} (\vec{AB} + \vec{BC}) = \frac{1}{2} \vec{AC}$ .....(1)

$\vec{SR} = \vec{SD} + \vec{DR} = \frac{1}{2} \vec{AD} + \frac{1}{2} \vec{DC} = \frac{1}{2} (\vec{AD} + \vec{DC}) = \frac{1}{2} \vec{AC}$ .....(2)

From (1) and (2), we have

$\vec{PQ} = \vec{SR}$

So, the sides PQ and SR are equal and parallel. Thus, PQRS is a parallelogram.

Now,

${|\vec{PQ}|}^{2} = \vec{PQ} . \vec{PQ} \Rightarrow {|\vec{PQ}|}^{2} = (\vec{PB} + \vec{BQ}) . (\vec{PB} + \vec{BQ}) \Rightarrow {|\vec{PQ}|}^{2} = {|\vec{PB}|}^{2} + 2 \vec{PB} . \vec{BQ} + {|\vec{BQ}|}^{2} \Rightarrow {|\vec{PQ}|}^{2} = {|\vec{PB}|}^{2} + 0 + {|\vec{BQ}|}^{2} (\vec{PB} ⊥ \vec{BQ}) \Rightarrow {|\vec{PQ}|}^{2} = {|\vec{PB}|}^{2} + {|\vec{BQ}|}^{2} . . . . . (3)$

Also,

${|\vec{PS}|}^{2} = \vec{PS} . \vec{PS} \Rightarrow {|\vec{PS}|}^{2} = (\vec{PA} + \vec{AS}) . (\vec{PA} + \vec{AS}) \Rightarrow {|\vec{PS}|}^{2} = {|\vec{PA}|}^{2} + 2 \vec{PA} . \vec{AS} + {|\vec{AS}|}^{2} \Rightarrow {|\vec{PS}|}^{2} = {|\vec{PB}|}^{2} + 0 + {|\vec{BQ}|}^{2} (\vec{PA} ⊥ \vec{AS}) \Rightarrow {|\vec{PS}|}^{2} = {|\vec{PB}|}^{2} + {|\vec{BQ}|}^{2} . . . . . (4)$

From (3) and (4), we have

${|\vec{PQ}|}^{2} = {|\vec{PS}|}^{2} \Rightarrow |\vec{PQ}| = |\vec{PS}|$

So, the adjacent sides of the parallelogram are equal. Hence, PQRS is a rhombus.

Page No 23.46:

Question 6:

Prove that the diagonals of a rhombus are perpendicular bisectors of each other.

Answer:

Let OABC be a rhombus, whose diagonals OB and AC intersect at D. Suppose O is the origin.

Let the position vector of A and C be $\vec{a}$ and $\vec{c}$ , respectively. Then,

$\vec{OA} = \vec{a}$ and $\vec{OC} = \vec{c}$

In ∆OAB,

$\vec{OB} = \vec{OA} + \vec{AB} = \vec{OA} + \vec{OC} = \vec{a} + \vec{c}$ $(\vec{AB} = \vec{OC})$

Position vector of mid-point of $\vec{OB} = \frac{1}{2} (\vec{a} + \vec{c})$

Position vector of mid-point of $\vec{AC} = \frac{1}{2} (\vec{a} + \vec{c})$ (Mid-point formula)

So, the mid-points of OB and AC coincide. Thus, the diagonals OB and AC bisect each other.

Now,

$\vec{OB} . \vec{AC} = (\vec{a} + \vec{c}) . (\vec{c} - \vec{a}) = (\vec{c} + \vec{a}) . (\vec{c} - \vec{a}) = {|\vec{c}|}^{2} - {|\vec{a}|}^{2} = {|\vec{OC}|}^{2} - {|\vec{OA}|}^{2} = 0 (|\vec{OC}| = |\vec{OA}|) \Rightarrow \vec{OB} ⊥ \vec{AC}$

Hence, the diagonals OB and AC are perpendicular to each other.

Thus, the diagonals of a rhombus are perpendicular bisectors of each other.

Page No 23.46:

Question 7:

Prove that the diagonals of a rectangle are perpendicular if and only if the rectangle is a square.

Answer:

Let ABCD be a rectangle. Take A as the origin.

Suppose the position vectors of points B and D be $\vec{a}$ and $\vec{b}$ , respectively.

Now,

$\vec{AC} = \vec{AB} + \vec{BC} = \vec{AB} + \vec{AD} = \vec{a} + \vec{b}$

Also, $\vec{BD} = \vec{a} - \vec{b}$

Since ABCD is rectangle, so $\vec{AB} ⊥ \vec{AD}$ .

$∴ \vec{a} . \vec{b} = 0$ .....(1)

Now, diagonals AC and BD are perpendicular

iff $\vec{AC} . \vec{BD} = 0$

$iff (\vec{a} + \vec{b}) . (\vec{a} - \vec{b}) = 0 iff {|\vec{a}|}^{2} - {|\vec{b}|}^{2} = 0 iff |\vec{a}| = |\vec{b}| iff |\vec{AB}| = |\vec{AD}|$

iff ABCD is a square

Thus, the diagonals of a rectangle are perpendicular if and only if the rectangle is a square.

Page No 23.46:

Question 8:

If AD is the median of ∆ABC, using vectors, prove that ${AB}^{2} + {AC}^{2} = 2 ({AD}^{2} + {CD}^{2})$ .

Answer:

Taking A as the origin, let the position vectors of B and C be $\vec{b}$ and $\vec{c}$ , respectively.

It is given that AD is the median of ∆ABC.

∴ Position vector of mid-point of BC = $\vec{AD} = \frac{\vec{b} + \vec{c}}{2}$ (Mid-point formula)

Now,

${AB}^{2} + {AC}^{2} = {|\vec{AB}|}^{2} + {|\vec{AC}|}^{2} = {|\vec{b}|}^{2} + {|\vec{c}|}^{2}$ .....(1)

Also,

$2 ({AD}^{2} + {CD}^{2}) = 2 ({|\vec{AD}|}^{2} + {|\vec{CD}|}^{2}) = 2 [(\frac{\vec{b} + \vec{c}}{2}) . (\frac{\vec{b} + \vec{c}}{2}) + (\frac{\vec{b} + \vec{c}}{2} - \vec{c}) . (\frac{\vec{b} + \vec{c}}{2} - \vec{c})] = 2 [(\frac{\vec{b} + \vec{c}}{2}) . (\frac{\vec{b} + \vec{c}}{2}) + (\frac{\vec{b} - \vec{c}}{2}) . (\frac{\vec{b} - \vec{c}}{2})] = \frac{{|\vec{b}|}^{2} + 2 \vec{b} . \vec{c} + {|\vec{c}|}^{2}}{2} + \frac{{|\vec{b}|}^{2} - 2 \vec{b} . \vec{c} + {|\vec{c}|}^{2}}{2} = \frac{2 {|\vec{b}|}^{2} + 2 {|\vec{c}|}^{2}}{2} = {|\vec{b}|}^{2} + {|\vec{c}|}^{2} . . . . . (2)$

From (1) and (2), we have

${AB}^{2} + {AC}^{2} = 2 ({AD}^{2} + {CD}^{2})$

Page No 23.46:

Question 9:

If the median to the base of a triangle is perpendicular to the base, then triangle is isosceles.

Answer:

Let ∆ABC be a triangle such that AD is the median. Taking A as the origin, let the position vectors of B and C be $\vec{b}$ and $\vec{c}$ , respectively. Then,

Position vector of D = $\frac{\vec{b} + \vec{c}}{2}$ (Mid-point formula)

Now,

$\vec{AD}$ = Position vector of D − Position vector of A = $\frac{\vec{b} + \vec{c}}{2}$

$\vec{BC}$ = Position vector of C − Position vector of B = $\vec{c} - \vec{b}$

Since $\vec{AD} ⊥ \vec{BC}$ ,

$∴ \vec{AD} . \vec{BC} = 0 \Rightarrow \frac{1}{2} (\vec{b} + \vec{c}) . (\vec{c} - \vec{b}) = 0 \Rightarrow (\vec{c} + \vec{b}) . (\vec{c} - \vec{b}) = 0 \Rightarrow {|\vec{c}|}^{2} - {|\vec{b}|}^{2} = 0 \Rightarrow |\vec{c}| = |\vec{b}| \Rightarrow AC = AB$

Hence, the ∆ABC is an isosceles triangle.

Page No 23.46:

Question 10:

In a quadrilateral ABCD, prove that ${AB}^{2} + {BC}^{2} + {CD}^{2} + {DA}^{2} = {AC}^{2} + {BD}^{2} + 4 {PQ}^{2}$ , where P and Q are middle points of diagonals AC and BD.

Answer:

Let ABCD be the quadrilateral. Taking A as the origin, let the position vectors of B, C and D be $\vec{b}, \vec{c}$ and $\vec{d}$ , respectively. Then,

Position vector of P = $\frac{\vec{c}}{2}$ (Mid-point formula)

Position vector of Q = $\frac{\vec{b} + \vec{d}}{2}$ (Mid-point formula)

Now,

${AB}^{2} + {BC}^{2} + {CD}^{2} + {DA}^{2} = {|\vec{AB}|}^{2} + {|\vec{BC}|}^{2} + {|\vec{CD}|}^{2} + {|\vec{DA}|}^{2} = {|\vec{b}|}^{2} + {|\vec{c} - \vec{b}|}^{2} + {|\vec{d} - \vec{c}|}^{2} + {|\vec{d}|}^{2} = {|\vec{b}|}^{2} + {|\vec{c}|}^{2} - 2 \vec{c} . \vec{b} + {|\vec{b}|}^{2} + {|\vec{d}|}^{2} - 2 \vec{d} . \vec{c} + {|\vec{c}|}^{2} + |\vec{d}| = 2 {|\vec{b}|}^{2} + 2 {|\vec{c}|}^{2} + 2 {|\vec{d}|}^{2} - 2 \vec{b} . \vec{c} - 2 \vec{c} . \vec{d} . . . . . (1)$

Also,

${AC}^{2} + {BD}^{2} + 4 {PQ}^{2} = {|\vec{AC}|}^{2} + {|\vec{BD}|}^{2} + 4 {|\vec{PQ}|}^{2} = {|\vec{c}|}^{2} + {|\vec{d} - \vec{b}|}^{2} + 4 {|\frac{\vec{b} + \vec{d}}{2} - \frac{\vec{c}}{2}|}^{2} = {|\vec{c}|}^{2} + {|\vec{d} - \vec{b}|}^{2} + {|\vec{b} + \vec{d}|}^{2} - 2 (\vec{b} + \vec{d}) . \vec{c} + {|\vec{c}|}^{2} = 2 {|\vec{c}|}^{2} + 2 {|\vec{d}|}^{2} + 2 {|\vec{b}|}^{2} - 2 \vec{b} . \vec{c} - 2 \vec{d} . \vec{c} = 2 {|\vec{b}|}^{2} + 2 {|\vec{c}|}^{2} + 2 {|\vec{d}|}^{2} - 2 \vec{b} . \vec{c} - 2 \vec{c} . \vec{d} . . . . . (2)$

From (1) and (2), we have

${AB}^{2} + {BC}^{2} + {CD}^{2} + {DA}^{2} = {AC}^{2} + {BD}^{2} + 4 {PQ}^{2}$

Page No 23.46:

Question 1:

The vectors $\vec{a} and \vec{b}$ satisfy the equations $2 \vec{a} + \vec{b} = \vec{p} and \vec{a} + 2 \vec{b} = \vec{q}, where \vec{p} = \hat{i} + \hat{j} and \vec{q} = \hat{i} - \hat{j} .$ If θ is the angle between $\vec{a} and \vec{b}$ , then
(a) $\cos θ = \frac{4}{5}$

(b) $\sin θ = \frac{1}{\sqrt{2}}$

(c) $\cos θ = - \frac{4}{5}$

(d) $\cos θ = - \frac{3}{5}$

Answer:

