Rd Sharma XII Vol 2 2020 for Class 12 Science Maths Chapter 7 - Scalar Triple Product

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Page No 25.16:

Question 1:

Evaluate the following:
(i) $[\hat{i} \hat{j} \hat{k}] + [\hat{j} \hat{k} \hat{i}] + [\hat{k} \hat{i} \hat{j}]$

(ii) $[2 \hat{i} \hat{j} \hat{k}] + [\hat{i} \hat{k} \hat{j}] + [\hat{k} \hat{j} 2 \hat{i}]$

Answer:

$(i) We have [\hat{i} \hat{j} \hat{k}] + [\hat{j} \hat{k} \hat{i}] + [\hat{k} \hat{i} \hat{j}] = [\hat{i} \hat{j} \hat{k}] + [\hat{i} \hat{j} \hat{k}] + [\hat{i} \hat{j} \hat{k}] (∵ [\vec{a} \vec{b} \vec{c}] = [\vec{b} \vec{c} \vec{a}] = [\vec{c} \vec{a} \vec{b}]) = 3 [\hat{i} \hat{j} \hat{k}] = 3 (\hat{i} \times \hat{j}) . \hat{k} (∵ [\vec{a} \vec{b} \vec{c}] = (\vec{a} \times \vec{b}) . \vec{c}) = 3 (\hat{k} . \hat{k}) = 3 (1) = 3$

$(ii) We have [2 \hat{i} \hat{j} \hat{k}] + [\hat{i} \hat{k} \hat{j}] + [\hat{k} \hat{j} 2 \hat{i}] = 2 [\hat{i} \hat{j} \hat{k}] + [\hat{i} \hat{k} \hat{j}] + 2 [\hat{k} \hat{j} \hat{i}] (∵ [l \vec{a} m \vec{b} n \vec{c}] = l m n [\vec{a} \vec{b} \vec{c}]) = 2 (\hat{i} \times \hat{j}) . \hat{k} + (\hat{i} \times \hat{k}) . \hat{j} + 2 (\hat{k} \times \hat{j}) . \hat{i} (∵ [\vec{a} \vec{b} \vec{c}] = (\vec{a} \times \vec{b}) . \vec{c}) = 2 (\hat{k} . \hat{k}) + (- \hat{j} . \hat{j}) + 2 (- \hat{i} . \hat{i}) = 2 (1) + (- 1) + 2 (- 1) = 2 - 1 - 2 = - 1$

Page No 25.16:

Question 2:

Find $[\vec{a} \vec{b} \vec{c}]$ , when
(i) $\vec{a} = 2 \hat{i} - 3 \hat{j}, \vec{b} = \hat{i} + \hat{j} - \hat{k} and \vec{c} = 3 \hat{i} - \hat{k}$
(ii) $\vec{a} = \hat{i} - 2 \hat{j} + 3 \hat{k}, \vec{b} = 2 \hat{i} + \hat{j} - \hat{k} and \vec{c} = \hat{j} + \hat{k}$
(iii) $\vec{a} = 2 \hat{i} + 3 \hat{j} + \hat{k}, \vec{b} = \hat{i} - 2 \hat{j} + \hat{k} and \vec{c} = - 3 \hat{i} + \hat{j} + 2 \hat{k}$

Answer:

(iii)
$given \vec{a} = 2 \hat{i} + 3 \hat{j} + \hat{k} \vec{b} = \hat{i} - 2 \hat{j} + \hat{k} \vec{c} = - 3 \hat{i} + \hat{j} + 2 \hat{k} then [\vec{a} \vec{b} \vec{c}] = |\begin{matrix} 2 & 3 & 1 \\ 1 & - 2 & 1 \\ - 3 & 1 & 2 \end{matrix}|$

= 2(−4 − 1) −3(2 + 3) + (1 + (−6))

= 2(−5) − 3(5) + 1(−5)

= −10 − 15 − 5

i.e $[\vec{a} \vec{b} \vec{c}] = - 30$

Page No 25.16:

Question 3:

Find the volume of the parallelopiped whose coterminous edges are represented by the vectors:
(i) $\vec{a} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}, \vec{b} = \hat{i} + 2 \hat{j} - \hat{k}, \vec{c} = 3 \hat{i} - \hat{j} + 2 \hat{k}$

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

(iii) $\vec{a} = 11 \hat{i}, \vec{b} = 2 \hat{j}, \vec{c} = 13 \hat{k}$

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

Answer:

$(i) Given : \vec{a} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k} \vec{b} = \hat{i} + 2 \hat{j} - \hat{k} \vec{c} = 3 \hat{i} - \hat{j} + 2 \hat{k} We know that the volume of a parallelopiped whose three adjacent edges are \vec{a}, \vec{b}, \vec{c} is equal to |[\vec{a} \vec{b} \vec{c}]| . Here, [\vec{a} \vec{b} \vec{c}] = |\begin{matrix} 2 & 3 & 4 \\ 1 & 2 & - 1 \\ 3 & - 1 & 2 \end{matrix}| = 2 (4 - 1) - 3 (2 + 3) + 4 (- 1 - 6) = - 37 Volume of the parallelopiped = |[\vec{a} \vec{b} \vec{c}]| = |- 37| = 37 cubic units$

$(ii) Given : \vec{a} = 2 \hat{i} - 3 \hat{j} + 4 \hat{k} \vec{b} = \hat{i} + 2 \hat{j} - \hat{k} \vec{c} = 3 \hat{i} - \hat{j} - 2 \hat{k} We know that the volume of a parallelopiped whose three adjacent edges are \vec{a}, \vec{b}, \vec{c} is equal to |[\vec{a} \vec{b} \vec{c}]| . Here, [\vec{a} \vec{b} \vec{c}] = |\begin{matrix} 2 & - 3 & 4 \\ 1 & 2 & - 1 \\ 3 & - 1 & - 2 \end{matrix}| = 2 (- 4 - 1) + 3 (- 2 + 3) + 4 (- 1 - 6) = - 35 Volume of the parallelopiped = |[\vec{a} \vec{b} \vec{c}]| = |- 35| = 35 cubic units$

$(iii) Given : \vec{a} = 11 \hat{i} \vec{b} = 2 \hat{j} \vec{c} = 13 \hat{k} We know that the volume of a parallelopiped whose three adjacent edges are \vec{a}, \vec{b}, \vec{c} is equal to |[\vec{a} \vec{b} \vec{c}]| . Here, [\vec{a} \vec{b} \vec{c}] = |\begin{matrix} 11 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 13 \end{matrix}| = 11 (26 - 0) - 0 (0 - 0) + 0 (0 - 0) = 286 Volume of the parallelopiped = |[\vec{a} \vec{b} \vec{c}]| = |286| = 286 cubic units$

$(iv) Given : \vec{a} = \hat{i} + \hat{j} + \hat{k} \vec{b} = \hat{i} - \hat{j} + k \vec{c} = \hat{i} + 2 j - \hat{k} We know that the volume of a parallelopiped whose three adjacent edges are \vec{a}, \vec{b}, \vec{c} is equal to |[\vec{a} \vec{b} \vec{c}]| . Here, [\vec{a} \vec{b} \vec{c}] = |\begin{matrix} 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 2 & - 1 \end{matrix}| = 1 (1 - 2) - 1 (- 1 - 1) + 1 (2 + 1) = 4 Volume of the parallelopiped = |[\vec{a} \vec{b} \vec{c}]| = |4| = 4 cubic units$

Page No 25.16:

Question 4:

Show the each of the following triads of vectors are coplanar:
(i) $\vec{a} = \hat{i} + 2 \hat{j} - \hat{k}, \vec{b} = 3 \hat{i} + 2 \hat{j} + 7 \hat{k}, \vec{c} = 5 \hat{i} + 6 \hat{j} + 5 \hat{k}$

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

(iii) $\hat{a} = \hat{i} - 2 \hat{j} + 3 \hat{k}, \hat{b} = - 2 \hat{i} + 3 \hat{j} - 4 \hat{k}, \hat{c} = \hat{i} - 3 \hat{j} + 5 \hat{k}$

Answer:

$(i) Given : \vec{a} = \hat{i} + 2 \hat{j} - \hat{k} \vec{b} = 3 \hat{i} + 2 \hat{j} + 7 \hat{k} \vec{c} = 5 \hat{i} + 6 \hat{j} + 5 \hat{k} We know that three vectors \vec{a}, \vec{b}, \vec{c} are coplanar iff their scalar triple product is zero, i . e . [\vec{a} \vec{b} \vec{c}] = 0 Here, [\vec{a} \vec{b} \vec{c}] = |\begin{matrix} 1 & 2 & - 1 \\ 3 & 2 & 7 \\ 5 & 6 & 5 \end{matrix}| = 1 (10 - 42) - 2 (15 - 35) - 1 (18 - 10) = 0 Hence, the given vectors are coplanar .$

$(ii) Given : \vec{a} = - 4 \hat{i} - 6 \hat{j} - 2 \hat{k} \vec{b} = - \hat{i} + 4 \hat{j} + 3 \hat{k} \vec{c} = - 8 \hat{i} - \hat{j} + 3 \hat{k} We know that three vectors \vec{a}, \vec{b}, \vec{c} are coplanar iff their scalar triple product is zero, i . e . [\vec{a} \vec{b} \vec{c}] = 0 . Here, [\vec{a} \vec{b} \vec{c}] = |\begin{matrix} - 4 & - 6 & - 2 \\ - 1 & 4 & 3 \\ - 8 & - 1 & 3 \end{matrix}| = - 4 (12 + 3) + 6 (- 3 + 24) - 2 (1 + 32) = 0 Hence, the given vectors are coplanar .$

$(iii) Given : \vec{a} = \hat{i} - 2 \hat{j} + 3 \hat{k} \vec{b} = - 2 \hat{i} + 3 \hat{j} - 4 \hat{k} \vec{c} = \hat{i} - 3 \hat{j} + 5 \hat{k} We know that three vectors \vec{a}, \vec{b}, \vec{c} are coplanar iff their scalar triple product is zero, i . e . [\vec{a} \vec{b} \vec{c}] = 0 . Here, [\vec{a} \vec{b} \vec{c}] = |\begin{matrix} 1 & - 2 & 3 \\ - 2 & 3 & - 4 \\ 1 & - 3 & 5 \end{matrix}| = 1 (15 - 12) + 2 (- 10 + 4) + 3 (6 - 3) = 0 Hence, the given vectors are coplanar .$

Page No 25.16:

Question 5:

Find the value of λ so that the following vectors are coplanar:
(i) $\vec{a} = \hat{i} - \hat{j} + \hat{k}, \vec{b} = 2 \hat{i} + \hat{j} - \hat{k}, \vec{c} = λ \hat{i} - \hat{j} + λ \hat{k}$