(c) $\cos θ = - \frac{4}{5}$

$Given that 2 \vec{a} + \vec{b} = \vec{p} . . . (1) \vec{a} + 2 \vec{b} = \vec{q} . . . (2) Solving these two we get \vec{a} = \frac{2 \vec{p} - \vec{q}}{3}, \vec{b} = \frac{2 \vec{q} - \vec{p}}{3} And we have \vec{p} = \hat{i} + \hat{j} and \vec{q} = \hat{i} - \hat{j} Substituting the values of \vec{p} and \vec{q,} we get \vec{a} = \frac{2 \vec{p} - \vec{q}}{3} = \frac{2 (\hat{i} + \hat{j}) - (\hat{i} - \hat{j})}{3} = \frac{\hat{i} + 3 \hat{j}}{3} \Rightarrow |\vec{a}| = \frac{1}{3} \sqrt{1 + 9} = \frac{\sqrt{10}}{3} \vec{b} = \frac{2 \vec{q} - \vec{p}}{3} = \frac{2 (\hat{i} - \hat{j}) - (\hat{i} + \hat{j})}{3} = \frac{\hat{i} - 3 \hat{j}}{3} \Rightarrow |\vec{b}| = \frac{1}{3} \sqrt{1 + 9} = \frac{\sqrt{10}}{3} \vec{a} . \vec{b} = \frac{1}{9} (1 - 9) = \frac{- 8}{9} We know that \vec{a} . \vec{b} = |\vec{a}| |\vec{b}| \cos θ \Rightarrow \frac{- 8}{9} = \frac{\sqrt{10}}{3} \times \frac{\sqrt{10}}{3} \cos θ \Rightarrow \frac{- 8}{9} = \frac{10}{9} \cos θ \Rightarrow \cos θ = \frac{- 8}{9} \times \frac{9}{10} = \frac{- 4}{5}$

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Question 2:

If $\vec{a} \cdot \hat{i} = \vec{a} \cdot (\hat{i} + \hat{j}) = \vec{a} \cdot (\hat{i} + \hat{j} + \hat{k}) = 1,$ then $\vec{a} =$
(a) $\vec{0}$

(b) $\hat{i}$

(c) $\hat{j}$

(d) $\hat{i} + \hat{j} + \hat{k}$

Answer:

(b) $\hat{i}$

$Let \vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k} \vec{a} . \hat{i} = a_{1} and \vec{a} . (\hat{i} + \hat{j}) = a_{1} + a_{2} and \vec{a} . (\hat{i} + \hat{j} + \hat{k}) = a_{1} + a_{2} + a_{3} Given, \vec{a} . \hat{i} = \vec{a} . (\hat{i} + \hat{j}) = \vec{a} . (\hat{i} + \hat{j} + \hat{k}) = 1 \Rightarrow a_{1} = a_{1} + a_{2} = a_{1} + a_{2} + a_{3} = 1 \Rightarrow a_{1} = 1, a_{2} = 0, a_{3} = 0 So, \vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k} = 1 \hat{i} + 0 \hat{j} + 0 \hat{k} = \hat{i}$

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Question 3:

If $\vec{a} + \vec{b} + \vec{c} = \vec{0}, |\vec{a}| = 3, |\vec{b}| = 5, |\vec{c}| = 7,$ then the angle between $\vec{a} and \vec{b}$ is
(a) $\frac{π}{6}$

(b) $\frac{2 π}{3}$

(c) $\frac{5 π}{3}$

(d) $\frac{π}{3}$

Answer:

(d) $\frac{π}{3}$

$Given, |\vec{a}| = 3, |\vec{b}| = 5 and |\vec{c}| = 7 . . . (i) Let θ be the angle between \vec{a} and \vec{b} . Given that \vec{a} + \vec{b} + \vec{c} = 0 \Rightarrow \vec{a} + \vec{b} = - \vec{c} \Rightarrow |\vec{a} + \vec{b}| = {|- \vec{c}|}^{2} \Rightarrow {|\vec{a}|}^{2} + {|\vec{b}|}^{2} + 2 \vec{a} . \vec{b} = {|\vec{c}|}^{2} \Rightarrow 2 \vec{a} . \vec{b} = {|\vec{c}|}^{2} - {|\vec{a}|}^{2} - {|\vec{b}|}^{2} \Rightarrow 2 \vec{a} . \vec{b} = 7^{2} - 3^{2} - 5^{2} [Using (i)] \Rightarrow 2 \vec{a} . \vec{b} = 15 \Rightarrow 2 |\vec{a}| |\vec{b}| \cos θ = 15 \Rightarrow 2 (3) (5) \cos θ = 15 [Using (i)] \Rightarrow \cos θ = \frac{1}{2} ∴ θ = \frac{π}{3}$

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Question 4:

Let $\vec{a} and \vec{b}$ be two unit vectors and α be the angle between them. Then, $\vec{a} + \vec{b}$ is a unit vector if
(a) $α = \frac{π}{4}$

(b) $α = \frac{π}{3}$

(c) $α = \frac{2 π}{3}$

(d) $α = \frac{π}{2}$

Answer:

(c) $α = \frac{2 π}{3}$

$\vec{a} and \vec{b} are unit vectors . \Rightarrow |\vec{a}| = |\vec{b}| = 1 . . . (1) Now, \vec{a} . \vec{b} = |\vec{a}| |\vec{b}| \cos α \Rightarrow \vec{a} . \vec{b} = \cos α . . . (2) [Using (1)] Given that |\vec{a} + \vec{b}| = 1 Squaring both sides, we get {|\vec{a} + \vec{b}|}^{2} = 1 \Rightarrow {|\vec{a}|}^{2} + {|\vec{b}|}^{2} + 2 \vec{a} . \vec{b} = 1 \Rightarrow 1 + 1 + 2 \cos α = 1 [Using (1) and (2)] \Rightarrow 2 + 2 \cos α = 1 \Rightarrow 2 \cos α = - 1 \Rightarrow 2 \cos α = - 1 \Rightarrow \cos α = \frac{- 1}{2} \Rightarrow α = \frac{2 π}{3}$

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Question 5:

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

$Let \vec{a} = (c o s α c o s β) \hat{i} + (c o s α s i n β) \hat{j} + (s i n α) \hat{k} |\vec{a}| = \sqrt{c o s^{2} α c o s^{2} β + c o s^{2} α s i n^{2} β + s i n^{2} α} = \sqrt{c o s^{2} α (c o s^{2} β + s i n^{2} β) + s i n^{2} α} = \sqrt{c o s^{2} α (1) + s i n^{2} α} = \sqrt{c o s^{2} α + s i n^{2} α} = \sqrt{1} = 1 So, \vec{a} is a unit vector.$

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Question 6:

If the position vectors of P and Q are $\hat{i} + 3 \hat{j} - 7 \hat{k} and 5 \hat{i} - 2 \hat{j} + 4 \hat{k}$ then the cosine of the angle between $\vec{P Q}$ and y-axis is
(a) $\frac{5}{\sqrt{162}}$

(b) $\frac{4}{\sqrt{162}}$

(c) $- \frac{5}{\sqrt{162}}$

(d) $\frac{11}{\sqrt{162}}$

Answer:

(c) $- \frac{5}{\sqrt{162}}$

$\vec{P Q} = \vec{O Q} - \vec{O P} = 5 \hat{i} - 2 \hat{j} + 4 \hat{k} - (\hat{i} + 3 \hat{j} - 7 \hat{k}) = 4 \hat{i} - 5 \hat{j} + 11 \hat{k} The unit vector along y- axis is \hat{j} . Let θ be the required angle . \cos θ = \frac{\vec{P Q} . \hat{j}}{|\vec{P Q}| |\hat{j}|} = \frac{- 5}{\sqrt{16 + 25 + 121} \sqrt{1}} = \frac{- 5}{\sqrt{162}}$

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Question 7:

If $\vec{a} and \vec{b}$ are unit vectors, then which of the following values of $\vec{a} . \vec{b}$ is not possible?
(a) $\sqrt{3}$

(b) $\sqrt{3} / 2$

(c) $1 / \sqrt{2}$

(d) −1/2

Answer:

(a) $\sqrt{3}$

$It is given that \vec{a} and \vec{b} are unit vectors. \Rightarrow |\vec{a}| = |\vec{b}| = 1 Now, \vec{a} . \vec{b} = |\vec{a}| |\vec{b}| \cos θ = (1) (1) \cos θ = \cos θ The range of \cos θ is [- 1, 1] . ∴ \sqrt{3} is not a possible value of \cos θ as it is greater than 1 .$

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Question 8:

If the vectors $\hat{i} - 2 x \hat{j} + 3 y \hat{k} and \hat{i} + 2 x \hat{j} - 3 y \hat{k}$ are perpendicular, then the locus of (x, y) is
(a) a circle
(b) an ellipse
(c) a hyperbola
(d) None of these

Answer:

(b) an ellipse

$Let, \vec{a} = \hat{i} - 2 x \hat{j} + 3 y \hat{k} and \vec{b} = \hat{i} + 2 x \hat{j} - 3 y \hat{k} It is given that the vectors are perpendicular. So, their dot product is zero. \vec{a} . \vec{b} = 0 \Rightarrow (\hat{i} - 2 x \hat{j} + 3 y \hat{k}) . (\hat{i} + 2 x \hat{j} - 3 y \hat{k}) = 0 \Rightarrow 1 - 4 x^{2} - 9 y^{2} = 0 \Rightarrow 4 x^{2} + 9 y^{2} = 1 Dividing both sides by 36, we get \frac{x^{2}}{9} + \frac{y^{2}}{4} = 1 This is an ellipse.$

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Question 9:

The vector component of $\vec{b}$ perpendicular to $\vec{a}$ is
(a) $(\vec{b} . \vec{c}) \vec{a}$

(b) $\frac{\vec{a} \times (\vec{b} \times \vec{a})}{{|\vec{a}|}^{2}}$

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

(d) None of these

Answer:

(b) $\frac{\vec{a} \times (\vec{b} \times \vec{a})}{{|\vec{a}|}^{2}}$

$The vector component of \vec{b} perpendicular to \vec{a} is \frac{\vec{a} \times (\vec{b} \times \vec{a})}{{|\vec{a}|}^{2}}$

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Question 10:

What is the length of the longer diagonal of the parallelogram constructed on $5 \vec{a} + 2 \vec{b} and \vec{a} - 3 \vec{b}$ if it is given that $|\vec{a}| = 2 \sqrt{2}, |\vec{b}| = 3$ and the angle between $\vec{a} and \vec{b}$ is π/4?
(a) 15
(b) $\sqrt{113}$
(c) $\sqrt{593}$
(d) $\sqrt{369}$

Answer:

(c) $\sqrt{593}$

$Let A B C D be a parallelogram in which side \vec{A B} = \vec{D C} = 5 \vec{a} + 2 \vec{b} and \vec{A D} = \vec{B C} = \vec{a} - 3 \vec{b} a nd diagonals are A C and B D . Now, \vec{A C} = \vec{A B} + \vec{B C} = (5 \vec{a} + 2 \vec{b}) + (\vec{a} - 3 \vec{b}) = 6 \vec{a} - \vec{b} ∴ |\vec{A C}| = |6 \vec{a} - \vec{b}| = \sqrt{{|6 \vec{a}|}^{2} + {|\vec{b}|}^{2} - 2 \times |6 \vec{a}| \times |\vec{b}| \cos θ} = \sqrt{36 {|\vec{a}|}^{2} + {|\vec{b}|}^{2} - 12 \times |\vec{a}| \times |\vec{b}| \cos \frac{π}{4}} = \sqrt{36 {|2 \sqrt{2}|}^{2} + {|3|}^{2} - 12 \times |2 \sqrt{2}| \times |3| \times \frac{1}{\sqrt{2}}} = \sqrt{288 + 9 - 72} = \sqrt{225} = 15 units \vec{B D} = \vec{B A} + \vec{B D} = - \vec{A B} + \vec{B D} = - (5 \vec{a} + 2 \vec{b}) + (\vec{a} - 3 \vec{b}) = - 4 \vec{a} - 5 \vec{b} ∴ |\vec{B D}| = |- 4 \vec{a} - 5 \vec{b}| = |4 \vec{a} + 5 \vec{b}| = \sqrt{{|4 \vec{a}|}^{2} + {|5 \vec{b}|}^{2} + 2 \times |4 \vec{a}| \times |5 \vec{b}| \cos θ} = \sqrt{16 {|\vec{a}|}^{2} + 25 {|\vec{b}|}^{2} + 40 \times |\vec{a}| \times |\vec{b}| \cos \frac{π}{4}} = \sqrt{16 {|2 \sqrt{2}|}^{2} + 25 {|3|}^{2} + 40 \times |2 \sqrt{2}| \times |3| \times \frac{1}{\sqrt{2}}} = \sqrt{128 + 225 + 240} = \sqrt{593} units Therefore, the larger diagonal = \sqrt{593}$