(ii) $\vec{a} = 2 \hat{i} - \hat{j} + \hat{k}, \vec{b} = \hat{i} + 2 \hat{j} - 3 \hat{k}, \vec{c} = λ \hat{i} + λ \hat{j} + 5 \hat{k}$

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

(iv) $\vec{a} = \hat{i} + 3 \hat{j}, \vec{b} = 5 \hat{k}, \vec{c} = λ \hat{i} - \hat{j}$

Answer:

$(i) Given : \vec{a} = \hat{i} - \hat{j} + \hat{k} \vec{b} = 2 \hat{i} + \hat{j} - \hat{k} \vec{c} = λ \hat{i} - j + λ \hat{k} We know that vectors \vec{a}, \vec{b}, \vec{c} are coplanar iff [\vec{a} \vec{b} \vec{c}] = 0 . It is given that \vec{a}, \vec{b}, \vec{c} are coplanar . ∴ [\vec{a} \vec{b} \vec{c}] = 0 \Rightarrow |\begin{matrix} 1 & - 1 & 1 \\ 2 & 1 & - 1 \\ λ & - 1 & λ \end{matrix}| = 0 \Rightarrow 1 (λ - 1) + 1 (2 λ + λ) + 1 (- 2 - λ) = 0 \Rightarrow λ - 1 + 3 λ - 2 - λ = 0 \Rightarrow 3 λ - 3 = 0 \Rightarrow λ = 1$

$(ii) Given : \vec{a} = 2 \hat{i} - \hat{j} + \hat{k} \vec{b} = \hat{i} + 2 \hat{j} - 3 \hat{k} \vec{c} = λ \hat{i} + λ j + 5 \hat{k} We know that vectors \vec{a}, \vec{b}, \vec{c} are coplanar iff [\vec{a} \vec{b} \vec{c}] = 0 . It is given that \vec{a}, \vec{b}, \vec{c} are coplanar . ∴ [\vec{a} \vec{b} \vec{c}] = 0 \Rightarrow |\begin{matrix} 2 & - 1 & 1 \\ 1 & 2 & - 3 \\ λ & λ & 5 \end{matrix}| = 0 \Rightarrow 2 (10 + 3 λ) + 1 (5 + 3 λ) + 1 (λ - 2 λ) = 0 \Rightarrow 8 λ + 25 = 0 \Rightarrow λ = - \frac{25}{8}$

$(iii) Given : \vec{a} = \hat{i} + 2 \hat{j} - 3 \hat{k} \vec{b} = 3 \hat{i} + λ \hat{j} + \hat{k} \vec{c} = \hat{i} + 2 \hat{j} + 2 \hat{k} We know that vectors \vec{a}, \vec{b}, \vec{c} are coplanar iff [\vec{a} \vec{b} \vec{c}] = 0 . It is given that \vec{a}, \vec{b}, \vec{c} are coplanar . ∴ [\vec{a} \vec{b} \vec{c}] = 0 \Rightarrow |\begin{matrix} 1 & 2 & - 3 \\ 3 & λ & 1 \\ 1 & 2 & 2 \end{matrix}| = 0 \Rightarrow 1 (2 λ - 2) - 2 (6 - 1) - 3 (6 - λ) = 0 \Rightarrow 5 λ - 30 = 0 \Rightarrow λ = 6$

$(iv) Given : \vec{a} = \hat{i} + 3 \hat{j} \vec{b} = 5 \hat{k} \vec{c} = λ \hat{i} - \hat{j} We know that vectors \vec{a}, \vec{b}, \vec{c} are coplanar iff [\vec{a} \vec{b} \vec{c}] = 0 . It is given that \vec{a}, \vec{b}, \vec{c} are coplanar . ∴ [\vec{a} \vec{b} \vec{c}] = 0 \Rightarrow |\begin{matrix} 1 & 3 & 0 \\ 0 & 0 & 5 \\ λ & - 1 & 0 \end{matrix}| = 0 \Rightarrow 1 (0 + 5) - 3 (0 - 5 λ) + 0 (0 - 0) = 0 \Rightarrow 5 + 15 λ = 0 \Rightarrow λ = - \frac{1}{3}$

Page No 25.17:

Question 6:

Show that the four points having position vectors $6 \hat{i} - 7 \hat{j}, 16 \hat{i} - 19 \hat{j} - 4 \hat{k}, 3 \hat{j} - 6 \hat{k}, 2 \hat{i} + 5 \hat{j} + 10 \hat{k}$ are not coplanar.

Answer:

$Let A, B, C and D be the given points . These points will be coplanar iff any one of the following triads of vectors are coplanar : \vec{A B}, \vec{A C}, \vec{A D}; \vec{A B}, \vec{B C}, \vec{C D}; \vec{B C}, \vec{B A}, \vec{B D,} etc . T o show that \vec{A B}, \vec{A C}, \vec{A D} are not coplanar, we have to p r o v e that their scaler triple product, i . e . [\vec{A B} \vec{A C} \vec{A D}] \neq 0 Now, \vec{P Q} = (Position vector of Q) - (Position vector of P) \vec{A B} = (16 \hat{i} - 19 \hat{j} - 4 \hat{k}) - (6 \hat{i} - 7 \hat{j}) = 10 \hat{i} - 12 \hat{j} - 4 \hat{k} \vec{A C} = (3 \hat{j} - 6 \hat{k}) - (6 \hat{i} - 7 \hat{j}) = - 6 \hat{i} + 10 \hat{j} - 6 \hat{k} \vec{A D} = (2 \hat{i} + 5 \hat{j} + 10 \hat{k}) - (6 \hat{i} - 7 \hat{j}) = - 4 \hat{i} + 12 \hat{j} + 10 \hat{k} ∴ [\vec{A B} \vec{A C} \vec{A D}] = |\begin{matrix} 10 & - 12 & - 4 \\ - 6 & 10 & - 6 \\ - 4 & 12 & 10 \end{matrix}| = 10 (100 + 72) + 12 (- 60 - 24) - 4 (- 72 + 40) = 840 \neq 0 Thus, the given points are not coplanar .$

Page No 25.17:

Question 7:

Show that the points A (−1, 4, −3), B (3, 2, −5), C (−3, 8, −5) and D (−3, 2, 1) are coplanar.

Answer:

$The points A, B, C and D will be coplanar iff any one of the following triads of vectors are coplanar : \vec{A B}, \vec{A C}, \vec{A D}; \vec{A B}, \vec{B C}, \vec{C D}; \vec{B C}, \vec{B A}, \vec{B D,} etc . T o show that \vec{A B}, \vec{A C}, \vec{A D} are coplanar, we have to prove that their scaler triple product, i . e . [\vec{A B} \vec{A C} \vec{A D}] = 0 Now, \vec{A B} = [3 - (- 1)] \hat{i} + (2 - 4) \hat{j} + [- 5 - (- 3)] \hat{k} = 4 \hat{i} - 2 \hat{j} - 2 \hat{k} \vec{A C} = [- 3 - (- 1)] \hat{i} + (8 - 4) \hat{j} + [- 5 - (- 3)] \hat{k} = - 2 \hat{i} + 4 \hat{j} - 2 \hat{k} \vec{A D} = [- 3 - (- 1)] \hat{i} + (2 - 4) \hat{j} + [1 - (- 3)] \hat{k} = - 2 \hat{i} - 2 \hat{j} + 4 \hat{k} ∴ [\vec{A B} \vec{A C} \vec{A D}] = |\begin{matrix} 4 & - 2 & - 2 \\ - 2 & 4 & - 2 \\ - 2 & - 2 & 4 \end{matrix}| = 4 (16 - 4) + 2 (- 8 - 4) - 2 (4 + 8) = 0 Thus, the given points are coplanar .$

Page No 25.17:

Question 8:

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

Answer:

DISCLAIMER: Given points are not coplanar.

$Let A, B, C, D be the given points . The given points will be coplanar iff any one of the following triads of vectors are coplanar : \vec{A B}, \vec{A C}, \vec{A D}; \vec{A B}, \vec{B C}, \vec{C D}; \vec{B C}, \vec{B A}, \vec{B D} etc . In order to show that \vec{A B}, \vec{A C}, \vec{A D} are coplanar, we will have to show that their scaler triple product i . e . [\vec{A B} \vec{A C} \vec{A D}] = 0 Using, \vec{P Q} = Position vector of Q - Position vector of P, we obtain Now, \vec{A B} = (16 \hat{i} - 19 \hat{j} - 4 \hat{k}) - (6 \hat{i} - 7 \hat{j}) = 10 \hat{i} - 12 \hat{j} - 4 \hat{k} \vec{A C} = (3 \hat{i} - 6 \hat{k}) - (6 \hat{i} - 7 \hat{j}) = - 3 \hat{i} + 7 \hat{j} - 6 \hat{k} and, \vec{A D} = (2 \hat{i} - 5 \hat{j} + 10 \hat{k}) - (6 \hat{i} - 7 \hat{j}) = - 4 \hat{i} + 2 \hat{j} + 10 \hat{k} ∴ [\vec{A B} \vec{A C} \vec{A D}] = |\begin{matrix} 10 & - 12 & - 4 \\ - 3 & 7 & - 6 \\ - 4 & 2 & 10 \end{matrix}| = 10 (70 + 12) + 12 (- 30 - 24) - 4 (- 6 + 28) = 84 Thus, the given points are not coplanar .$

Page No 25.17:

Question 9:

Find the value of λ for which the four points with position vectors $- \hat{j} - \hat{k}, 4 \hat{i} + 5 \hat{j} + λ \hat{k}, 3 \hat{i} + 9 \hat{j} + 4 \hat{k} and - 4 \hat{i} + 4 \hat{j} + 4 \hat{k}$ are coplanar.

Answer:

$Let A, B, C and D be the given points . Then, \vec{A B} = (4 \hat{i} + 5 \hat{j} + λ \hat{k}) - (0 \hat{i} - \overset{⏜}{j} - \overset{⏜}{k}) = 4 \hat{i} + 6 \hat{j} + (λ + 1) \hat{k} \vec{A C} = (3 \hat{i} + 9 \hat{j} + 4 \hat{k}) - (0 \hat{i} - \hat{j} - \hat{k}) = 3 \hat{i} + 10 \hat{j} + 5 \hat{k} \vec{A D} = (- 4 \hat{i} + 4 \hat{j} + 4 \hat{k}) - (0 \hat{i} - \hat{j} - \hat{k}) = - 4 \hat{i} + 5 \hat{j} + 5 \hat{k} The given points are coplanar iff vector s \vec{A B}, \vec{A C}, \vec{A D} are coplanar . Now, \vec{A B}, \vec{A C}, \vec{A D} are coplanar . \Rightarrow [\begin{matrix} \vec{A B} & \vec{A C} & \vec{A D} \end{matrix}] = 0 \Rightarrow |\begin{matrix} 4 & 6 & (λ + 1) \\ 3 & 10 & 5 \\ - 4 & 5 & 5 \end{matrix}| = 0 \Rightarrow 4 (50 - 25) - 6 (15 + 20) + (λ + 1) (15 + 40) = 0 \Rightarrow 100 - 210 + 55 λ + 55 = 0 \Rightarrow 55 λ = 55 \Rightarrow λ = 1$

Page No 25.17:

Question 10:

Prove that: $(\vec{a} - \vec{b}) \cdot \{(\vec{b} - \vec{c}) \times (\vec{c} - \vec{a})\} = 0$

Answer:

$We have (\vec{a} - \vec{b}) . \{(\vec{b} - \vec{c}) \times (\vec{c} - \vec{a})\} = (\vec{a} - \vec{b}) . (\vec{b} \times \vec{c} - \vec{b} \times \vec{a} - \vec{c} \times \vec{c} + \vec{c} \times \vec{a}) (By distributive law) = (\vec{a} - \vec{b}) . (\bar{b} \times \vec{c} - \vec{b} \times \vec{a} + \vec{c} \times \vec{a}) (∵ \vec{c} \times \vec{c} = 0) = (\vec{a} - \vec{b}) . (\bar{b} \times \vec{c} + \vec{a} \times \vec{b} + \vec{c} \times \vec{a}) = \vec{a} . (\vec{b} \times \vec{c}) + \vec{a} . (\overset{⇀}{a} \times \vec{b}) + \vec{a} . (\vec{c} \times \vec{a}) - \vec{b} . (\vec{b} \times \vec{c}) - \vec{b} . (\vec{a} \times \vec{b}) - \vec{b} . (\vec{c} \times \vec{a}) (By distributive law) = [\begin{matrix} \vec{a} & \vec{b} & \vec{c} \end{matrix}] + [\begin{matrix} \vec{a} & \vec{a} & \vec{b} \end{matrix}] + [\begin{matrix} \vec{a} & \vec{c} & \vec{a} \end{matrix}] - [\begin{matrix} \vec{b} & \vec{b} & \vec{c} \end{matrix}] - [\vec{b} \vec{a} \vec{b}] - [\begin{matrix} \vec{b} & \vec{c} & \vec{a} \end{matrix}] = [\begin{matrix} \vec{a} & \vec{b} & \vec{c} \end{matrix}] + 0 + 0 - 0 - 0 - [\begin{matrix} \vec{a} & \vec{b} & \vec{c} \end{matrix}] (∵ [\begin{matrix} \vec{a} & \vec{b} & \vec{c} \end{matrix}] = [\begin{matrix} \vec{b} & \vec{c} & \vec{a} \end{matrix}]) = 0$

Page No 25.17:

Question 11:

$\vec{a,} \vec{b} and \vec{c}$ are the position vectors of points A, B and C respectively, prove that: $\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}$ is a vector perpendicular to the plane of triangle ABC.

Answer:

$We know that if any vector is perpendicular to all three sides of ∆ A B C, it must be perpendicular to the plane of ∆ A B C . Now, \vec{AB} = \vec{b} - \vec{a}, \vec{BC} = \vec{c} - \vec{b}, \vec{CA} = \vec{a} - \vec{c} (Position vectors of A, B and C are \vec{a}, \vec{b}, \vec{c}) We have \vec{AB} . (\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}) = (\vec{b} - \vec{a}) . (\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}) = \vec{b} . (\vec{a} \times \vec{b}) + \vec{b} . (\vec{b} \times \vec{c}) + \vec{b} . (\vec{c} \times \vec{a}) - \vec{a} . (\vec{a} \times \vec{b}) - \vec{a} . (\vec{b} \times \vec{c}) - \vec{a} . (\vec{c} \times \vec{a}) (By distributive law) = [\vec{b} \vec{a} \vec{b}] + [\vec{b} \vec{b} \vec{c}] + [\vec{b} \vec{c} \vec{a}] - [\vec{a} \vec{a} \vec{b}] - [\vec{a} \vec{b} \vec{c}] - [\vec{a} \vec{c} \vec{a}] = 0 + 0 + [\vec{b} \vec{c} \vec{a}] - 0 - [\vec{a} \vec{b} \vec{c}] - 0 = 0 (∵ [\vec{b} \vec{c} \vec{a}] = [\vec{a} \vec{b} \vec{c}]) \vec{BC} . (\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}) = (\vec{c} - \vec{b}) . (\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}) = \vec{c} . (\vec{a} \times \vec{b}) + \vec{c} . (\vec{b} \times \vec{c}) + \vec{c} . (\vec{c} \times \vec{a}) - \vec{b} . (\vec{a} \times \vec{b}) - \vec{b} . (\vec{b} \times \vec{c}) - \vec{b} . (\vec{c} \times \vec{a}) (By distributive law) = [\vec{c} \vec{a} \vec{b}] + [\vec{c} \vec{b} \vec{c}] + [\vec{c} \vec{c} \vec{a}] - [\vec{b} \vec{a} \vec{b}] - [\vec{b} \vec{b} \vec{c}] - [\vec{b} \vec{c} \vec{a}] = [\vec{c} \vec{a} \vec{b}] + 0 + 0 - 0 - 0 - [\vec{b} \vec{c} \vec{a}] = 0 (∵ [\vec{c} \vec{a} \vec{b}] = [\vec{b} \vec{c} \vec{a}]) Similarly, \vec{CA} . (\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}) = (\vec{a} - \vec{c}) . (\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}) = \vec{a} . (\vec{a} \times \vec{b}) + \vec{a} . (\vec{b} \times \vec{c}) + \vec{a} (\vec{c} \times \vec{a}) - \vec{c} . (\vec{a} \times \vec{b}) - \vec{c} . (\vec{b} \times \vec{c}) - \vec{c} . (\vec{c} \times \vec{a}) (By distributive law) = [\vec{a} \vec{a} \vec{b}] + [\vec{a} \vec{b} \vec{c}] + [\vec{a} \vec{c} \vec{a}] - [\vec{c} \vec{a} \vec{b}] - [\vec{c} \vec{b} \vec{c}] - [\vec{c} \vec{c} \vec{a}] = 0 + [\vec{a} \vec{b} \vec{c}] + 0 - [\vec{c} \vec{a} \vec{b}] - 0 - 0 = 0 (∵ [\vec{c} \vec{a} \vec{b}] = [\vec{a} \vec{b} \vec{c}]) Hence, vector \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} is perpendicular to all sides of ∆ ABC and also perpendicular to the plane of ∆ ABC .$

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Question 12:

$Let \vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = \hat{i} and \hat{c} = c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k} . Then,$
(i) If c₁ = 1 and c₂ = 2, find c₃ which makes $\vec{a,} \vec{b} and \vec{c}$ coplanar.

(ii) If c₂ = −1 and c₃ = 1, show that no value of c₁ can make $\vec{a,} \vec{b} and \vec{c}$ coplanar.

Answer:

$(i) If c_{1} = 1 and c_{2} = 2, then \vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = \hat{i} and \hat{c} = \hat{i} + 2 \hat{j} + c_{3} \hat{k} . We know that vector s \vec{a}, \vec{b}, \vec{c} are coplanar iff [\vec{a} \vec{b} c] = 0 . It is given that \vec{a}, \vec{b}, \vec{c} are coplanar . ∴ [\vec{a} \vec{b} c] = 0 \Rightarrow |\begin{matrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 2 & c_{3} \end{matrix}| = 0 \Rightarrow 1 (0 - 0) - 1 (c_{3} - o) + 1 (2 - 0) = 0 \Rightarrow - c_{3} + 2 = 0 \Rightarrow c_{3} = 2$

$(ii) If c_{2} = - 1 and c_{3} = 1, then \vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = \hat{i} and \hat{c} = c_{1} \hat{i} - \hat{j} + \hat{k .} We know that vector s \vec{a}, \vec{b}, \vec{c} are coplanar iff [\vec{a} \vec{b} c] = 0 . For \vec{a}, \vec{b}, \vec{c} to be coplanar : \Rightarrow [\vec{a} \vec{b} c] = 0 \Rightarrow |\begin{matrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ c_{1} & - 1 & 1 \end{matrix}| = 0 \Rightarrow 1 (0 - 0) - 1 (1 - 0) + 1 (- 1 - 0) = 0 \Rightarrow - 1 - 1 = 0 \Rightarrow - 2 = 0 But this is never possible, whatever be the value of c_{1} . Thus, no vaue of c_{1} can make \vec{a}, \vec{b} and \vec{c} coplanar .$

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Question 13:

Find λ for which the points A (3, 2, 1), B (4, λ, 5), C (4, 2, −2) and D (6, 5, −1) are coplanar.

Answer:

$The points A, B, C and D will be coplanar iff any one of the following traces of vectors are coplanar : \vec{A B}, \vec{A C}, \vec{A D}; \vec{A B}, \vec{B C}, \vec{C D}; \vec{B C}, \vec{B A}, \vec{B D}, etc . It is given that \vec{A B}, \vec{A C}, \vec{A D} are coplanar . Thus, their scaler triple product [\vec{A B} \vec{A C} \vec{A D}] is equal to zero . Now, Direction ratios of the \vec{P Q} = (Direction ratios of vector Q) - (Direction ratios of the vector P) Direction ratios of vector \vec{A B} = (4 - 3), (λ - 2), (5 - 1), i . e . 1, λ - 2, 4 Direction ratios of vector \vec{A C} = (4 - 3), (2 - 2), (- 2 - 1), i . e . 1, 0, - 3 Direction ratios of vector \vec{A D} = (6 - 3), (5 - 2), (- 1 - 1), i . e . 3, 3, - 2 ∴ [\vec{A B} \vec{A C} \vec{A D}] = |\begin{matrix} 1 & λ - 2 & 4 \\ 1 & 0 & - 3 \\ 3 & 3 & - 2 \end{matrix}| = 1 [0 - (- 9)] - (λ - 2) [- 2 - (- 9)] + 4 (3 - 0) = 0 \Rightarrow 7 λ = 35 \Rightarrow λ = 5$

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Question 14:

If four points A, B, C and D with position vectors 4 $\hat{i} +$ 3 $\hat{j} +$ 3 $\hat{k}$ , 5 $\hat{i} +$ $x \hat{j} +$ 7 $\hat{k}$ , 5 $\hat{i} +$ 3 $\hat{j}$ and $7 \hat{i} + 6 \hat{j} + \hat{k}$ respectively are coplanar, then find the value of x.