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Question 11:

If $\vec{a}$ is a non-zero vector of magnitude 'a' and λ is a non-zero scalar, then λ $\vec{a}$ is a unit vector if
(a) λ = 1
(b) λ = −1
(c) a = |λ|
(d) $a = \frac{1}{|λ|}$

Answer:

(d) $a = \frac{1}{|λ|}$

$Given that |\vec{a}| = a; Now, |λ \vec{a}| = 1 \Rightarrow |λ| |\vec{a}| = 1 \Rightarrow |λ| a = 1 \Rightarrow a = \frac{1}{|λ|}$

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Question 12:

If θ is the angle between two vectors $\vec{a} and \vec{b}, then \vec{a} \cdot \vec{b} \geq 0$ only when
(a) $0 < θ < \frac{π}{2}$

(b) $0 \leq θ \leq \frac{π}{2}$

(c) 0 < θ < π

(d) 0 ≤ θ ≤ π

Answer:

(b) $0 \leq θ \leq \frac{π}{2}$

$\vec{a} . \vec{b} \geq 0 \Rightarrow |\vec{a}| |\vec{b}| \cos θ \geq 0 \Rightarrow \cos θ \geq 0 \Rightarrow 0 \leq θ \leq \frac{π}{2}$

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Question 13:

The values of x for which the angle between $\vec{a} = 2 x^{2} \hat{i} + 4 x \hat{j} + \hat{k}, \vec{b} = 7 \hat{i} - 2 \hat{j} + x \hat{k}$ is obtuse and the angle between $\vec{b}$ and the z-axis is acute and less than $\frac{π}{6}$ are
(a) $x > \frac{1}{2} or x < 0$

(b) $0 < x < \frac{1}{2}$

(c) $\frac{1}{2} < x < 15$

(d) ϕ

Answer:

(b) $0 < x < \frac{1}{2}$

$\vec{a} = 2 x^{2} \hat{i} + 4 x \hat{j} + \hat{k}, \vec{b} = 7 \hat{i} - 2 \hat{j} + x \hat{k}$

Let the angle between vector a and vector b be A.

$∴ \cos A = \frac{\vec{a} . \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{(2 x^{2} \hat{i} + 4 x \hat{j} + \hat{k}) . (7 \hat{i} - 2 \hat{j} + x \hat{k})}{|2 x^{2} \hat{i} + 4 x \hat{j} + \hat{k}| |7 \hat{i} - 2 \hat{j} + x \hat{k}|} = \frac{14 x^{2} - 8 x + x}{\sqrt{4 x^{4} + 16 x^{2} + 1} \sqrt{49 + 4 + x^{2}}} = \frac{14 x^{2} - 7 x}{\sqrt{4 x^{4} + 16 x^{2} + 1} \sqrt{53 + x^{2}}} Now, ∠ A is an obtuse angle . ∴ \cos A < 0 \Rightarrow \frac{14 x^{2} - 7 x}{\sqrt{4 x^{4} + 16 x^{2} + 1} \sqrt{53 + x^{2}}} < 0 \Rightarrow 14 x^{2} - 7 x < 0 \Rightarrow 2 x^{2} - x < 0 \Rightarrow x (2 x - 1) < 0 \Rightarrow x < 0 & 2 x - 1 > 0 or x > 0 & 2 x - 1 < 0 \Rightarrow x < 0 & x > \frac{1}{2} or x > 0 & x < \frac{1}{2} \Rightarrow x > 0 & x < \frac{1}{2} (As there cannot be any number less than zero and greater than 1 / 2) \Rightarrow x \in (0, \frac{1}{2}) . . . (i)$

$Let the equation of the z - axis be z \hat{k} . And let the angle between \vec{b} and z - axis be B . ∴ \cos B = \frac{(7 \hat{i} - 2 \hat{j} + x \hat{k}) . (z \hat{k})}{|7 \hat{i} - 2 \hat{j} + x \hat{k}| |z \hat{k}|} = \frac{x z}{z \sqrt{49 + 4 + x^{2}}} = \frac{x}{\sqrt{53 + x^{2}}} Now, angle B is acute and less than π / 6 . ∴ 0 < \frac{x}{\sqrt{53 + x^{2}}} < \cos \frac{π}{6} \Rightarrow 0 < x < \frac{\sqrt{3}}{2} \sqrt{53 + x^{2}} . . . (ii) From (i) and (ii) we get 0 < x < \frac{1}{2}$

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Question 14:

If $\vec{a}, \vec{b}, \vec{c}$ are any three mutually perpendicular vectors of equal magnitude a, then $|\vec{a} + \vec{b} + \vec{c}|$ is equal to
(a) a
(b) $\sqrt{2} a$
(c) $\sqrt{3} a$
(d) 2a
(e) None of these

Answer:

(c) $\sqrt{3} a$

$Given that So, |\vec{a}| = |\vec{b}| = |\vec{c}| = a . . . (i) Since they are mutually perpendicular, \vec{a} . \vec{b} = \vec{b} . \vec{c} = \vec{c} . \vec{a} = 0 . . . (ii) Now, {|\vec{a} + \vec{b} + \vec{c}|}^{2} = {|\vec{a}|}^{2} + {|\vec{b}|}^{2} + {|\vec{c}|}^{2} + 2 \vec{a} . \vec{b} + 2 \vec{b} . \vec{c} + 2 \vec{c} . \vec{a} = a^{2} + a^{2} + a^{2} + 0 + 0 + 0 [Using (i) and (ii)] = 3 a^{2} ∴ |\vec{a} + \vec{b} + \vec{c}| = \sqrt{3} a$

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Question 15:

If the vectors $3 \hat{i} + λ \hat{j} + \hat{k} and 2 \hat{i} - \hat{j} + 8 \hat{k}$ are perpendicular, then λ is equal to
(a) −14
(b) 7
(c) 14
(d) $\frac{1}{7}$

Answer:

(c) 14

$It is given that vectors 3 \hat{i} + λ \hat{j} + \hat{k} and 2 \hat{i} - \hat{j} + 8 \hat{k} are perpendicular. So, their dot product is zero. (3 \hat{i} + λ \hat{j} + \hat{k}) . (2 \hat{i} - \hat{j} + 8 \hat{k}) = 0 \Rightarrow 6 - λ + 8 = 0 \Rightarrow 14 - λ = 0 ∴ λ = 14$

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Question 16:

The projection of the vector $\hat{i} + \hat{j} + \hat{k}$ along the vector of $\hat{j}$ is
(a) 1
(b) 0
(c) 2
(d) −1
(e) −2

Answer:

(a) 1

$Let \vec{a} = \hat{i} + \hat{j} + \hat{k} and \vec{b} = \hat{j} The projection of \vec{a} on \vec{b} is \frac{\vec{a} . \vec{b}}{|\vec{b}|} = \frac{(\hat{i} + \hat{j} + \hat{k}) . \hat{j}}{|\hat{j}|} = \frac{0 + 1 + 0}{1} = 1$

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Question 17:

The vectors $2 \hat{i} + 3 \hat{j} - 4 \hat{k}$ and $a \hat{i} + b \hat{j} + c \hat{k}$ are perpendicular if
(a) a = 2, b = 3, c = −4
(b) a = 4, b = 4, c = 5
(c) a = 4, b = 4, c = −5
(d) a = −4, b = 4, c = −5

Answer:

(b) a = 4, b = 4, c = 5

$It is given that vectors 2 \hat{i} + 3 \hat{j} - 4 \hat{k} and a \hat{i} + b \hat{j} + c \hat{k} are perpendicular. So, their dot product is zero. \Rightarrow 2 a + 3 b - 4 c = 0 (b) a = 4; b = 4; c = 5 \Rightarrow 2 (4) + 3 (4) - 4 (5) = 0 8 + 12 - 20 = 0 0 = 0, which is true.$

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Question 18:

If $|\vec{a}| = |\vec{b}|, then (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) =$
(a) positive
(b) negative
(c) 0
(d) None of these

Answer:

(c) 0

$Given that |\vec{a}| = |\vec{a}| \Rightarrow (\vec{a} + \vec{b}) . (\vec{a} - \vec{b}) = {|\vec{a}|}^{2} - {|\vec{b}|}^{2} = {|\vec{a}|}^{2} - {|\vec{a}|}^{2} = 0$

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Question 19:

If $\vec{a} and \vec{b}$ are unit vectors inclined at an angle θ, then the value of $|\vec{a} - \vec{b}|$ is
(a) $2 \sin \frac{θ}{2}$

(b) 2 sin θ

(c) $2 \cos \frac{θ}{2}$

(d) 2 cos θ

Answer:

(a) $2 \sin \frac{θ}{2}$

$\vec{a} . \vec{b} = |\vec{a}| |\vec{b}| \cos θ = 1 \times 1 \cos θ (Because \vec{a} and \vec{b} are unit vectors) = \cos θ . . . (i) {|\vec{a} - \vec{b}|}^{2} = {|\vec{a}|}^{2} + {|\vec{b}|}^{2} - 2 \vec{a} . \vec{b} = 1 + 1 - 2 \cos θ [Using (i)] = 2 - 2 \cos θ = 2 (1 - \cos θ) = 2 (2 \sin^{2} \frac{θ}{2}) = 4 \sin^{2} \frac{θ}{2} ∴ |\vec{a} - \vec{b}| = 2 \sin \frac{θ}{2}$

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Question 20:

If $\vec{a} and \vec{b}$ are unit vectors, then the greatest value of $\sqrt{3} |\vec{a} + \vec{b}| + |\vec{a} - \vec{b}|$ is
(a) 2
(b) $2 \sqrt{2}$
(c) 4
(d) None of these

Answer:

(c) 4

$We have \sqrt{3} |\vec{a} + \vec{b}| + |\vec{a} - \vec{b}| = \sqrt{3} \times \sqrt{{|\vec{a}|}^{2} + {|\vec{b}|}^{2} + 2 |\vec{a}| |\vec{b}| \cos θ} + \sqrt{{|\vec{a}|}^{2} + {|\vec{b}|}^{2} - 2 |\vec{a}| |\vec{b}| \cos θ} = \sqrt{3} \times \sqrt{1^{2} + 1^{2} + 2 \times 1 \times 1 \cos θ} + \sqrt{1^{2} + 1^{2} - 2 \times 1 \times 1 \cos θ} (As \vec{a} and \vec{b} unit vectors) = \sqrt{3} \times \sqrt{2 + 2 \cos θ} + \sqrt{2 - 2 \cos θ} = \sqrt{3} \times \sqrt{2 (1 + \cos θ)} + \sqrt{2 (1 - \cos θ)} = \sqrt{3} \times \sqrt{2 \times 2 \cos^{2} \frac{θ}{2}} + \sqrt{2 \times 2 \sin^{2} \frac{θ}{2}} = 2 \sqrt{3} \cos \frac{θ}{2} + 2 \sin \frac{θ}{2} = 2 (\sqrt{3} \cos \frac{θ}{2} + \sin \frac{θ}{2}) = 2 \times 2 (\frac{\sqrt{3}}{2} \cos \frac{θ}{2} + \frac{1}{2} \sin \frac{θ}{2}) = 2 \times 2 (\sin \frac{π}{3} \cos \frac{θ}{2} + \cos \frac{π}{3} \sin \frac{θ}{2}) = 4 \sin (\frac{π}{3} + \frac{θ}{2}) Now, m aximum value of \sin α = 1 \Rightarrow Maximum value of \sin (\frac{π}{3} + \frac{θ}{2}) = 1 \Rightarrow Maximum value of 4 \sin (\frac{π}{3} + \frac{θ}{2}) = 4 ∴ Maximum value of \sqrt{3} |\vec{a} + \vec{b}| + |\vec{a} - \vec{b}| = 4$

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Question 21:

If the angle between the vectors $x \hat{i} + 3 \hat{j} - 7 \hat{k} and x \hat{i} - x \hat{j} + 4 \hat{k}$ is acute, then x lies in the interval
(a) (−4, 7)
(b) [−4, 7]
(c) R −[−4, 7]
(d) R −(4, 7)

Answer:

(c) R −[−4, 7]

$Let θ be the angle between \vec{a} and \vec{b} . \cos θ = \frac{\vec{a} . \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{x^{2} - 3 x - 28}{\sqrt{x^{2} + 3^{2} + 49} \sqrt{x^{2} + x^{2} + 4^{2}}} For θ to be acute, \cos θ > 0 \Rightarrow x^{2} - 3 x - 28 > 0 \Rightarrow (x - 7) (x + 4) > 0 \Rightarrow x \in (- \infty, - 4) \cup (7, \infty) \Rightarrow x \in R - [- 4, 7]$

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Question 22:

If $\vec{a} and \vec{b}$ are two unit vectors inclined at an angle θ, such that $|\vec{a} + \vec{b}| < 1,$ then
(a) $θ < \frac{π}{3}$

(b) $θ > \frac{2 π}{3}$

(c) $\frac{π}{3} < θ < \frac{2 π}{3}$

(d) $\frac{2 π}{3} < θ < π$

Answer:

(d) $\frac{2 π}{3} < θ < π$

$We have |\vec{a} + \vec{b}| < 1 \Rightarrow \sqrt{{|\vec{a}|}^{2} + {|\vec{b}|}^{2} + 2 |\vec{a}| \times |\vec{b}| \cos θ} < 1 \Rightarrow \sqrt{1^{2} + 1^{2} + 2 \times 1 \times 1 \times \cos θ} < 1 \Rightarrow \sqrt{2 + 2 \cos θ} < 1 \Rightarrow \sqrt{2 (1 + \cos θ)} < 1 \Rightarrow \sqrt{2 \times 2 \cos^{2} \frac{θ}{2}} < 1 \Rightarrow 2 |\cos \frac{θ}{2}| < 1 \Rightarrow |\cos \frac{θ}{2}| < \frac{1}{2} \Rightarrow \frac{π}{3} < \frac{θ}{2} < \frac{2 π}{3} \Rightarrow \frac{2 π}{3} < θ < \frac{4 π}{3} But here θ cannot be more than π . ∴ \frac{2 π}{3} < θ < π$

Page No 23.48:

Question 23:

Let $\vec{a}, \vec{b}, \vec{c}$ be three unit vectors, such that $|\vec{a} + \vec{b} + \vec{c}|$ =1 and $\vec{a}$ is perpendicular to $\vec{b}$ . If $\vec{c}$ makes angles α and β with $\vec{a} and \vec{b}$ respectively, then cos α + cos β =
(a) $- \frac{3}{2}$

(b) $\frac{3}{2}$

(c) 1

(d) −1

Answer:

(d) −1

$Given that \vec{a}, \vec{b} and \vec{c} are unit vectors. So, |\vec{a}| =1, |\vec{b}| =1and |\vec{c}| =1. Since \vec{a} and \vec{b} are mutually perpendicular, \vec{a} . \vec{b} = 0 Now, |\vec{a} + \vec{b} + \vec{c}| = 1 \Rightarrow {|\vec{a} + \vec{b} + \vec{c}|}^{2} = 1 \Rightarrow {|\vec{a}|}^{2} + {|\vec{b}|}^{2} + {|\vec{c}|}^{2} + 2 \vec{a} . \vec{b} + 2 \vec{b} . \vec{c} + 2 \vec{c} . \vec{a} = 1 \Rightarrow 1 + 1 + 1 + 2 (0) + 2 |\vec{a}| |\vec{b}| \cos β + 2 |\vec{c}| |\vec{a}| \cos α = 1 \Rightarrow 3 + 2 (\cos α + \cos β) = 1 \Rightarrow 2 (\cos α + \cos β) = - 2 \Rightarrow \cos α + \cos β = - 1$

Page No 23.48:

Question 24:

The orthogonal projection of $\vec{a} on \vec{b}$ is
(a) $\frac{(\vec{a} \cdot \vec{b}) \vec{a}}{{|\vec{a}|}^{2}}$

(b) $\frac{(\vec{a} \cdot \vec{b}) \vec{b}}{{|\vec{b}|}^{2}}$

(c) $\frac{\vec{a}}{|\vec{a}|}$

(d) $\frac{\vec{b}}{|\vec{b}|}$

Answer:

(b) $\frac{(\vec{a} \cdot \vec{b}) \vec{b}}{{|\vec{b}|}^{2}}$

$The orthogonal projection of \vec{a} on \vec{b} is \frac{(\vec{a} . \vec{b}) \vec{b}}{{|\vec{b}|}^{2}}$

Page No 23.48:

Question 25:

If θ is an acute angle and the vector (sin θ) $\hat{i}$ + (cos θ) $\hat{j}$ is perpendicular to the vector $\hat{i} - \sqrt{3} \hat{j},$ then θ =
(a) $\frac{π}{6}$

(b) $\frac{π}{5}$

(c) $\frac{π}{4}$

(d) $\frac{π}{3}$

Answer:

(d) $\frac{π}{3}$

$The given vectors are perpendicular. So, their dot product is zero. [(\sin θ) \hat{i} + (\cos θ) \hat{j}] . (\hat{i} - \sqrt{3} \hat{j}) = 0 \Rightarrow \sin θ - \sqrt{3} \cos θ = 0 \Rightarrow \sin θ = \sqrt{3} \cos θ \Rightarrow \tan θ = \sqrt{3} \Rightarrow θ = \frac{π}{3} (Because θ is acute)$

Page No 23.48:

Question 26:

The angle between two vectors $\vec{a} and \vec{b}$ with magnitudes $\sqrt{3}$ and 4 respectively and $\vec{a} \cdot \vec{b} = 2 \sqrt{3}, is$
$(a) \frac{π}{6} (b) \frac{π}{3} (c) \frac{π}{2} (d) \frac{5 π}{2}$

Answer:

Given:
$|\vec{a}| = \sqrt{3} |\vec{b}| = 4 \vec{a} \cdot \vec{b} = 2 \sqrt{3}$

$\vec{a} \cdot \vec{b} = 2 \sqrt{3} \Rightarrow |\vec{a}| |\vec{b}| \cos θ = 2 \sqrt{3} \Rightarrow \sqrt{3} \times 4 \times \cos θ = 2 \sqrt{3} \Rightarrow \cos θ = \frac{2 \sqrt{3}}{4 \sqrt{3}} \Rightarrow \cos θ = \frac{1}{2} \Rightarrow θ = \frac{π}{3}$

Hence, the correct option is (b).

Page No 23.49:

Question 27:

If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a} + \vec{b} + \vec{c} = \vec{0},$ then the value of $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} is$
(a) 1
(b) 3
(c) $- \frac{3}{2}$
(d) none of these

Answer:

Given:
$\vec{a}, \vec{b}, \vec{c}$ are unit vectors
$\vec{a} + \vec{b} + \vec{c} = \vec{0}$

$\vec{a} + \vec{b} + \vec{c} = \vec{0} \Rightarrow {|\vec{a} + \vec{b} + \vec{c}|}^{2} = 0 \Rightarrow (\vec{a} + \vec{b} + \vec{c}) . (\vec{a} + \vec{b} + \vec{c}) = 0 \Rightarrow \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} + \vec{c} \cdot \vec{c} + 2 \vec{a} \cdot \vec{b} + 2 \vec{b} \cdot \vec{c} + 2 \vec{c} \cdot \vec{a} = 0 \Rightarrow {|\vec{a}|}^{2} + {|\vec{b}|}^{2} + {|\vec{c}|}^{2} + 2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \Rightarrow 1 + 1 + 1 + 2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 (∵ \vec{a}, \vec{b} and \vec{c} are unit vectors) \Rightarrow 2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = - 3 \Rightarrow \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = - \frac{3}{2}$

Hence, the correct option is (c).

Page No 23.49:

Question 28:

If $\vec{a}, \vec{b,} \vec{c}$ are three vectors such that $\vec{a} + \vec{b} + \vec{c} = \vec{0} and |\vec{a}| = 2, |\vec{b}| = 3, |\vec{c}| = 5,$ then the value of $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \cdot \vec{c} \cdot \vec{a} is$
(a) 0
(b) 1
(c) -19
(d) 38

Answer:

Given:
$|\vec{a}| = 2, |\vec{b}| = 3, |\vec{c}| = 5$
$\vec{a} + \vec{b} + \vec{c} = \vec{0}$

$\vec{a} + \vec{b} + \vec{c} = \vec{0} \Rightarrow {|\vec{a} + \vec{b} + \vec{c}|}^{2} = 0 \Rightarrow (\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) = 0 \Rightarrow \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} + \vec{c} \cdot \vec{c} + 2 \vec{a} \cdot \vec{b} + 2 \vec{b} \cdot \vec{c} + 2 \vec{c} \cdot \vec{a} = 0 \Rightarrow {|\vec{a}|}^{2} + {|\vec{b}|}^{2} + {|\vec{c}|}^{2} + 2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \Rightarrow 2^{2} + 3^{2} + 5^{2} + 2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 (∵ |\vec{a}| = 2, |\vec{b}| = 3, |\vec{c}| = 5) \Rightarrow 4 + 9 + 25 + 2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \Rightarrow 38 + 2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \Rightarrow 2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = - 38 \Rightarrow \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = - \frac{38}{2} \Rightarrow \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = - 19$

Hence, the correct option is (c).

Page No 23.49:

Question 29:

If $\vec{a} and \vec{b}$ are unit vectors, then what is the angle between $\vec{a} and \vec{b} for \sqrt{3} \vec{a} - \vec{b}$ to be a unit vector?
$(a) \frac{π}{6} (b) \frac{π}{4} (c) \frac{π}{3} (d) \frac{π}{2}$

Answer:

Given:
$\vec{a} and \vec{b}$ are unit vectors
$\sqrt{3} \vec{a} - \vec{b}$ is a unit vector

$if \sqrt{3} \vec{a} - \vec{b} is a unit vector \Rightarrow |\sqrt{3} \vec{a} - \vec{b}| = 1 \Rightarrow {|\sqrt{3} \vec{a} - \vec{b}|}^{2} = 1 \Rightarrow (\sqrt{3} \vec{a} - \vec{b}) \cdot (\sqrt{3} \vec{a} - \vec{b}) = 1 \Rightarrow \sqrt{3} \vec{a} \cdot \sqrt{3} \vec{a} + \vec{b} \cdot \vec{b} - 2 \sqrt{3} \vec{a} \cdot \vec{b} = 1 \Rightarrow 3 \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} - 2 \sqrt{3} \vec{a} \cdot \vec{b} = 1 \Rightarrow 3 {|\vec{a}|}^{2} + {|\vec{b}|}^{2} - 2 \sqrt{3} (|\vec{a}| |\vec{b}| \cos θ) = 1 \Rightarrow 3 (1) + (1) - 2 \sqrt{3} ((1) (1) \cos θ) = 1 (∵ |\vec{a}| = 1, |\vec{b}| = 1) \Rightarrow 4 - 2 \sqrt{3} (\cos θ) = 1 \Rightarrow 2 \sqrt{3} (\cos θ) = 4 - 1 \Rightarrow 2 \sqrt{3} (\cos θ) = 3 \Rightarrow \cos θ = \frac{3}{2 \sqrt{3}} \Rightarrow \cos θ = \frac{\sqrt{3}}{2} \Rightarrow θ = \frac{π}{6}$

Hence, the correct option is (a).

Page No 23.49:

Question 1:

If $\vec{a} and \vec{b}$ are two unit vectors such that $|\vec{a} + \vec{b}|$ is also a unit vector, then $|\overset{⇀}{a} - \overset{⇀}{b}|______________________.$

Answer:

Given:
$\vec{a} and \vec{b}$ are unit vectors
$|\vec{a} + \vec{b}|$ is also a unit vector

$if (\vec{a} + \vec{b}) is a unit vector \Rightarrow |\vec{a} + \vec{b}| = 1 \Rightarrow {|\vec{a} + \vec{b}|}^{2} = 1 \Rightarrow (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = 1 \Rightarrow \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} + 2 \vec{a} \cdot \vec{b} = 1 \Rightarrow {|\vec{a}|}^{2} + {|\vec{b}|}^{2} + 2 \vec{a} \cdot \vec{b} = 1 \Rightarrow 1 + 1 + 2 \vec{a} \cdot \vec{b} = 1 (∵ |\vec{a}| = 1, |\vec{b}| = 1) \Rightarrow 2 \vec{a} \cdot \vec{b} = - 1 . . . (1) Now, {|\vec{a} - \vec{b}|}^{2} = (\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b}) = \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} - 2 \vec{a} \cdot \vec{b} = {|\vec{a}|}^{2} + {|\vec{b}|}^{2} - (- 1) (From (1)) = {|\vec{a}|}^{2} + {|\vec{b}|}^{2} + 1 = 1 + 1 + 1 (∵ |\vec{a}| = 1, |\vec{b}| = 1) = 3 \Rightarrow |\vec{a} - \vec{b}| = \sqrt{3}$

Hence, $|\vec{a} - \vec{b}| = \sqrt{3}$ .