Answer:

Let $\vec{OA} = 4 \hat{i} + 3 \hat{j} + 3 \hat{k}, \vec{OB} = 5 \hat{i} + x \hat{j} + 7 \hat{k}, \vec{OC} = 5 \hat{i} + 3 \hat{j}$ and $\vec{OD} = 7 \hat{i} + 6 \hat{j} + \hat{k}$ .

$∴ \vec{AB} = (5 \hat{i} + x \hat{j} + 7 \hat{k}) - (4 \hat{i} + 3 \hat{j} + 3 \hat{k}) = \hat{i} + (x - 3) \hat{j} + 4 \hat{k}$

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

$\vec{AD} = (7 \hat{i} + 6 \hat{j} + \hat{k}) - (4 \hat{i} + 3 \hat{j} + 3 \hat{k}) = 3 \hat{i} + 3 \hat{j} - 2 \hat{k}$

Since the given four points are coplanar, so the vectors $\vec{AB}, \vec{AC}$ and $\vec{AD}$ are also coplanar.

$∴ [\begin{matrix} \vec{AB} & \vec{AC} & \vec{AD} \end{matrix}] = 0$

$\Rightarrow |\begin{matrix} 1 & x - 3 & 4 \\ 1 & 0 & - 3 \\ 3 & 3 & - 2 \end{matrix}| = 0 \Rightarrow 1 (0 + 9) - (x - 3) (- 2 + 9) + 4 (3 - 0) = 0 \Rightarrow 9 - 7 x + 21 + 12 = 0 \Rightarrow 7 x = 42 \Rightarrow x = 6$
Thus, the value of x is 6.

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Question 1:

If $\vec{a}$ lies in the plane of vectors $\vec{b} and \vec{c}$ , then which of the following is correct?
(a) $[\vec{a} \vec{b} \vec{c}] = 0$

(b) $[\vec{a} \vec{b} \vec{c}] = 1$

(c) $[\vec{a} \vec{b} \vec{c}] = 3$

(d) $[\vec{b} \vec{c} \vec{a}] = 1$

Answer:

(a) $[\vec{a} \vec{b} \vec{c}] = 0$

$If \vec{a} lies in the plane of vectors \vec{b} and \vec{c}, then \vec{a}, \vec{b}, \vec{c} will lie in the same plane, i . e . they will be coplanar . ∴ [\vec{a} \vec{b} \vec{c}] = 0$

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Question 2:

The value of $[\vec{a} - \vec{b}, \vec{b} - \vec{c}, \vec{c} - \vec{a}], where |\vec{a}| = 1, |\vec{b}| = 5, |\vec{c}| = 3, is$
(a) 0
(b) 1
(c) 6
(d) none of these

Answer:

(a) 0

$We have [\vec{a} - \vec{b}, \vec{b} - \vec{c}, \vec{c} - \vec{a}] = ((\vec{a} - \vec{b}) \times (\vec{b} - \vec{c})) . (\vec{c} - \vec{a}) (By definition of scalar triple product) = ((\vec{a} - \vec{b}) \times \vec{b} - (\vec{a} - \vec{b}) \times \vec{c}) . (\vec{c} - \vec{a}) = (\vec{a} \times \vec{b} - \vec{b} \times \vec{b} - \vec{a} \times \vec{c} + \vec{b} \times \vec{c}) . (\vec{c} - \vec{a}) = (\vec{a} \times \vec{b} - 0 - \vec{a} \times \vec{c} + \vec{b} \times \vec{c}) . (\vec{c} - \vec{a}) = (\vec{a} \times \vec{b}) . (\vec{c} - \vec{a}) - (\vec{a} \times \vec{c}) . (\vec{c} - \vec{a}) + (\vec{b} \times \vec{c}) . (\vec{c} - \vec{a}) = (\vec{a} \times \vec{b}) . \vec{c} - (\vec{a} \times \vec{b}) . \vec{a} - (\vec{a} \times \vec{c}) . \vec{c} + (\vec{a} \times \vec{c}) . \vec{a} + (\vec{b} \times \vec{c}) . \vec{c} - (\vec{b} \times \vec{c}) . \vec{a} = [\vec{a} \vec{b} \vec{c}] - [\vec{a} \vec{b} \vec{a}] - [\vec{a} \vec{c} \vec{c}] + [\vec{a} \vec{c} \vec{a}] + [\vec{b} \vec{c} \vec{c}] - [\vec{b} \vec{c} \vec{a}] = [\vec{a} \vec{b} \vec{c}] - 0 - 0 + 0 + 0 - [\vec{a} \vec{b} \vec{c}] (∵ [\vec{a} \vec{b} \vec{c}] = [\vec{b} \vec{c} \vec{a}] = [\vec{c} \vec{a} \vec{b}]) = 0$

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Question 3:

If $\vec{a,} \vec{b,} \vec{c}$ are three non-coplanar mutually perpendicular unit vectors, then $[\vec{a} \vec{b} \vec{c}],$ is
(a) ± 1
(b) 0
(c) −2
(d) 2

Answer:

$(a) \pm 1 We have [\vec{a} \vec{b} \vec{c}] = (\vec{a} \times \vec{b}) . \vec{c} = |\vec{a} \times \vec{b}| |\vec{c}| \cos 0^{°} or |\vec{a} \times \vec{b}| |\vec{c}| co s 180^{°} (∵ \vec{a}, \vec{b,} \vec{c} are perpendicular to each other) = |\vec{a} \times \vec{b}| or - |\vec{a} \times \vec{b}| (∵ |\vec{c}| = 1, co s 0^{°} = 1 and \cos 180^{°} = - 1) = |\vec{a}| |\vec{b}| \sin 90^{°} or - |\vec{a}| |\vec{b}| \sin 90^{°} (∵ \overset{}{\vec{a}} is perpendicular to \vec{b}) = 1 or - 1 (∵ |\vec{a}| = 1 and |\vec{b}| = 1) = \pm 1$

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Question 4:

If $\vec{r} \cdot \vec{a} = \vec{r} \cdot \vec{b} = \vec{r} \cdot \vec{c} = 0$ for some non-zero vector $\vec{r},$ then the value of $[\vec{a} \vec{b} \vec{c}],$ is
(a) 2
(b) 3
(c) 0
(d) none of these

Answer:

(c) 0

$If \vec{r} . \vec{a} = 0 for some non - zero vector \vec{r}, then either \vec{a} is a zero - vector or it is perpendicular to \vec{r} . If one of \vec{a}, \vec{b}, \bar{c} is zero, then [\vec{a} \vec{b} \vec{c}] = 0 If all \vec{a}, \vec{b} and \vec{c} are non - zero, then they must be coplanar as they are perpendicular to vector \vec{r} . ∴ [\vec{a} \vec{b} \vec{c}] = 0$

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Question 5:

For any three vectors $\vec{a,} \vec{b,} \vec{c}$ the expression $(\vec{a} - \vec{b}) . \{(\vec{b} - \vec{c}) \times (\vec{c} - \vec{a})\}$ equals
(a) $[\vec{a} \vec{b} \vec{c}]$

(b) $2 [\vec{a} \vec{b} \vec{c}]$

(c) ${[\vec{a} \vec{b} \vec{c}]}^{2}$

(d) none of these

Answer:

(d) none of these

$We have (\vec{a} - \vec{b}) . [(\vec{b} - \vec{c}) \times (\vec{c} - \vec{a})] = (\vec{a} - \vec{b}) . [(\vec{b} - \vec{c}) \times c - (\vec{b} - \vec{c}) \times \vec{a}] = (\vec{a} - \vec{b}) . (\vec{b} \times \vec{c} - \vec{c} \times \vec{c} - \vec{b} \times \vec{a} + \vec{c} \times \vec{a}) = (\vec{a} - \vec{b}) . (\vec{b} \times \vec{c} - 0 - \vec{b} \times \vec{a} + \vec{c} \times \vec{a}) = (\vec{a} - \vec{b}) . (\vec{b} \times \vec{c}) - (\vec{a} - \vec{b}) . (\vec{b} \times \vec{a}) + (\vec{a} - \vec{b}) . (\vec{c} \times \vec{a}) = \vec{a} . (\vec{b} \times \vec{c}) - \vec{b .} (\vec{b} \times \vec{c}) - \vec{a} . (\vec{b} \times \vec{a}) + \vec{b} . (\vec{b} \times \vec{a}) + \vec{a} . (\vec{c} \times \vec{a}) - \vec{b} . (\vec{c} \times \vec{a}) = [\vec{a} \vec{b} \vec{c}] - 0 - 0 + 0 + 0 - [\vec{b} \vec{c} \vec{a}] (∵ [\vec{b} \vec{b} \vec{c}] = [\vec{a} \vec{b} \vec{a}] = [\vec{b} \vec{b} \vec{a}] = 0) = [\vec{a} \vec{b} \vec{c}] - [\vec{a} \vec{b} \vec{c}] (∵ [\vec{a} \vec{b} \vec{c}] = [\vec{b} \vec{c} \vec{a}] = [\vec{c} \vec{a} \vec{b}]) = 0$

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Question 6:

If $\vec{a,} \vec{b,} \vec{c}$ are non-coplanar vectors, then $\frac{\vec{a} \cdot (\vec{b} \times \vec{c})}{(\vec{c} \times \vec{a}) \cdot \vec{b}} + \frac{\vec{b} \cdot (\vec{a} \times \vec{c})}{\vec{c} \cdot (\vec{a} \times \vec{b})}$ is equal to
(a) 0
(b) 2
(c) 1
(d) none of these

Answer:

(a) 0

$We have \frac{\vec{a} . (\vec{b} \times \vec{c})}{(\vec{c} \times \vec{a}) . \vec{b}} + \frac{\vec{b} . (\vec{a} \times \vec{c})}{\vec{c} . (\vec{a} \times \vec{b})} = \frac{[\vec{a} \vec{b} \vec{c}]}{[\vec{c} \vec{a} \vec{b}]} + \frac{[\vec{b} \vec{a} \vec{c}]}{[\vec{c} \vec{a} \vec{b}]} (By definition of scalar tiple product) = \frac{[\vec{a} \vec{b} \vec{c}]}{[\vec{a} \vec{b} \vec{c}]} + \frac{- [\vec{a} \vec{b} \vec{c}]}{[\vec{a} \vec{b} \vec{c}]} (Change in cyclic order of vectors changes the sign of the scalar triple product) = 1 - 1 = 0$

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Question 7:

Let $\vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}, \vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k} and \vec{c} = c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k}$ be three non-zero vectors such that $\vec{c}$ is a unit vector perpendicular to both $\vec{a} and \vec{b}$ . If the angle between $\vec{a} and \vec{b}$ is $\frac{π}{6},$ then ${|\begin{matrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{matrix}|}^{2}$ is equal to