Page No 23.49:

Question 2:

$If |\vec{a}| = 3, |\vec{b}| = 4 and \vec{a} + λ \vec{b}$ is perpendicular to $\vec{a} - λ \vec{b}, then λ =____________________.$

Answer:

Given:
$|\vec{a}| = 3, |\vec{b}| = 4$
$\vec{a} + λ \vec{b}$ is perpendicular to $\vec{a} - λ \vec{b}$

$Since, (\vec{a} + λ \vec{b}) is perpendicular to (\vec{a} - λ \vec{b}) Therefore, (\vec{a} + λ \vec{b}) \cdot (\vec{a} - λ \vec{b}) = 0 \Rightarrow \vec{a} \cdot \vec{a} - λ \vec{b} \cdot λ \vec{b} = 0 \Rightarrow {|\vec{a}|}^{2} - λ^{2} {|\vec{b}|}^{2} = 0 \Rightarrow {(3)}^{2} - λ^{2} {(4)}^{2} = 0 (∵ |\vec{a}| = 3, |\vec{b}| = 4) \Rightarrow 9 - 16 λ^{2} = 0 \Rightarrow 16 λ^{2} = 9 \Rightarrow λ^{2} = \frac{9}{16} \Rightarrow λ = \pm \frac{3}{4}$

Hence, $λ = \pm \frac{3}{4}$ .

Page No 23.49:

Question 3:

$If \vec{a} = 2 \hat{i} - 7 \hat{j} + \hat{k}, \vec{b} = \hat{i} + 3 \hat{j} - 5 \hat{k} and \vec{a} . m \vec{b} = 120, then m =_____________.$

Answer:

Given:
$\vec{a} = 2 \hat{i} - 7 \hat{j} + \hat{k} \vec{b} = \hat{i} + 3 \hat{j} - 5 \hat{k}$
$\vec{a} \cdot m \vec{b} = 120$

$\vec{a} \cdot m \vec{b} = 120 \Rightarrow m (\vec{a} \cdot \vec{b}) = 120 \Rightarrow m ((2) (1) + (- 7) (3) + (1) (- 5)) = 120 \Rightarrow m (2 - 21 - 5) = 120 \Rightarrow m (- 24) = 120 \Rightarrow m = - \frac{120}{24} \Rightarrow m = - \frac{10}{2} \Rightarrow m = - 5$

Hence, $m = - 5$ .

Page No 23.49:

Question 4:

The non-zero vectors $\vec{a}, \vec{b} and \vec{c} are related by \vec{a} = 8 \vec{b} and \vec{c} = - 7 \vec{b}$ , then the angle between $\vec{a} and \vec{c}$ is ____________.

Answer:

Given:
$\vec{a} = 8 \vec{b} and \vec{c} = - 7 \vec{b}$

$Let θ be the angle between \vec{a} and \vec{c} . \vec{a} \cdot \vec{c} = (8 \vec{b}) \cdot (- 7 \vec{b}) \Rightarrow |\vec{a}| |\vec{c}| \cos θ = 8 \times (- 7) \times \vec{b} \cdot \vec{b} \Rightarrow |8 \vec{b}| |- 7 \vec{b}| \cos θ = - 56 \times {|\vec{b}|}^{2} \Rightarrow 56 {|\vec{b}|}^{2} \cos θ = - 56 \times {|\vec{b}|}^{2} \Rightarrow \cos θ = - 1 \Rightarrow θ = π$

Hence, the angle between $\vec{a} and \vec{c}$ is $π$ .

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Question 5:

If $\vec{a} and \vec{b}$ are mutually perpendicular unit vectors, then $|\vec{a} + \vec{b}| =____________.$

Answer:

Given:
$\vec{a} and \vec{b}$ are unit vectors
$\vec{a} and \vec{b}$ are mutually perpendicular vectors

${|\vec{a} + \vec{b}|}^{2} = (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} + 2 \vec{a} \cdot \vec{b} = {|\vec{a}|}^{2} + {|\vec{b}|}^{2} + 2 \vec{a} \cdot \vec{b} = 1^{2} + 1^{2} + 2 (0) (∵ |\vec{a}| = 1, |\vec{b}| = 1, \vec{a} \cdot \vec{b} = 0) = 1 + 1 = 2 \Rightarrow |\vec{a} + \vec{b}| = \sqrt{2}$

Hence, $|\vec{a} + \vec{b}| = \sqrt{2}$ .

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Question 6:

If the angle between the vectors $\hat{i} + \hat{k} and \hat{i} - \hat{j} + α \vec{k} i s \frac{π}{3}, then α =_____________________.$

Answer:

Given:
The angle between the vectors $\hat{i} + \hat{k} and \hat{i} - \hat{j} + α \vec{k} is \frac{π}{3}$

$Let \vec{a} = \hat{i} + \hat{k} and \vec{b} = \hat{i} - \hat{j} + α \vec{k} Angle between \vec{a} and \vec{b} is θ = \frac{π}{3} |\vec{a}| = \sqrt{{(1)}^{2} + {(1)}^{2}} = \sqrt{2} |\vec{b}| = \sqrt{{(1)}^{2} + {(- 1)}^{2} + {(α)}^{2}} = \sqrt{2 + α^{2}} \vec{a} \cdot \vec{b} = (1) (1) + (0) (- 1) + (1) (α) = 1 + α Now, \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos θ \Rightarrow 1 + α = \sqrt{2} \sqrt{2 + α^{2}} \cos \frac{π}{3} \Rightarrow 1 + α = \sqrt{4 + 2 α^{2}} \times \frac{1}{2} \Rightarrow 2 + 2 α = \sqrt{4 + 2 α^{2}} Squaring both sides, we get \Rightarrow {(2 + 2 α)}^{2} = 4 + 2 α^{2} \Rightarrow 4 + 4 α^{2} + 8 α = 4 + 2 α^{2} \Rightarrow 2 α^{2} + 8 α = 0 \Rightarrow 2 α (α + 4) = 0 \Rightarrow α = 0, - 4$

Hence, $α = 0, - 4$ .

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Question 7:

If $\vec{a}, \vec{b} and \vec{c}$ are unit vectors such that $\vec{a} + \vec{b} - \vec{c} = \vec{0}$ , then the angle between $\vec{a} and \vec{b}$ is ___________.

Answer:

Given:
$\vec{a}, \vec{b} and \vec{c}$ are unit vectors
$\vec{a} + \vec{b} - \vec{c} = \vec{0}$

$\vec{a} + \vec{b} - \vec{c} = \vec{0} \Rightarrow \vec{a} + \vec{b} = \vec{c} \Rightarrow {(\vec{a} + \vec{b})}^{2} = {(\vec{c})}^{2} \Rightarrow (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = 1 (∵ |\vec{c}| = 1) \Rightarrow \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} + 2 \vec{a} \cdot \vec{b} = 1 \Rightarrow {|\vec{a}|}^{2} + {|\vec{b}|}^{2} + 2 (|\vec{a}| |\vec{b}| \cos θ) = 1 \Rightarrow 1^{2} + 1^{2} + 2 ((1) (1) \cos θ) = 1 (∵ |\vec{a}| = 1, |\vec{b}| = 1) \Rightarrow 1 + 1 + 2 \cos θ = 1 \Rightarrow 2 \cos θ = - 1 \Rightarrow \cos θ = - \frac{1}{2} \Rightarrow θ = π - \frac{π}{3} \Rightarrow θ = \frac{2 π}{3}$

Hence, the angle between $\vec{a} and \vec{b}$ is $\frac{2 π}{3}$ .

Page No 23.49:

Question 8:

If $\vec{a} and \vec{b}$ are unit vectors such that $|\vec{a} + \vec{b}| = \sqrt{3},$ then the angle between $\vec{a} and \vec{b}$ is ___________.

Answer:

Given:
$\vec{a} and \vec{b}$ are unit vectors
$|\vec{a} + \vec{b}| = \sqrt{3}$

$|\vec{a} + \vec{b}| = \sqrt{3} \Rightarrow {|\vec{a} + \vec{b}|}^{2} = {(\sqrt{3})}^{2} \Rightarrow (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = 3 \Rightarrow \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} + 2 \vec{a} \cdot \vec{b} = 3 \Rightarrow {|\vec{a}|}^{2} + {|\vec{b}|}^{2} + 2 (|\vec{a}| |\vec{b}| \cos θ) = 3 \Rightarrow 1^{2} + 1^{2} + 2 ((1) (1) \cos θ) = 3 (∵ |\vec{a}| = 1, |\vec{b}| = 1) \Rightarrow 1 + 1 + 2 \cos θ = 3 \Rightarrow 2 \cos θ = 1 \Rightarrow \cos θ = \frac{1}{2} \Rightarrow θ = \frac{π}{3}$

Hence, the angle between $\vec{a} and \vec{b}$ is $\frac{π}{3}$ .

Page No 23.49:

Question 9:

If $\vec{a} and \vec{b}$ are two non-zero vectors, then the projection of $\vec{a} on \vec{b}$ is _____________.

Answer:

The projection of $\vec{a} on \vec{b}$ is given by $(\vec{a} \cdot \hat{b}) = \frac{1}{|\vec{b}|} (\vec{a} \cdot \vec{b})$ .

Hence, the projection of $\vec{a} on \vec{b}$ is $\frac{1}{|\vec{b}|} (\vec{a} \cdot \vec{b})$ .

Page No 23.49:

Question 10:

Let $\vec{a}, \vec{b}$ be unit vectors such that $\vec{a} - \sqrt{2} \vec{b}$ is also a unit vector. Then the angle between $\vec{a} and \vec{b}$ is _____________.

Answer:

Given:
$\vec{a} and \vec{b}$ are unit vectors
$\vec{a} - \sqrt{2} \vec{b}$ is also a unit vector

$\vec{a} - \sqrt{2} \vec{b} is a unit vector \Rightarrow |\vec{a} - \sqrt{2} \vec{b}| = 1 \Rightarrow {|\vec{a} - \sqrt{2} \vec{b}|}^{2} = {(1)}^{2} \Rightarrow (\vec{a} - \sqrt{2} \vec{b}) \cdot (\vec{a} - \sqrt{2} \vec{b}) = 1 \Rightarrow \vec{a} \cdot \vec{a} + \sqrt{2} \vec{b} \cdot \sqrt{2} \vec{b} - 2 \sqrt{2} \vec{a} \cdot \vec{b} = 1 \Rightarrow {|\vec{a}|}^{2} + 2 {|\vec{b}|}^{2} - 2 \sqrt{2} (|\vec{a}| |\vec{b}| \cos θ) = 1 \Rightarrow 1^{2} + 2 {(1)}^{2} - 2 \sqrt{2} ((1) (1) \cos θ) = 1 (∵ |\vec{a}| = 1, |\vec{b}| = 1) \Rightarrow 1 + 2 - 2 \sqrt{2} \cos θ = 1 \Rightarrow - 2 \sqrt{2} \cos θ = - 2 \Rightarrow \cos θ = \frac{1}{\sqrt{2}} \Rightarrow θ = \frac{π}{4}$

Hence, the angle between $\vec{a} and \vec{b}$ is $\frac{π}{4}$ .

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Question 11:

$If |\vec{a}| = 1, |\vec{b}| = 3 a n d |\vec{a} - \vec{b}| = \sqrt{7},$ then the angle between $\vec{a} and \vec{b}$ is ______________.