(a) 0
(b) 1
(c) $\frac{1}{4} {|\vec{a}|}^{2} {|\vec{b}|}^{2}$

(d) $\frac{3}{4} {|\vec{a}|}^{2} {|\vec{b}|}^{2}$

Answer:

(c) $\frac{1}{4} {|\vec{a}|}^{2} {|\vec{b}|}^{2}$

$We have {|\begin{matrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{matrix}|}^{2} = {[(\vec{a} \times \vec{b}) . \vec{c}]}^{2} (By definition of scalar triple product) = {[|(\vec{a} \times \vec{b})| |\vec{c}| c os 0 °]}^{2} (∵ \vec{a} \times \vec{b} is parallel to vector \vec{c} as \vec{c} is perpendicular to both \vec{a} and \vec{b}) = {(|\vec{a}| |\vec{b}| s in \frac{π}{6})}^{2} (∵ |\vec{c}| = 1 and angle between \vec{a} and \vec{b} is \frac{π}{6}) = {|\vec{a}|}^{2} {|\vec{b}|}^{2} {(\frac{1}{2})}^{2} = \frac{1}{4} {|\vec{a}|}^{2} {|\vec{b}|}^{2}$

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Question 8:

If $\vec{a} = 2 \hat{i} - 3 \hat{j} + 5 \hat{k}, \vec{b} = 3 \hat{i} - 4 \hat{j} + 5 \hat{k} and \vec{c} = 5 \hat{i} - 3 \hat{j} - 2 \hat{k},$ then the volume of the parallelopiped with conterminous edges $\vec{a} + \vec{b,} \vec{b} + \vec{c,} \vec{c} + \vec{a}$ is
(a) 2
(b) 1
(c) −1
(d) 0

Answer:

$We have \vec{a} + \vec{b} = (2 \hat{i} - 3 \hat{j} + 5 \hat{k}) + (3 \hat{i} - 4 \hat{j} + 5 \hat{k}) = 5 \hat{i} - 7 \hat{j} + 10 \hat{k} \vec{b} + \vec{c} = (3 \hat{i} - 4 \hat{j} + 5 \hat{k}) + (5 \hat{i} - 3 \hat{j} - 2 \hat{k}) = 8 \hat{i} - 7 \hat{j} + 3 \hat{k} \vec{c} + \vec{a} = (5 \hat{i} - 3 \hat{j} - 2 \hat{k}) + (2 \hat{i} - 3 \hat{j} + 5 \hat{k}) = 7 \hat{i} - 6 \hat{j} + 3 \hat{k} We know that the volume of a parallelopiped whose three adjacent adges are \vec{a} + \vec{b}, \vec{b} + \vec{c} and \vec{c} + \vec{a} is equal to |[\vec{a} + \vec{b} \vec{b} + \vec{c} \vec{c} + \vec{a}]| . We have [\vec{a} + \vec{b} \vec{b} + \vec{c} \vec{c} + \vec{a}] = |\begin{matrix} 5 & - 7 & 10 \\ 8 & - 7 & 3 \\ 7 & - 6 & 3 \end{matrix}| = 5 (- 21 + 18) + 7 (24 - 21) + 10 (- 48 + 49) = (5 \times - 3) + (7 \times 3) + (10 \times 1) = 16 ∴ Volume of parallelopiped = |16| = 16$

Disclaimer: None of the given options is correct.

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Question 9:

If $[2 \vec{a} + 4 \vec{b} \vec{c} \vec{d}] = λ [\vec{a} \vec{c} \vec{d}] + μ [\vec{b} \vec{c} \vec{d}],$ then λ + μ =
(a) 6
(b) −6
(c) 10
(d) 8

Answer:

(a) 6

$We have [2 \vec{a} + 4 \vec{b} \vec{c} \vec{d}] = λ [\vec{a} \vec{c} \vec{d}] + μ [\vec{b} \vec{c} \vec{d}] \Rightarrow [(2 \vec{a} + 4 \vec{b}) \times \vec{c}] . \vec{d} = λ [\vec{a} \vec{c} \vec{d}] + μ [\vec{b} \vec{c} \vec{d}] (By definition of scalar triple product) \Rightarrow [(2 \vec{a} \times \vec{c}) + (4 \vec{b} \times \vec{c})] . \vec{d} = λ [\vec{a} \vec{c} \vec{d}] + μ [\vec{b} \vec{c} \vec{d}] \Rightarrow (2 \vec{a} \times \vec{c}) . \vec{d} + (4 \vec{b} \times \vec{c}) . \vec{d} = λ [\vec{a} \vec{c} \vec{d}] + μ [\vec{b} \vec{c} \vec{d}] \Rightarrow [2 \vec{a} \vec{c} \vec{d}] + [4 \vec{b} \vec{c} \vec{d}] = λ [\vec{a} \vec{c} \vec{d}] + μ [\vec{b} \vec{c} \vec{d}] \Rightarrow 2 [\vec{a} \vec{c} \vec{d}] + 4 [\vec{b} \vec{c} \vec{d}] = λ [\vec{a} \vec{c} \vec{d}] + μ [\vec{b} \vec{c} \vec{d}] (∵ [λ \vec{a} \vec{b} \vec{c}] = λ [\vec{a} \vec{b} \vec{c}] for any scalar λ) Comparing both sides, we get λ = 2 μ = 4 ∴ λ + μ = 2 + 4 = 6$

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Question 10:

$[\vec{a} \vec{b} \vec{a} \times \vec{b}] + {(\vec{a} . \vec{b})}^{2} =$

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

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

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

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

Answer:

$(a) {|\vec{a}|}^{2} {|\vec{b}|}^{2} We have [\vec{a} \vec{b} \vec{a} \times \vec{b}] + {(\vec{a} . \vec{b})}^{2} = (\vec{a} \times \vec{b}) . (\vec{a} \times \vec{b}) + {(\vec{a} . \vec{b})}^{2} = {|(\vec{a} \times \vec{b})|}^{2} + {(\vec{a} . \vec{b})}^{2} = {(|\vec{a}| |\vec{b}| \sin θ)}^{2} + {(|\vec{a}| {|\vec{b}|}^{} \cos θ)}^{2} = {|\vec{a}|}^{2} {|\vec{b}|}^{2} \sin^{2} θ + {|\vec{a}|}^{2} {|\vec{b}|}^{2} \cos^{2} θ = {|\vec{a}|}^{2} {|\vec{b}|}^{2} (\sin^{2} θ + \cos^{2} θ) = {|\vec{a}|}^{2} {|\vec{b}|}^{2}$

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Question 11:

If the vectors $4 \hat{i} + 11 \hat{j} + m \hat{k}, 7 \hat{i} + 2 \hat{j} + 6 \hat{k} and \hat{i} + 5 \hat{j} + 4 \hat{k}$ are coplanar, then m =
(a) 0
(b) 38
(c) −10
(d) 10

Answer:

$(d) 10 Let : \vec{a} = 4 \hat{i} + 11 \hat{j} + m \hat{k} \vec{b} = 7 \hat{i} + 2 \hat{j} + 6 \hat{k} \vec{c} = \hat{i} + 5 \hat{j} + 4 \hat{k} We know that vectors \vec{a}, \vec{b} and \vec{c} are coplanar iff their scalar triple product is zero, i . e . [\vec{a} \vec{b} \vec{c}] = 0 \Rightarrow |\begin{matrix} 4 & 11 & m \\ 7 & 2 & 6 \\ 1 & 5 & 4 \end{matrix}| = 0 \Rightarrow 4 (8 - 30) - 11 (28 - 6) + m (35 - 2) = 0 \Rightarrow - 88 - 242 + 33 m = 0 \Rightarrow 33 m = 330 ∴ m = 10$

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Question 12:

For non-zero vectors $\vec{a,} \vec{b} and \vec{c}$ the relation $|(\vec{a} \times \vec{b}) \cdot \vec{c}| = |\vec{a}| |\vec{b}| |\vec{c}|$ holds good, if
(a) $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = 0$

(b) $\vec{a} \cdot \vec{b} = 0 = \vec{c} \cdot \vec{a}$

(c) $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = 0$

(d) $\vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = 0$

Answer:

$(c) \vec{a} . \vec{b} = \vec{b} . \vec{c} = \vec{c} . \vec{a} = 0 We have |(\vec{a} \times \vec{b}) . \vec{c}| = |(\vec{a} \times \vec{b})| |\vec{c}| |cosθ| = |(\vec{a} \times \vec{b})| |\vec{c}| (If θ = 0 ° or 180 °, i . e . vectors \vec{a} \times \vec{b} and \vec{c} are parallel) = |(|\vec{a}| |\vec{b}| \sin α)| |\vec{c}| = |\vec{a}| |\vec{b}| |\vec{c}| (If α = 90 °, i . e . vectors \vec{a} and \vec{b} are perpendicular) ∴ |(\vec{a} \times \vec{b}) . \vec{c}| = |\vec{a}| |\vec{b}| |\vec{c}| (If vectors \vec{a}, \vec{b}, \vec{c} are perpendicular to each other) Thus, the relation |(\vec{a} \times \vec{b}) . \vec{c}| = |\vec{a}| |\vec{b}| |\vec{c}| holds good if \vec{a} . \vec{b} = 0, \vec{b} . \vec{c} = 0 and \vec{c} . \vec{a} = 0 .$

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Question 13:

$(\vec{a} + \vec{b}) \cdot (\vec{b} + \vec{c}) \times (\vec{a} + \vec{b} + \vec{c}) =$

(a) 0

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

(c) $2 [\vec{a} \vec{b} \vec{c}]$

(d) $[\vec{a} \vec{b} \vec{c}]$

Answer:

$(d) [\vec{a} \vec{b} \vec{c}] We have (\vec{a} + \vec{b}) . (\vec{b} + \vec{c}) \times (\vec{a} + \vec{b} + \vec{c}) = (\vec{a} + \vec{b}) . [(\vec{b} + \vec{c}) \times \vec{a} + (\vec{b} + \vec{c}) \times \vec{b} + (\vec{b} + \vec{c}) \times \vec{c}] = (\vec{a} + \vec{b}) . (\vec{b} \times \vec{a} + \vec{c} \times \vec{a} + \vec{b} \times \vec{b} + \vec{c} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{c}) = (\vec{a} + \vec{b}) . (\vec{b} \times \vec{a} + \vec{c} \times \vec{a} + 0 + \vec{c} \times \vec{b} + \vec{b} \times \vec{c} + 0) = (\vec{a} + \vec{b}) . (\vec{b} \times \vec{a} + \vec{c} \times \vec{a} - \vec{b} \times \vec{c} + \vec{b} \times \vec{c}) = (\vec{a} + \vec{b}) . (\vec{b} \times \vec{a} + \vec{c} \times \vec{a}) = \vec{a} . (\vec{b} \times \vec{a}) + \vec{b} . (\vec{b} \times \vec{a}) + \vec{a} . (\vec{c} \times \vec{a}) + \vec{b} . (\vec{c} \times \vec{a}) = 0 + 0 + 0 + [\vec{b} \vec{c} \vec{a}] = [\vec{b} \vec{c} \vec{a}] = [\vec{a} \vec{b} \vec{c}]$