Answer:

Given:
$|\vec{a}| = 1 |\vec{b}| = 3 |\vec{a} - \vec{b}| = \sqrt{7}$

$|\vec{a} - \vec{b}| = \sqrt{7} \Rightarrow {|\vec{a} - \vec{b}|}^{2} = {(\sqrt{7})}^{2} \Rightarrow (\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b}) = 7 \Rightarrow \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} - 2 \vec{a} \cdot \vec{b} = 7 \Rightarrow {|\vec{a}|}^{2} + {|\vec{b}|}^{2} - 2 (|\vec{a}| |\vec{b}| \cos θ) = 7 \Rightarrow 1^{2} + {(3)}^{2} - 2 ((1) (3) \cos θ) = 7 (∵ |\vec{a}| = 1, |\vec{b}| = 3) \Rightarrow 1 + 9 - 6 \cos θ = 7 \Rightarrow - 6 \cos θ = - 3 \Rightarrow \cos θ = \frac{1}{2} \Rightarrow θ = \frac{π}{3}$

Hence, the angle between $\vec{a} and \vec{b}$ is $\frac{π}{3}$ .

Page No 23.50:

Question 1:

What is the angle between vectors $\vec{a} and \vec{b}$ with magnitudes 2 and $\sqrt{3}$ respectively? Given $\vec{a} . \vec{b} = \sqrt{3} .$

Answer:

$Let θ be the angle between \vec{a} and \vec{b} . Given that |\vec{a}| = 2, |\vec{b}| = \sqrt{3} and \vec{a} . \vec{b} = \sqrt{3} We know that \vec{a} . \vec{b} = |\vec{a}| |\vec{b}| \cos θ \Rightarrow \sqrt{3} = (2) (\sqrt{3}) \cos θ \Rightarrow \cos θ = \frac{\sqrt{3}}{2 \sqrt{3}} \Rightarrow \cos θ = \frac{1}{2} \Rightarrow θ = \cos^{- 1} (\frac{1}{2}) = \frac{π}{3}$

Page No 23.50:

Question 2:

$\vec{a} and \vec{b}$ are two vectors such that $\vec{a} . \vec{b} = 6, |\vec{a}| = 3 and |\vec{b}| = 4 .$ Write the projection of $\vec{a} on \vec{b}$ .

Answer:

$We have \vec{a} . \vec{b} = 6 and |\vec{b}| = 4 The projection of \vec{a} on \vec{b} is \frac{\vec{a} . \vec{b}}{|\vec{b}|} = \frac{6}{4} = \frac{3}{2}$

Page No 23.50:

Question 3:

Find the cosine of the angle between the vectors $4 \hat{i} - 3 \hat{j} + 3 \hat{k} and 2 \hat{i} - \hat{j} - \hat{k} .$

Answer:

$Let, \vec{a} = 4 \hat{i} - 3 \hat{j} + 3 \hat{k} and \vec{b} = 2 \hat{i} - \hat{j} - \hat{k} Let θ be the angle between \vec{a} and \vec{b} . |\vec{a}| = \sqrt{{(4)}^{2} + {(- 3)}^{2} + {(3)}^{2}} = \sqrt{34} |\vec{b}| = \sqrt{{(2)}^{2} + {(- 1)}^{2} + {(- 1)}^{2}} = \sqrt{6} ∴ \vec{a} . \vec{b} = 8 + 3 - 3 = 8 \cos θ = \frac{\vec{a} . \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{8}{\sqrt{34} \sqrt{6}} = \frac{8}{2 \sqrt{51}} = \frac{4}{\sqrt{51}}$

Page No 23.50:

Question 4:

If the vectors $3 \hat{i} + m \hat{j} + \hat{k} and 2 \hat{i} - \hat{j} - 8 \hat{k}$ are orthogonal, find m.

Answer:

$It is given that the vectors are othgonal. So, their dot product is zero. (3 \hat{i} + m \hat{j} + \hat{k}) . (2 \hat{i} - \hat{j} - 8 \hat{k}) = 0 \Rightarrow 6 - m - 8 = 0 \Rightarrow - m - 2 = 0 \Rightarrow m = - 2$

Page No 23.50:

Question 5:

If the vectors $3 \hat{i} - 2 \hat{j} - 4 \hat{k} and 18 \hat{i} - 12 \hat{j} - m \hat{k}$ are parallel, find the value of m.

Answer:

$The given vectors are parallel. ∴ 3 \hat{i} - 2 \hat{j} - 4 \hat{k} = t (18 \hat{i} - 12 \hat{j} - m \hat{k}) \Rightarrow 3 \hat{i} - 2 \hat{j} - 4 \hat{k} = 18 t \hat{i} - 12 t \hat{j} - t m \hat{k} Comparing both sides, we get 18 t = 3, - 12 t = - 2, - 4 = - t m \Rightarrow t = \frac{1}{6} Substituting the value of m in -4=- t m, we get - 4 = - m (\frac{1}{6}) ∴ m = 24$

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Question 6:

If $\vec{a} and \vec{b}$ are vectors of equal magnitude, write the value of $(\vec{a} + \vec{b}) . (\vec{a} - \vec{b}) .$

Answer:

$We have |\vec{a}| = |\vec{b}| . . . (i) Now, (\vec{a} + \vec{b}) . (\vec{a} - \vec{b}) = {|\vec{a}|}^{2} - {|\vec{b}|}^{2} = {|\vec{a}|}^{2} - {|\vec{a}|}^{2} [Using (i)] = 0$

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Question 7:

If $\vec{a} and \vec{b}$ are two vectors such that $(\vec{a} + \vec{b}) . (\vec{a} - \vec{b}) = 0,$ find the relation between the magnitudes of $\vec{a} and \vec{b}$ .

Answer:

$Given that (\vec{a} + \vec{b}) . (\vec{a} - \vec{b}) = 0 \Rightarrow {|\vec{a}|}^{2} - {|\vec{b}|}^{2} = 0 \Rightarrow {|\vec{a}|}^{2} = {|\vec{b}|}^{2} ∴ |\vec{a}| = |\vec{b}|$

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Question 8:

For any two vectors $\vec{a} and \vec{b}$ , write when $|\vec{a} + \vec{b}| = |\vec{a}| + |\vec{b}|$ holds.

Answer:

$Given that |\vec{a} + \vec{b}| = |\vec{a}| + |\vec{b}| Squaring both sides,we get {|\vec{a} + \vec{b}|}^{2} = {(|\vec{a}| + |\vec{b}|)}^{2} \Rightarrow {|\vec{a}|}^{2} + {|\vec{b}|}^{2} + 2 \vec{a} . \vec{b} = {|\vec{a}|}^{2} + {|\vec{b}|}^{2} + 2 |\vec{a}| |\vec{b}| \Rightarrow \vec{a} . \vec{b} = |\vec{a}| |\vec{b}| \Rightarrow |\vec{a}| |\vec{b}| \cos θ = |\vec{a}| |\vec{b}| (where θ is the angle between \vec{a} and \vec{b}) \Rightarrow \cos θ = 1 \Rightarrow θ = 0^{o} \Rightarrow \vec{a} and \vec{b} are parallel.$

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Question 9:

For any two vectors $\vec{a} and \vec{b}$ , write when $|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|$ holds.

Answer:

$Given that |\vec{a} + \vec{b}| = |\vec{a} - \vec{b}| Squaring both sides, we get {|\vec{a} + \vec{b}|}^{2} = {|\vec{a} - \vec{b}|}^{2} \Rightarrow {|\vec{a}|}^{2} + {|\vec{b}|}^{2} + 2 \vec{a} . \vec{b} = {|\vec{a}|}^{2} + {|\vec{b}|}^{2} - 2 \vec{a} . \vec{b} \Rightarrow 4 \vec{a} . \vec{b} = 0 \Rightarrow \vec{a} . \vec{b} = 0 \Rightarrow \vec{a} and \vec{b} are perpendicular.$

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Question 10:

If $\vec{a} and \vec{b}$ are two vectors of the same magnitude inclined at an angle of 60° such that $\vec{a} . \vec{b} = 8,$ write the value of their magnitude.

Answer:

$Given that |\vec{a}| = |\vec{b}| and \vec{a} and \vec{b} are inclined at an angle of 60 ° Also, given that \vec{a} . \vec{b} = 8 \Rightarrow |\vec{a}| |\vec{b}| \cos 60 ° = 8 \Rightarrow |\vec{a}| |\vec{a}| (\frac{1}{2}) = 8 \Rightarrow {|\vec{a}|}^{2} = 16 \Rightarrow |\vec{a}| = 4$

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Question 11:

If $\vec{a} . \vec{a} = 0 and \vec{a} . \vec{b} = 0,$ what can you conclude about the vector $\vec{b}$ ?

Answer:

$Given that \vec{a} . \vec{a} = 0 \Rightarrow {|\vec{a}|}^{2} = 0 \Rightarrow |\vec{a}| = 0 . . . (1) Also, given that \vec{a} . \vec{b} = 0 \Rightarrow |\vec{a}| |\vec{b}| \cos θ = 0 (where θ is the angle between \vec{a} and \vec{b}) \Rightarrow 0 |\vec{b}| \cos θ = 0 [From (1)] \Rightarrow 0 = 0 So, it means that for any vector \vec{b}, the given equation \vec{a} . \vec{b} = 0 is satisfied.$

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Question 12:

If $\vec{b}$ is a unit vector such that $(\vec{a} + \vec{b}) . (\vec{a} - \vec{b}) = 8, find |\vec{a}| .$

Answer:

$Given that \vec{b} is a unit vector. ∴ |\vec{b}| = 1 And (\vec{a} + \vec{b}) . (\vec{a} - \vec{b}) = 8 (Given) \Rightarrow {|\vec{a}|}^{2} - {|\vec{b}|}^{2} = 8 \Rightarrow {|\vec{a}|}^{2} - 1^{2} = 8 \Rightarrow {|\vec{a}|}^{2} = 9 ∴ |\vec{a}| = 3$

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Question 13:

If $\hat{a}, \hat{b}$ are unit vectors such that $\hat{a} + \hat{b}$ is a unit vector, write the value of $|\hat{a} - \hat{b}| .$

Answer:

$Given that \hat{a} and \hat{b} are unit vectors such that \hat{a} + \hat{b} is a unit vector. \Rightarrow |\hat{a}| = |\hat{b}| = |\hat{a} + \hat{b}| = 1 . . . (1) Now, |\hat{a} + \hat{b}| = 1 Squaring both sides, we get {|\hat{a}|}^{2} + {|\hat{b}|}^{2} + 2 \hat{a} . \hat{b} = 1 \Rightarrow 1 + 1 + 2 \hat{a} . \hat{b} = 1 [From (1)] \Rightarrow \hat{a} . \hat{b} = \frac{- 1}{2} . . . (2) Now, {|\hat{a} - \hat{b}|}^{2} = {|\hat{a}|}^{2} + {|\hat{b}|}^{2} - 2 \hat{a} . \hat{b} = 1 + 1 - 2 (\frac{- 1}{2}) = 3 [From (1) and (2)] ∴ |\hat{a} - \hat{b}| = \sqrt{3}$

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Question 14:

If $|\vec{a}| = 2, |\vec{b}| = 5 and \vec{a} . \vec{b} = 2, find |\vec{a} - \vec{b}| .$

Answer:

$We have |\vec{a}| = 2, |\vec{b}| = 5 and \vec{a} . \vec{b} = 2 Now, {|\vec{a} - \vec{b}|}^{2} = {|\vec{a}|}^{2} + {|\vec{b}|}^{2} - 2 \vec{a} . \vec{b} = 2^{2} + 5^{2} - 2 (2) = 4 + 25 - 4 = 25 ∴ |\vec{a} - \vec{b}| = \sqrt{25} = 5$

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Question 15:

If $\vec{a} = \hat{i} - \hat{j} and \vec{b} = - \hat{j} + \hat{k},$ find the projection of $\vec{a} on \vec{b}$ .