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Question 14:

If $\vec{a,} \vec{b,} \vec{c}$ are three non-coplanar vectors, then $(\vec{a} + \vec{b} + \vec{c}) . [(\vec{a} + \vec{b}) \times (\vec{a} + \vec{c})]$ equals

(a) 0

(b) $[\vec{a} \vec{b} \vec{c}]$

(c) $2 [\vec{a} \vec{b} \vec{c}]$

(d) $- [\vec{a} \vec{b} \vec{c}]$

Answer:

$(d) - [\vec{a} \vec{b} \vec{c}] We have (\vec{a} + \vec{b} + \vec{c}) . [(\vec{a} + \vec{b}) \times (\vec{a} + \vec{c})] = (\vec{a} + \vec{b} + \vec{c}) . [(\vec{a} + \vec{b}) \times \vec{a} + (\vec{a} + \vec{b}) \times \vec{c}] (By definition of cross poduct) = (\vec{a} + \vec{b} + \vec{c}) . [\vec{a} \times \vec{a} + \vec{b} \times \vec{a} + \vec{a} \times \vec{c} + \vec{b} \times \vec{c}] = (\vec{a} + \vec{b} + \vec{c}) . [0 + \vec{b} \times \vec{a} + \vec{a} \times \vec{c} + \vec{b} \times \vec{c}] = \vec{a} . (\vec{b} \times \vec{a}) + \vec{a} . (\vec{a} \times \vec{c}) + \vec{a} . (\vec{b} \times \vec{c}) + \vec{b} . (\vec{b} \times \vec{a}) + \vec{b} . (\vec{a} \times \vec{c}) + \vec{b} . (\vec{b} \times \vec{c}) + \vec{c} . (\vec{b} \times \vec{a}) + \vec{c} . (\vec{a} \times \vec{c}) + \vec{c} . (\vec{b} \times \vec{c}) = [\vec{a} \vec{b} \vec{a}] + [\vec{a} \vec{a} \vec{c}] + [\vec{a} \vec{b} \vec{c}] + [\vec{b} \vec{b} \vec{a}] + [\vec{b} \vec{a} \vec{c}] + [\vec{b} \vec{b} \vec{c}] + [\vec{c} \vec{b} \vec{a}] + [\vec{c} \vec{a} \vec{c}] + [\vec{c} \vec{b} \vec{c}] = 0 + 0 + [\vec{a} \vec{b} \vec{c}] + 0 + [\vec{b} \vec{a} \vec{c}] + 0 + [\vec{c} \vec{b} \vec{a}] + 0 + 0 = [\vec{a} \vec{b} \vec{c}] - [\vec{a} \vec{b} \vec{c}] - [\vec{a} \vec{b} \vec{c}] (∵ [\vec{b} \vec{a} \vec{c}] = - [\vec{c} \vec{a} \vec{b}], [\vec{b} \vec{a} \vec{c}] = - [\vec{a} \vec{b} \vec{c}]) = - [\vec{a} \vec{b} \vec{c}]$

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Question 15:

$(\vec{a} + 2 \vec{b} - \vec{c}) \cdot \{(\vec{a} - \vec{b}) \times (\vec{a} - \vec{b} - \vec{c})\}$ is equal to

(a) $[\vec{a} \vec{b} \vec{c}]$

(b) $2 [\vec{a} \vec{b} \vec{c}]$

(c) $3 [\vec{a} \vec{b} \vec{c}]$

(d) 0

Answer:

$(c) 3 [\vec{a} \vec{b} \vec{c}] We have (\vec{a} + 2 \vec{b} - \vec{c}) . \{(\vec{a} - \vec{b}) \times (\vec{a} - \vec{b} - \vec{c})\} = (\vec{a} + 2 \vec{b} - \vec{c}) . \{(\vec{a} - \vec{b}) \times \vec{a} - (\vec{a} - \vec{b}) \times \vec{b} - (\vec{a} - \vec{b}) \times \vec{c}\} = (\vec{a} + 2 \vec{b} - \vec{c}) . \{\vec{a} \times \vec{a} - \vec{b} \times \vec{a} - \vec{a} \times \vec{b} + \vec{b} \times \vec{b} - \vec{a} \times \vec{c} + \vec{b} \times \vec{c}\} = (\vec{a} + 2 \vec{b} - \vec{c}) . \{0 - \vec{b} \times \vec{a} - \vec{a} \times \vec{b} + 0 - \vec{a} \times \vec{c} + \vec{b} \times \vec{c}\} = (\vec{a} + 2 \vec{b} - \vec{c}) . \{- \vec{a} \times \vec{c} + \vec{b} \times \vec{c}\} (∵ \vec{a} \times \vec{b} = - \vec{b} \times \vec{a}) = - \vec{a} . (\vec{a} \times \vec{c}) + \vec{a} . (\vec{b} \times \vec{c}) - 2 \vec{b} . (\vec{a} \times \vec{c}) + 2 \vec{b} . (\vec{b} \times \vec{c}) + \vec{c} . (\vec{a} \times \vec{c}) - \vec{c} . (\vec{b} \times \vec{c}) = 0 + [\vec{a} \vec{b} \vec{c}] - 2 [\vec{b} \vec{a} \vec{c}] + 0 + 0 - 0 (∵ [λ \vec{a} \vec{b} \vec{c}] = λ [\vec{a} \vec{b} \vec{c}] for any scalar λ) = 3 [\vec{a} \vec{b} \vec{c}] (∵ - [\vec{b} \vec{a} \vec{c}] = [\vec{a} \vec{b} \vec{c}])$

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Question 16:

The vectors $λ \hat{i} + \hat{j} + 2 \hat{k}, \hat{i} + λ \hat{j} - \hat{k} and 2 \hat{i} - \hat{j} + λ \hat{k}$ are coplanar, if λ =
(a) –2
(b) 0
(c) 1
(d) –1

Answer:

$λ \hat{i} + \hat{j} + 2 \hat{k}, \hat{i} + λ \hat{j} - \hat{k} and 2 \hat{i} - \hat{j} + λ \hat{k} are coplanar if |\begin{matrix} λ & 1 & 2 \\ 1 & λ & - 1 \\ 2 & - 1 & λ \end{matrix}| = 0 i . e λ (λ^{2} - 1) - 1 (λ + 2) + 2 (- 1 - 2 λ) = 0 i . e λ^{3} - λ - λ - 2 - 2 - 4 λ = 0 i . e λ^{3} - 6 λ - 4 = 0 \{\begin{cases} Since for λ = - 2 \\ {(- 2)}^{3} - 6 (- 2) - 4 \\ = - 8 + 12 - 4 = 0 \end{cases} i . e (λ + 2) (λ^{2} - 2 λ - 2) = 0 i . e λ = - 2$
Hence, the correct answer is option A.

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Question 1:

$[\hat{i} \hat{k} \hat{j}] + [\hat{k} \hat{j} \hat{i}] + [\hat{j} \hat{k} \hat{i}] =_______________.$

Answer:

$[\hat{i} \hat{k} \hat{j}] + [\hat{k} \hat{j} \hat{i}] + [\hat{j} \hat{k} \hat{i}] = (\hat{i} \times \hat{k}) \cdot \hat{j} + (\hat{k} \times \hat{j}) . \hat{i} + (\hat{j} \times \hat{k}) . \hat{i} = - \hat{j} \cdot \hat{j} + (- \hat{i}) . \hat{i} + \hat{i} . \hat{i} = - 1 - 1 + 1 i . e [\hat{i} \hat{k} \hat{j}] + [\hat{k} \hat{j} \hat{i}] + [\hat{j} \hat{k} \hat{i}] = - 1 (∵ \hat{i} . \hat{i} = 1 = \hat{j} \cdot \hat{j})$

Page No 25.19:

Question 2:

If $\vec{a}, \vec{b}, \vec{c}$ are three vectors such that $[\vec{a}, \vec{b}, \vec{c}] = 10, then [\vec{a} + \vec{b} \vec{b} + \vec{c} \vec{c} + \vec{a}]$ = _______________.

Answer:

$for vectors \vec{a}, \vec{b} and \vec{c,} given [\vec{a} \vec{b} \vec{c}] = 10 then [\vec{a} + \vec{b} \vec{b} + \vec{c} \vec{c} + \vec{a}] = \{(\vec{a} + \vec{b}) \times (\vec{b} + \vec{c})\} \cdot (\vec{c} + \vec{a}) (defination) = (\vec{a} \times \vec{b} + \vec{a} \times \vec{c} + \vec{b} \times \vec{b} + \vec{b} \times \vec{c}) \cdot (\vec{c} + \vec{a}) [By distribution law] = (\vec{a} \times \vec{b} + \vec{a} \times \vec{c} + \vec{b} \times \vec{c}) \cdot (\vec{c} + \vec{a}) [∵ \vec{b} \times \vec{b} = 0] = (\vec{a} \times \vec{b}) \cdot \vec{c} + (\vec{a} \times \vec{c}) \cdot \vec{c} + (\vec{b} \times \vec{c}) \cdot \vec{c} + (\vec{a} \times \vec{b}) \cdot \vec{a} + (\vec{a} \times \vec{c}) \cdot \vec{a} + (\vec{b} \times \vec{c}) \cdot \vec{a}$
$= [\vec{a} \vec{b} \vec{c}] + [\vec{a} \vec{c} \vec{c}] + [\vec{b} \vec{c} \vec{c}] + [\vec{a} \vec{b} \vec{a}] + [\vec{a} \vec{c} \vec{a}] + [\vec{b} \vec{c} \vec{a}] [Since scalar triple product when any two vectors are same is zero .] = [\vec{a} \vec{b} \vec{c}] + 0 + 0 + 0 + 0 + [\vec{a} \vec{b} \vec{c}] = 2 [\vec{a} \vec{b} \vec{c}] = 2 \times 10 = 20 i . e [\vec{a} + \vec{b} \vec{b} + \vec{c} \vec{c} + \vec{a}] = 20$

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Question 3:

If $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar vectors, then $\frac{(\vec{b} \times \vec{c}), (\vec{a} + \vec{b} + \vec{c})}{[\vec{a} \vec{b} \vec{c}]}$ = _______________.