Answer:

$We have \vec{a} = \hat{i} - \hat{j} a n d \vec{b} = - \hat{j} + \hat{k} The projection of \vec{a} on \vec{b} is \frac{\vec{a} . \vec{b}}{|\vec{b}|} = \frac{(\hat{i} - \hat{j}) . (- \hat{j} + \hat{k})}{|- \hat{j} + \hat{k}|} = \frac{0 + 1 + 0}{\sqrt{1 + 1}} = \frac{1}{\sqrt{2}}$

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Question 16:

For any two non-zero vectors, write the value of $\frac{{|\vec{a} + \vec{b}|}^{2} + {|\vec{a} - \vec{b}|}^{2}}{{|\vec{a}|}^{2} + {|\vec{b}|}^{2}} .$

Answer:

$We have \frac{{|\vec{a} + \vec{b}|}^{2} + {|\vec{a} - \vec{b}|}^{2}}{{|\vec{a}|}^{2} + {|\vec{b}|}^{2}} = \frac{{|\vec{a}|}^{2} + {|\vec{b}|}^{2} + 2 \vec{a} . \vec{b} + {|\vec{a}|}^{2} + {|\vec{b}|}^{2} - 2 \vec{a} . \vec{b}}{{|\vec{a}|}^{2} + {|\vec{b}|}^{2}} = \frac{2 ({|\vec{a}|}^{2} + {|\vec{b}|}^{2})}{{|\vec{a}|}^{2} + {|\vec{b}|}^{2}} = 2$

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Question 17:

Write the projections of $\vec{r} = 3 \hat{i} - 4 \hat{j} + 12 \hat{k}$ on the coordinate axes.

Answer:

$We have \vec{r} = 3 \hat{i} - 4 \hat{j} + 12 \hat{k} Projection of \vec{r} on x -axis = \frac{\vec{r} . \hat{i}}{|\hat{i}|} = \frac{3}{1} =3 Projection of \vec{r} on y -axis = \frac{\vec{r} . \hat{j}}{|\hat{j}|} = \frac{- 4}{1} =-4 Projection of \vec{r} on z -axis = \frac{\vec{r} . \hat{k}}{|\hat{k}|} = \frac{12}{1} =12$

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Question 18:

Write the component of $\vec{b}$ along $\vec{a}$ .

Answer:

$Component of \vec{b} on \vec{a} is \{\frac{\vec{a} . \vec{b}}{|a|}\} \hat{a} = \{\frac{(\vec{a} . \vec{b})}{{|\vec{a}|}^{2}}\} \vec{a} = \frac{(\vec{a} . \vec{b}) \vec{a}}{{|\vec{a}|}^{2}}$

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Question 19:

Write the value of $(\vec{a} . \hat{i}) \hat{i} + (\vec{a} . \hat{j}) \hat{j} + (\vec{a} . \hat{k}) \hat{k},$ where $\vec{a}$ is any vector.

Answer:

$Let \vec{a} = a_{1} \vec{i} + a_{2} \vec{j} + a_{3} \vec{k} Now, (\vec{a} . \vec{i}) \vec{i} + (\vec{a} . \vec{j}) \vec{j} + (\vec{a} . \vec{k}) \vec{k} = a_{1} \vec{i} + a_{2} \vec{j} + a_{3} \vec{k} = \vec{a}$

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Question 20:

Find the value of θ ∈(0, π/2) for which vectors $\vec{a} = (\sin θ) \hat{i} + (\cos θ) \hat{j} and \vec{b} = \hat{i} - \sqrt{3} \hat{j} + 2 \hat{k}$ are perpendicular.

Answer:

$We have \vec{a} = (\sin θ) \hat{i} + (\cos θ) \hat{j} and \vec{b} = \hat{i} - \sqrt{3} \hat{j} + 2 \hat{k} It is given that the vectors are perpendicular. \Rightarrow \vec{a} . \vec{b} = 0 \Rightarrow \sin θ - \sqrt{3} \cos θ = 0 \Rightarrow \sin θ = \sqrt{3} \cos θ \Rightarrow \tan θ = \sqrt{3} \Rightarrow θ = \frac{π}{3}$

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Question 21:

Write the projection of $\hat{i} + \hat{j} + \hat{k}$ along the vector $\hat{j}$ .

Answer:

$Projection of \vec{a} on \vec{b} = \frac{\vec{a} . \vec{b}}{|\vec{b}|} Projection of \vec{i} + \vec{j} + \vec{k} along \vec{j} = \frac{(\vec{i} + \vec{j} + \vec{k}) . \vec{j}}{|\vec{j}|} = \frac{1}{1} = 1$

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Question 22:

Write a vector satisfying $\vec{a} . \hat{i} = \vec{a} . (\hat{i} + \hat{j}) = \vec{a} . (\hat{i} + \hat{j} + \hat{k}) = 1 .$

Answer:

$Let \vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k} \vec{a} . \hat{i} = a_{1} \vec{a} . (\hat{i} + \hat{j}) = a_{1} + a_{2} \vec{a} . (\hat{i} + \hat{j} + \hat{k}) = a_{1} + a_{2} + a_{3} Given that \vec{a} . \hat{i} = \vec{a} . (\hat{i} + \hat{j}) = \vec{a} . (\hat{i} + \hat{j} + \hat{k}) = 1 \Rightarrow a_{1} = a_{1} + a_{2} = a_{1} + a_{2} + a_{3} = 1 \Rightarrow a_{1} = 1; a_{1} + a_{2} = 1; a_{1} + a_{2} + a_{3} = 1 \Rightarrow a_{1} = 1; 1 + a_{2} = 1; 1 + a_{2} + a_{3} = 1 \Rightarrow a_{1} = 1; a_{2} = 0; 1 + 0 + a_{3} = 1 \Rightarrow a_{1} = 1; a_{2} = 0; a_{3} = 0 So, \vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k} = 1 \hat{i} + 0 \hat{j} + 0 \hat{k} = \hat{i}$

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Question 23:

If $\vec{a} and \vec{b}$ are unit vectors, find the angle between $\vec{a} + \vec{b} and \vec{a} - \vec{b} .$

Answer:

$We have |\vec{a}| = 1 and |\vec{b}| = 1 . . . (i) Now, (\vec{a} + \vec{b}) . (\vec{a} - \vec{b}) = {|\vec{a}|}^{2} - {|\vec{b}|}^{2} = 1^{2} - 1^{2} [Using (i)] = 0 \Rightarrow \vec{a} + \vec{b} and \vec{a} - \vec{b} are perpendicular. ∴ Angle between (\vec{a} + \vec{b}) and (\vec{a} - \vec{b}) {is 90}^{0} .$

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Question 24:

If $\vec{a} and \vec{b}$ are mutually perpendicular unit vectors, write the value of $|\vec{a} + \vec{b}| .$

Answer:

$\vec{a} and \vec{b} are unit vectors and they are perpendicular. \Rightarrow |\vec{a}| = |\vec{b}| = 1; \vec{a} . \vec{b} = 0 . . . (i) Now, {|\vec{a} + \vec{b}|}^{2} = {|\vec{a}|}^{2} + {|\vec{b}|}^{2} + 2 \vec{a} . \vec{b} = 1 + 1 + 2 (0) [Using (i)] = 2 ∴ |\vec{a} + \vec{b}| = \sqrt{2}$

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Question 25:

If $\vec{a}, \vec{b} and \vec{c}$ are mutually perpendicular unit vectors, write the value of $|\vec{a} + \vec{b} + \vec{c}| .$

Answer:

$Given that \vec{a}, \vec{b} and \vec{c} are unit vectors. So, |\vec{a}| =1, |\vec{b}| =1 and |\vec{c}| =1 ... (i) Since they are mutually perpendicular, \vec{a} . \vec{b} = \vec{b} . \vec{c} = \vec{c} . \vec{a} = 0 ... (ii) Now, {|\vec{a} + \vec{b} + \vec{c}|}^{2} = {|\vec{a}|}^{2} + {|\vec{b}|}^{2} + {|\vec{c}|}^{2} + 2 \vec{a} . \vec{b} + 2 \vec{b} . \vec{c} + 2 \vec{c} . \vec{a} = 1 + 1 + 1 + 0 + 0 + 0 [Using (i) and (ii)] = 3 ∴ |\vec{a} + \vec{b} + \vec{c}| = \sqrt{3}$

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Question 26:

Find the angle between the vectors $\vec{a} = \hat{i} - \hat{j} + \hat{k} and \vec{b} = \hat{i} + \hat{j} - \hat{k} .$

Answer:

$We have \vec{a} = \hat{i} - \hat{j} + \hat{k} and \vec{b} = i + j - k Let θ be the angle between \vec{a} and \vec{b} . |\vec{a}| = \sqrt{{(1)}^{2} + {(- 1)}^{2} + {(1)}^{2}} = \sqrt{3} |\vec{b}| = \sqrt{{(1)}^{2} + {(1)}^{2} + {(- 1)}^{2}} = \sqrt{3} and \vec{a} . \vec{b} = 1 - 1 - 1 = - 1 Now, \cos θ = \frac{\vec{a} . \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{- 1}{\sqrt{3} \sqrt{3}} = \frac{- 1}{3} ∴ θ = \cos^{- 1} (\frac{- 1}{3})$

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Question 27:

For what value of λ are the vectors $\vec{a} = 2 \hat{i} + λ \hat{j} + \hat{k} and \vec{b} = \hat{i} - 2 \hat{j} + 3 \hat{k}$ perpendicular to each other?

Answer:

$We have \vec{a} = 2 \hat{i} + λ \hat{j} + \hat{k} and \vec{b} = \hat{i} - 2 \hat{j} + 3 \hat{k} Given that \vec{a} and \vec{b} are perpendicular. \Rightarrow \vec{a} . \vec{b} = 0 \Rightarrow (2 \hat{i} + λ \hat{j} + \hat{k}) . (\hat{i} - 2 \hat{j} + 3 \hat{k}) = 0 \Rightarrow 2 - 2 λ + 3 = 0 \Rightarrow 5 - 2 λ = 0 ∴ λ = \frac{5}{2}$

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Question 28:

Find the projection of $\vec{a} on \vec{b} if \vec{a} \cdot \vec{b} = 8 and \vec{b} = 2 \hat{i} + 6 \hat{j} + 3 \hat{k} .$

Answer:

$We have \vec{a} . \vec{b} = 8 and \vec{b} = 2 \hat{i} + 6 \hat{j} + 3 \hat{k} The projection of \vec{a} on \vec{b} is \frac{\vec{a} . \vec{b}}{|\vec{b}|} = \frac{8}{\sqrt{4 + 36 + 9}} = \frac{8}{7}$

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Question 29:

Write the value of p for which $\vec{a} = 3 \hat{i} + 2 \hat{j} + 9 \hat{k} and \vec{b} = \hat{i} + p \hat{j} + 3 \hat{k}$ are parallel vectors.

Answer:

$We have \vec{a} = 3 \hat{i} + 2 \hat{j} + 9 \hat{k} and \vec{b} = \hat{i} + p \hat{j} + 3 \hat{k} Given that \vec{a} and \vec{b} are parallel. \Rightarrow \vec{a} = t \vec{b} for some t . \Rightarrow 3 \hat{i} + 2 \hat{j} + 9 \hat{k} = t (\hat{i} + p \hat{j} + 3 \hat{k}) \Rightarrow 3 \hat{i} + 2 \hat{j} + 9 \hat{k} = t \hat{i} + p t \hat{j} + 3 t \hat{k} Comparing both sides, we get 3 = t, 2 = p t and 9 = 3 t \Rightarrow t = 3 and p t = 2 \Rightarrow 3 t = 2 ∴ t = \frac{2}{3}$

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Question 30:

Find the value of λ if the vectors $2 \hat{i} + λ \hat{j} + 3 \hat{k} and 3 \hat{i} + 2 \hat{j} - 4 \hat{k}$ are perpendicular to each other.

Answer:

$Given : 2 \hat{i} + λ \hat{j} + 3 \hat{k} and 3 i + 2 j - 4 k are perpendicular to each other . So, their dot product is zero. (2 \hat{i} + λ \hat{j} + 3 \hat{k}) . (3 i + 2 j - 4 k) \Rightarrow 6 + 2 λ - 12 = 0 \Rightarrow 2 λ - 6 = 0 \Rightarrow λ = 3$

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Question 31:

If $|\vec{a}| = 2, |\vec{b}| = 3 and \vec{a} \cdot \vec{b} = 3,$ find the projection of $\vec{b} on \vec{a}$ .

Answer:

$We have |\vec{a}| = 2 and \vec{a} . \vec{b} = 3 So,the projection of \vec{b} on \vec{a} is (\frac{\vec{a} . \vec{b}}{|\vec{a}|}) = \frac{3}{2}$

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Question 32:

Write the angle between two vectors $\vec{a} and \vec{b}$ with magnitudes $\sqrt{3}$ and 2 respectively if $\vec{a} \cdot \vec{b} = \sqrt{6} .$

Answer:

$Let θ be the angle between \vec{a} and \vec{b} . Given, |\vec{a}| = \sqrt{3}; |\vec{b}| = 2; \vec{a} . \vec{b} = \sqrt{6} We know that \vec{a} . \vec{b} = |\vec{a}| |\vec{b}| \cos θ \Rightarrow \sqrt{6} = (\sqrt{3}) (2) \cos θ \Rightarrow \cos θ = \frac{\sqrt{6}}{2 \sqrt{3}} = \frac{1}{\sqrt{2}} \Rightarrow θ = \cos^{- 1} (\frac{1}{\sqrt{2}}) = \frac{π}{4}$

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Question 33:

Write the projection of the vector $\hat{i} + 3 \hat{j} + 7 \hat{k}$ on the vector $2 \hat{i} - 3 \hat{j} + 6 \hat{k}$ .

Answer:

We know that projection of $\vec{a}$ on $\vec{b}$ = $\frac{\vec{a} . \vec{b}}{|\vec{b}|}$ .

Let $\vec{a} = \hat{i} + 3 \hat{j} + 7 \hat{k}$ and $\vec{b} = 2 \hat{i} - 3 \hat{j} + 6 \hat{k}$ .

∴ Projection of $\vec{a}$ on $\vec{b}$

$= \frac{(\hat{i} + 3 \hat{j} + 7 \hat{k}) . (2 \hat{i} - 3 \hat{j} + 6 \hat{k})}{|2 \hat{i} - 3 \hat{j} + 6 \hat{k}|} = \frac{1 \times 2 + 3 \times (- 3) + 7 \times 6}{\sqrt{2^{2} + {(- 3)}^{2} + 6^{2}}} = \frac{2 - 9 + 42}{\sqrt{49}} = \frac{35}{7} = 5$

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Question 34:

Find λ when the projection of $\vec{a} = λ \hat{i} + \hat{j} + 4 \hat{k} on \vec{b} = 2 \hat{i} + 6 \hat{j} + 3 \hat{k}$ is 4 units.

Answer:

$We have a = λ \hat{i} + \hat{j} + 4 \hat{k} and b = 2 \hat{i} + 6 \hat{j} + 3 \hat{k} The projection of \vec{a} on \vec{b} is \frac{\vec{a} . \vec{b}}{|\vec{b}|} Given that \frac{\vec{a} . \vec{b}}{|\vec{b}|} = 4 \Rightarrow \frac{(λ \hat{i} + \hat{j} + 4 \hat{k}) . (2 \hat{i} + 6 \hat{j} + 3 \hat{k})}{|2 \hat{i} + 6 \hat{j} + 3 \hat{k}|} \Rightarrow \frac{2 λ + 6 + 12}{\sqrt{4 + 36 + 9}} = 4 \Rightarrow \frac{2 λ + 18}{7} = 4 \Rightarrow 2 λ + 18 = 28 \Rightarrow 2 λ = 10 ∴ λ = 5$

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Question 35:

For what value of λ are the vectors $\vec{a} = 2 \hat{i} + λ \hat{j} + \hat{k} and \vec{b} = \hat{i} - 2 \hat{j} + 3 \hat{k}$ perpendicular to each other?

Answer:

$We have \vec{a} = 2 \hat{i} + λ \hat{j} + \hat{k} and \vec{b} = \hat{i} - 2 \hat{j} + 3 \hat{k} Given, \vec{a} and \vec{b} are perpendicular. So, their dot product is zero. \Rightarrow (2 \hat{i} + λ \hat{j} + \hat{k}) . (\hat{i} - 2 \hat{j} + 3 \hat{k}) = 0 \Rightarrow 2 - 2 λ + 3 = 0 \Rightarrow - 2 λ + 5 = 0 ∴ λ = \frac{5}{2}$

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Question 36:

Write the projection of the vector $7 \hat{i} + \hat{j} - 4 \hat{k}$ on the vector $2 \hat{i} + 6 \hat{j} + 3 \hat{k} .$

Answer:

$Let \vec{a} = 7 \hat{i} + \hat{j} - 4 \hat{k}; \vec{b} = 2 \hat{i} + 6 \hat{j} + 3 \hat{k} The projection of \vec{a} on \vec{b} is (\frac{\vec{a} . \vec{b}}{|\vec{b}|}) = \frac{(7 \hat{i} + \hat{j} - 4 \hat{k}) . (2 \hat{i} + 6 \hat{j} + 3 \hat{k})}{|2 \hat{i} + 6 \hat{j} + 3 \hat{k}|} = \frac{14 + 6 - 12}{\sqrt{4 + 36 + 9}} = \frac{8}{7}$

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Question 37:

Write the value of λ so that the vectors $\vec{a} = 2 \hat{i} + λ \hat{j} + \hat{k} and \vec{b} = \hat{i} - 2 \hat{j} + 3 \hat{k}$ are perpendicular to each other.

Answer:

$We have \vec{a} = 2 \hat{i} + λ \hat{j} + \hat{k} and \vec{b} = \hat{i} - 2 \hat{j} + 3 \hat{k} The given vectors are perpendicular. So, their dot product is zero. (2 \hat{i} + λ \hat{j} + \hat{k}) . (\hat{i} - 2 \hat{j} + 3 \hat{k}) \Rightarrow 2 - 2 λ + 3 = 0 \Rightarrow 5 - 2 λ = 0 \Rightarrow - 2 λ = - 5 \Rightarrow λ = \frac{5}{2}$

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Question 38:

Write the projection of $\vec{b} + \vec{c} on \vec{a} when \vec{a} = 2 \hat{i} - 2 \hat{j} + \hat{k}, \vec{b} = \hat{i} + 2 \hat{j} - 2 \hat{k} and \vec{c} = 2 \hat{i} - \hat{j} + 4 \hat{k} .$

Answer:

$Given that \vec{a} = 2 \hat{i} - 2 \hat{j} + \hat{k} \vec{b} = \hat{i} + 2 \hat{j} - 2 \hat{k} \vec{c} = 2 \hat{i} - \hat{j} + 4 \hat{k} \vec{b} + \vec{c} = \hat{i} + 2 \hat{j} - 2 \hat{k} + 2 \hat{i} - \hat{j} + 4 \hat{k} = 3 \hat{i} + \hat{j} + 2 \hat{k} Projection of \vec{b} + \vec{c} on \vec{a} is \frac{(\vec{b} + \vec{c}) . \vec{a}}{|\vec{a}|} = \frac{(3 \hat{i} + \hat{j} + 2 \hat{k}) . (2 \hat{i} - 2 \hat{j} + \hat{k})}{2 \hat{i} - 2 \hat{j} + \hat{k}} = \frac{6 - 2 + 2}{\sqrt{4 + 4 + 1}} = \frac{6}{3} = 2$

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Question 39:

If $\vec{a}$ and $\vec{b}$ are perpendicular vectors, $|\vec{a} + \vec{b}| = 3$ and $|\vec{a}| = 5$ , find the value of $|\vec{b}|$ . [CBSE 2014]

Answer:

Disclaimer: $|\vec{a} + \vec{b}| = 13$ has been taken in order to solve the question.

It is given that $\vec{a}$ and $\vec{b}$ are perpendicular vectors.

$∴ \vec{a} . \vec{b} = 0$ .....(1)

$|\vec{a} + \vec{b}| = 13 \Rightarrow {|\vec{a} + \vec{b}|}^{2} = 169 \Rightarrow {|\vec{a}|}^{2} + 2 \vec{a} . \vec{b} + {|\vec{b}|}^{2} = 169 \Rightarrow 25 + 2 \times 0 + {|\vec{b}|}^{2} = 169 [Using (1)]$
$\Rightarrow {|\vec{b}|}^{2} = 169 - 25 = 144 \Rightarrow |\vec{b}| = 12$

Thus, the value of $|\vec{b}|$ is 12.

Page No 23.52:

Question 40:

If the vectors $\vec{a}$ and $\vec{b}$ are such that $|\vec{a}| = 3, |\vec{b}| = \frac{2}{3}$ and $\vec{a} \times \vec{b}$ is a unit vector, then write the angle between $\vec{a}$ and $\vec{b}$ . [CBSE 2014]

Answer:

Let the angle between $\vec{a}$ and $\vec{b}$ be $θ$ .

It is given that $\vec{a} \times \vec{b}$ is a unit vector.

$∴ |\vec{a} \times \vec{b}| = 1 \Rightarrow |\vec{a}| |\vec{b}| \sin θ = 1 \Rightarrow 3 \times \frac{2}{3} \times \sin θ = 1 \Rightarrow \sin θ = \frac{1}{2} \Rightarrow θ = \frac{π}{6}$

Thus, the angle between $\vec{a}$ and $\vec{b}$ is $\frac{π}{6}$ .

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Question 41:

If $\vec{a}$ and $\vec{b}$ are two unit vectors such that $\vec{a} + \vec{b}$ is also a unit vector, then find the angle between $\vec{a}$ and $\vec{b}$ . [CBSE 2014]

Answer:

Let the angle between $\vec{a}$ and $\vec{b}$ be $θ$ .

It is given that $|\vec{a}| = |\vec{b}| = |\vec{a} + \vec{b}| = 1$ .

$|\vec{a} + \vec{b}| = 1 \Rightarrow {|\vec{a} + \vec{b}|}^{2} = 1 \Rightarrow {|\vec{a}|}^{2} + 2 |\vec{a}| |\vec{b}| \cos θ + {|\vec{b}|}^{2} = 1 \Rightarrow 1 + 2 \times 1 \times 1 \times \cos θ + 1 = 1 \Rightarrow 2 \cos θ = - 1 \Rightarrow \cos θ = - \frac{1}{2} = \cos \frac{2 π}{3} \Rightarrow θ = \frac{2 π}{3}$

Thus, the angle between $\vec{a}$ and $\vec{b}$ is $\frac{2 π}{3}$ .

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Question 42:

If $\vec{a}$ and $\vec{b}$ are unit vectors, then find the angle between $\vec{a}$ and $\vec{b}$ , given that $(\sqrt{3} \vec{a} - \vec{b})$ is a unit vector. [CBSE 2014]

Answer:

Let the angle between $\vec{a}$ and $\vec{b}$ be $θ$ .

It is given that $|\vec{a}| = |\vec{b}| = |\sqrt{3} \vec{a} - \vec{b}| = 1$ .

$|\sqrt{3} \vec{a} - \vec{b}| = 1 \Rightarrow {|\sqrt{3} \vec{a} - \vec{b}|}^{2} = 1 \Rightarrow {|\sqrt{3} \vec{a}|}^{2} - 2 \sqrt{3} \vec{a} . \vec{b} + {|\vec{b}|}^{2} = 1 \Rightarrow 3 {|\vec{a}|}^{2} - 2 \sqrt{3} |\vec{a}| |\vec{b}| \cos θ + {|\vec{b}|}^{2} = 1$
$\Rightarrow 3 \times 1 - 2 \sqrt{3} \times 1 \times 1 \times \cos θ + 1 = 1 \Rightarrow 2 \sqrt{3} \cos θ = 3 \Rightarrow \cos θ = \frac{\sqrt{3}}{2} = \cos \frac{π}{6} \Rightarrow θ = \frac{π}{6}$

Thus, the angle between $\vec{a}$ and $\vec{b}$ is $\frac{π}{6}$ .

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Question 43:

Find the magnitude of each of the two vectors $\vec{a} and \vec{b}$ , having the same magnitude such that the angle between them is 60° and their scalar product is $\frac{9}{2} .$

Answer:

Let the two vectors be $\vec{a} and \vec{b}$ .
Since the vectors have same magnitude so,
$|\vec{a}| = |\vec{b}|$
Scalar product of two vectors = $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos θ$
$\Rightarrow \frac{9}{2} = |\vec{a}| |\vec{b}| \cos 60 ° \Rightarrow \frac{9}{2} = |\vec{a}| |\vec{b}| \times \frac{1}{2} \Rightarrow |\vec{a}| |\vec{b}| = 9 \Rightarrow |\vec{a}| |\vec{a}| = 9 (From (i)) \Rightarrow {|\vec{a}|}^{2} = 9 \Rightarrow |\vec{a}| = |\vec{b}| = 3$

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