Answer:

$for non ‐ coplanar vectors \vec{a}, \vec{b} and \vec{c} \frac{(\vec{b} \times \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c})}{[\vec{a} \vec{b} \vec{c}]} = \frac{(\vec{b} \times \vec{c}) \cdot \vec{a} + (\vec{b} \times \vec{c}) \cdot \vec{b} + (\vec{b} \times \vec{c}) \cdot \vec{c}}{[\vec{a} \vec{b} \vec{c}]} [By distributive law] = \frac{(\vec{b} \times \vec{c}) \cdot \vec{a} + 0 + 0}{[\vec{a} \vec{b} \vec{c}]} [∵ Scalar triple product when two vectors are same is zero] = \frac{[\vec{a} \vec{b} \vec{c}]}{[\vec{a} \vec{b} \vec{c}]} [By defination] i . e \frac{(\vec{b} \times \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c})}{[\vec{a} \vec{b} \vec{c}]} = 1 .$

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Question 4:

For any three vectors $\vec{a}, \vec{b}, \vec{c}$ the value of $\vec{a} . \{(\vec{b} + \vec{c}) \times (\vec{a} + \vec{b} + \vec{c})\}$ is _____________.

Answer:

For any three vectors $\vec{a}, \vec{b}, \vec{c}$ ;
$\vec{a} \cdot \{(\vec{b} + \vec{c}) \times (\vec{a} + \vec{b} + \vec{c})\}$
$= \vec{a} \cdot \{(\vec{b} + \vec{c}) \times \vec{a} + (\vec{b} + \vec{c}) \times (\vec{b} + \vec{c})\} [By distributive law] = \vec{a} \cdot \{(\vec{b} + \vec{c}) \times \vec{a} + 0\} [∵ \vec{x} \times \vec{x} = 0] = \vec{a} \cdot \{(\vec{b} + \vec{c}) \times \vec{a}\} = \vec{a} \cdot \{\vec{b} \times \vec{a} + \vec{c} \times \vec{a}\} [By distributive law] = \vec{a} \cdot (\vec{b} \times \vec{a}) + \vec{a} . (\vec{c} \times \vec{a}) = 0 + 0 [Since two vectors are same] ∴ \vec{a} \cdot \{(\vec{b} + \vec{c}) \times (\vec{a} + \vec{b} + \vec{c})\} = 0$

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Question 5:

The value of $[\vec{a} - \vec{b} \vec{b} - \vec{c} \vec{c} - \vec{a}], where |\vec{a}| = 1, |\vec{b}| = 5 and |\vec{c}| = 3,$ is _________________.

Answer:

$for two vectors \vec{a}, \vec{b} and \vec{c}, given |\vec{a}| = 1, |\vec{b}| = 5 and |\vec{c}| = 3 Then [\vec{a} - \vec{b} \vec{b} - \vec{c} \vec{c} - \vec{a}] = \{(\vec{a} - \vec{b}) \times (\vec{b} - \vec{c})\} \cdot (\vec{c} - \vec{a}) [By defination] = \{(\vec{a} \times \vec{b}) - (\vec{a} \times \vec{c}) - (\vec{b} \times \vec{b}) + (\vec{b} \times \vec{c})\} \cdot (\vec{c} - \vec{a}) [By distributive law] = \{(\vec{a} \times \vec{b}) - (\vec{a} \times \vec{c}) - 0 + (\vec{b} \times \vec{c})\} . (\vec{c} - \vec{a}) [∵ \vec{b} \times \vec{b} = 0] = (\vec{a} \times \vec{b}) \cdot \vec{c} - (\vec{a} \times \vec{c}) \cdot \vec{c} + (\vec{b} \times \vec{c}) \cdot \vec{c} - (\vec{a} \times \vec{b}) \cdot \vec{a} + (\vec{a} \times \vec{c}) \cdot \vec{a} - (\vec{b} \times \vec{c}) \cdot \vec{a} [By distributive law] = (\vec{a} \times \vec{b}) \cdot \vec{c} - 0 + 0 - 0 + 0 - (\vec{b} \times \vec{c}) \cdot \vec{a} = [\vec{a} \vec{b} \vec{c}] - [\vec{b} \vec{c} \vec{a}] = [\vec{a} \vec{b} \vec{c}] - [\vec{a} \vec{b} \vec{c}] = 0$

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Question 6:

For any two vectors $[\vec{a} \vec{b} \vec{a} \times \vec{b}] + (\vec{a} . \vec{b})$ = ________________.

Answer:

check answer
For any two vectors $[\vec{a} \vec{b} \vec{a} \times \vec{b}] + (\vec{a} . \vec{b})$
= $(\vec{a} \times \vec{b}) . (\vec{a} \times \vec{b}) + (\vec{a} . \vec{b})$

Page No 25.19:

Question 7:

If non-coplanar vectors $\vec{a}, \vec{b}, \vec{c}$ from a parallelopiped of volume 6 cubic units, then the values of $[\vec{a} + \vec{b} \vec{b} + \vec{c} \vec{c} + \vec{a}]$ are _______________.

Answer:

$For non ‐ coplanar vectors \vec{a}, \vec{b} and \vec{c} . Given volume of parallelopiped is 6 cubic units Then [\vec{a} + \vec{b} \vec{b} + \vec{c} \vec{c} + \vec{a}] = \{(\vec{a} + \vec{b}) \times (\vec{b} + \vec{c})\} \cdot (\vec{c} + \vec{a}) (By defination) = \{(\vec{a} \times \vec{b}) + (\vec{a} \times \vec{c}) + (\vec{b} \times \vec{b}) + (\vec{b} \times \vec{c})\} \cdot (\vec{c} + \vec{a}) [By distributive law] = \{(\vec{a} \times \vec{b}) + (\vec{a} \times \vec{c}) + 0 + (\vec{b} \times \vec{c})\} \cdot (\vec{c} + \vec{a}) [\vec{b} \times \vec{b} = 0] = (\vec{a} \times \vec{b}) \cdot \vec{c} + (\vec{a} \times \vec{c}) \cdot \vec{c} + (\vec{b} \times \vec{c}) \cdot \vec{c} + (\vec{a} \times \vec{b}) \cdot \vec{a} + (\vec{a} \times \vec{c}) \cdot \vec{a} + (\vec{b} \times \vec{c}) \cdot \vec{a} = (\vec{a} \times \vec{b}) \cdot \vec{c} + 0 + 0 + 0 + 0 + (\vec{b} \times \vec{c}) \cdot \vec{a} [∵ two vectors of scalar triple product is zero] = + [\vec{a} \vec{b} \vec{c}] + [\vec{a} \vec{b} \vec{c}] = \pm 2 [\vec{a} \vec{b} \vec{c}] = \pm 2 \times 6 = \pm 12 i . e [\vec{a} + \vec{b} \vec{b} + \vec{c} \vec{c} + \vec{a}] = \pm 12$

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Question 8:

If three non-coplanar vectors $\vec{a}, \vec{b}, \vec{c}$ from a parallelopiped of volume 8 cubic units, then the values of $[3 \vec{a} 4 \vec{b} 5 \vec{c}]$ are _________________.

Answer:

Volume of parallelopiped is 8 cubic units $[3 \vec{a} 4 \vec{b} 5 \vec{c}]$
$= \pm 3 \times 4 \times 5 [\vec{a} \vec{b} \vec{c}] = \pm 60 [\vec{a} \vec{b} \vec{c}] = \pm 60 \times 8 = \pm 480 i . e [3 \vec{a} 4 \vec{b} 5 \vec{c}] = \pm 480$

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Question 9:

If $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar vectors, then vectors $\vec{a} - \vec{b}, \vec{b} - \vec{c} and \vec{c} - \vec{a}$ from a parallelopiped whose volume is ______________.

Answer:

$for vectors \vec{a}, \vec{b} and \vec{c}, [\vec{a} - \vec{b} \vec{b} - \vec{c} \vec{c} - \vec{a}] = \{(\vec{a} - \vec{b}) \times (\vec{b} - \vec{c})\} \cdot (\vec{c} - \vec{a}) [By defination] = \{(\vec{a} - \vec{b}) \times (\vec{b} - \vec{c})\} \cdot (\vec{c} - \vec{a}) = \{(\vec{a} \times \vec{b}) - (\vec{a} \times \vec{c}) - (\vec{b} \times \vec{b}) + (\vec{b} \times \vec{c})\} \cdot (\vec{c} - \vec{a}) (By distributive law) = \{(\vec{a} \times \vec{b}) - (\vec{a} \times \vec{c}) - 0 + (\vec{b} \times \vec{c})\} \cdot (\vec{c} - \vec{a}) [∵ \vec{b} \times \vec{b} = 0] = (\vec{a} \times \vec{b}) \cdot \vec{c} - (\vec{a} \times \vec{c}) \cdot \vec{c} + (\vec{b} \times \vec{c}) \cdot \vec{c} - (\vec{a} \times \vec{b}) \cdot \vec{a} + (\vec{a} \times \vec{c}) \cdot \vec{a} - (\vec{b} \times \vec{c}) \cdot \vec{a} = (\vec{a} \times \vec{b}) \cdot \vec{c} - 0 + 0 - 0 + 0 - (\vec{b} \times \vec{c}) \cdot \vec{a} = [\vec{a} \vec{b} \vec{c}] - [\vec{a} \vec{b} \vec{c}] = 0 i . e [\vec{a} - \vec{b} \vec{b} - \vec{c} \vec{c} - \vec{a}] = 0 i . e volume of parallelopiped is 0 .$

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Question 10:

Let $\vec{a}, \vec{b}, \vec{c}$ be three non-coplanar vectors such that $[\vec{a} \vec{b} \vec{c}] = 24 and |\vec{a} \times \vec{b}| = 6 .$ Then the height the parallelopiped formed by $\vec{a}, \vec{b}, \vec{c} with \vec{a} and \vec{b}$ as two adjacent edges of the base, is ____________.

Answer:

Given $[\vec{a} \vec{b} \vec{c}] = 24$

$|\vec{a} \times \vec{b}| = 6$

when $\vec{a} and \vec{b}$ are two adjacent edges of the base

$∴$ Area of Base is $|\vec{a} \times \vec{b}| = 6$ .

i.e c will represent height

Since Volume is given by $[\vec{a} \vec{b} \vec{c}] = 24$

$i . e (\vec{a} \times \vec{b}) \cdot \vec{c} = 24 i . e |\vec{a} \times \vec{b}| \cdot |\vec{c}| = 24 i . e 6 \times |\vec{c}| = 24 i . e |\vec{c}| = 4 .$

i.e height of the parallelepiped is 4.

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Question 11:

If $\vec{r} \cdot \vec{a} = 0, \vec{r} \cdot \vec{b} = 0 and \vec{r} \cdot \vec{c} = 0$ for some non-zero vector $\vec{r}, then the value of \vec{a} . (\vec{b} \times \vec{c})$ is ____________.

Answer:

Given for same non-zero vector $\vec{r},$
$\vec{r} \cdot \vec{a} = 0 = \vec{r} \cdot \vec{b} = \vec{r} \cdot \vec{c} i . e \vec{r} ⊥ \vec{a}, \vec{r} ⊥ \vec{b} and \vec{r} ⊥ \vec{c} i . e \vec{a}, \vec{b} and \vec{c} are coplanar ∴ \vec{a} \cdot (\vec{b} \times \overset{⇀}{c}) = [\vec{a} \vec{b} \overset{⇀}{c}] \vec{a} \cdot (\vec{b} \times \overset{⇀}{c}) = 0$

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Question 1:

Write the value of $[2 \hat{i} 3 \hat{j} 4 \hat{k}] .$

Answer:

$We have [2 \hat{i} 3 \hat{j} 4 \hat{k}] = (2 \hat{i} \times 3 \hat{j}) \cdot 4 \hat{k} (∵ [\vec{a} \vec{b} \vec{c}] = (\vec{a} \times \vec{b}) . \vec{c}) = 6 \hat{k} \cdot 4 \hat{k} = 24$

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Question 2:

Write the value of $[\hat{i} + \hat{j} \hat{j} + \hat{k} \hat{k} + \hat{i}] .$

Answer:

$We have [\hat{i} + \hat{j} \hat{j} + \hat{k} \hat{k} + \hat{i}] = [(\hat{i} + \hat{j}) \times (\hat{j} + \hat{k})] \cdot (\hat{k} + \hat{i}) (∵ [\vec{a} \vec{b} \vec{c}] = (\vec{a} \times \vec{b}) . \vec{c}) = [\hat{i} \times \hat{j} + \hat{i} \times \hat{k} + \hat{j} \times \hat{j} + \hat{j} \times \hat{k}] \cdot (\hat{k} + \hat{i}) = [\hat{k} - \hat{j} + \hat{i}] \cdot (\hat{k} + \hat{i}) = [(\hat{k} \cdot \hat{k}) + (\hat{k} \cdot \hat{i}) - (\hat{j} \cdot \hat{k}) - (\hat{j} \cdot \hat{i}) + (\hat{i} \cdot \hat{k}) + (\hat{i} \cdot \hat{i})] = 1 + 0 - 0 - 0 + 0 + 1 = 2$

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Question 3:

Write the value of $[\hat{i} - \hat{j} \hat{j} - \hat{k} \hat{k} - \hat{i}] .$

Answer:

$We have [\hat{i} - \hat{j} \hat{j} - \hat{k} \hat{k} - \hat{i}] = [(\hat{i} - \hat{j}) \times (\hat{j} - \hat{k})] \cdot (\hat{k} - \hat{i}) (∵ [\vec{a} \vec{b} \vec{c}] = (\vec{a} \times \vec{b}) . \vec{c}) = [(\hat{i} \times \hat{j}) - (\hat{i} \times \hat{k}) - (\hat{j} \times \hat{j}) + (\hat{j} \times \hat{k})] \cdot (\hat{k} - \hat{i}) = [\hat{k} + \hat{j} + \hat{i}] \cdot (\hat{k} - \hat{i}) = [(\hat{k} \cdot \hat{k}) - (\hat{k} \cdot \hat{i}) + (\hat{j} \cdot \hat{k}) - (\hat{j} \cdot \hat{i}) + (\hat{i} \cdot \hat{k}) - (\hat{i} \cdot \hat{i})] = 1 - 0 + 0 - 0 + 0 - 1 = 0$

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Question 4:

Find the values of 'a' for which the vectors $\vec{α} = \hat{i} + 2 \hat{j} + \hat{k}, \vec{β} = a \hat{i} + \hat{j} + 2 \hat{k} and \vec{γ} = \hat{i} + 2 \hat{j} + a \hat{k}$ are coplanar.

Answer:

$Given : \vec{α} = \hat{i} + 2 \hat{j} + \hat{k} \vec{β} = a \hat{i} + \hat{j} + 2 \hat{k} \vec{γ} = \hat{i} + 2 \hat{j} + a \hat{k}$

$We know that three vectors \vec{α}, \vec{β}, \vec{γ} are coplanar iff their scalar product is zero . ∴ [\vec{α} \vec{β} \vec{γ}] = 0 \Rightarrow |\begin{matrix} 1 & 2 & 1 \\ a & 1 & 2 \\ 1 & 2 & a \end{matrix}| = 0 \Rightarrow 1 (a - 4) - 2 (a^{2} - 2) + 1 (2 a - 1) = 0 \Rightarrow - 2 a^{2} + 3 a - 1 = 0 \Rightarrow 2 a^{2} - 3 a + 1 = 0 \Rightarrow (a - 1) (2 a - 1) = 0 \Rightarrow a = 1, \frac{1}{2}$

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Question 5:

Find the volume of the parallelopiped with its edges represented by the vectors $\hat{i} + \hat{j}, \hat{i} + 2 \hat{j} and \hat{i} + \hat{j} + π k .$

Answer:

$Let : \vec{a} = \hat{i} + \hat{j} \vec{b} = \hat{i} + 2 \hat{j} \vec{c} = \hat{i} + \hat{j} + π \hat{k} We know that the volume of a parallelopiped whose three adjecent edges are \vec{a}, \vec{b} and \vec{c} is equal to |[\vec{a} \vec{b} \vec{c}]| . We have [\vec{a} \vec{b} \vec{c}] = |\begin{matrix} 1 & 1 & 0 \\ 1 & 2 & 0 \\ 1 & 1 & π \end{matrix}| = 1 (2 π - 0) - 1 (π - 0) + 0 (1 - 2) = 2 π - π = π ∴ Volume = |[\vec{a} \vec{b} \vec{c}]| = |π| = π cubic units$

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Question 6:

If $\vec{a,} \vec{b}$ are non-collinear vectors, then find the value of $[\vec{a} \vec{b} \hat{i}] \hat{i} + [\vec{a} \vec{b} \hat{j}] \hat{j} + [\vec{a} \vec{b} \hat{k}] \hat{k} .$

Answer:

$For any vector \vec{r}, we have (\vec{r} \cdot \hat{i}) \hat{i} + (\vec{r} \cdot \hat{j}) \hat{j} + (\vec{r} \cdot \hat{k}) \hat{k} = \vec{r} Replacing \vec{r} by \vec{a} \times \vec{b}, we have [(\vec{a} \times \vec{b}) \cdot \hat{i}] \hat{i} + [(\vec{a} \times \vec{b}) \cdot \hat{j}] \hat{j} + [(\vec{a} \times \vec{b}) \cdot \hat{k}] \hat{k} = \vec{a} \times \vec{b} ∴ [\vec{a} \vec{b} \hat{i}] \hat{i} + [\vec{a} \vec{b} \hat{j}] \hat{j} + [\vec{a} \vec{b} \hat{k}] \hat{k} = \vec{a} \times \vec{b}$

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Question 7:

If the vectors (sec² A) $\hat{i} + \hat{j} + \hat{k}, \hat{i} + (\sec^{2} B) \hat{j} + \hat{k}, \hat{i} + \hat{j} + (\sec^{2} C) \hat{k}$ are coplanar, then find the value of cosec² A + cosec² B + cosec² C.

Answer:

$Let : \vec{a} = (\sec^{2} A) \hat{i} + \hat{j} + \hat{k}, \vec{b} = \hat{i} + (\sec^{2} B) \hat{j} + \hat{k} and \vec{c} = \hat{i} + \hat{j} + (\sec^{2} C) \hat{k} W e know that three vectors are coplanar iff their scaler triple product is zero . i . e ., [\vec{a} \vec{b} \vec{c}] = 0 Here, [\vec{a} \vec{b} \vec{c}] = 0 \Rightarrow |\begin{matrix} \sec^{2} A & 1 & 1 \\ 1 & \sec^{2} B & 1 \\ 1 & 1 & \sec^{2} C \end{matrix}| = 0 \Rightarrow \sec^{2} A [(\sec^{2} B \times \sec^{2} C) - 1] - 1 (\sec^{2} C - 1) + 1 (1 - \sec^{2} B) = 0 \Rightarrow \sec^{2} A \sec^{2} B \sec^{2} C - \sec^{2} A - \sec^{2} C + 1 + 1 - \sec^{2} B = 0 \Rightarrow (1 + \tan^{2} A) (1 + \tan^{2} B) (1 + \tan^{2} C) - (1 + \tan^{2} A) - (1 + \tan^{2} C) + 1 + 1 - (1 + \tan^{2} B) = 0$

$\Rightarrow 1 + \tan^{2} A + \tan^{2} B + \tan^{2} C + \tan^{2} A \tan^{2} B + \tan^{2} B \tan^{2} C + \tan^{2} C \tan^{2} A + \tan^{2} A \tan^{2} B \tan^{2} C - 1 - \tan^{2} A - 1 - \tan^{2} C + 1 + 1 - 1 - \tan^{2} B = 0 \Rightarrow \tan^{2} A \tan^{2} B + \tan^{2} B \tan^{2} C + \tan^{2} C \tan^{2} A + \tan^{2} A \tan^{2} B \tan^{2} C = 0 \Rightarrow \tan^{2} A \tan^{2} B + \tan^{2} B \tan^{2} C + \tan^{2} C \tan^{2} A = - \tan^{2} A \tan^{2} B \tan^{2} C \Rightarrow \frac{\tan^{2} A \tan^{2} B + \tan^{2} B \tan^{2} C + \tan^{2} C \tan^{2} A}{\tan^{2} A \tan^{2} B \tan^{2} C} = - 1 \Rightarrow \cot^{2} C + \cot^{2} A + \cot^{2} B = - 1 \Rightarrow {cosec}^{2} C - 1 + {cosec}^{2} A - 1 + {cosec}^{2} B - 1 = - 1 ∴ {cosec}^{2} A + {cosec}^{2} B + {cosec}^{2} C = 2$

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Question 8:

For any two vectors $\vec{a} and \vec{b}$ of magnitudes 3 and 4 respectively, write the value of $[\vec{a} \vec{b} \vec{a} \times \vec{b}] + {(\vec{a} \cdot \vec{b})}^{2} .$

Answer:

$We have [\vec{a} \vec{b} \vec{a} \times \vec{b}] + {(\vec{a} . \vec{b})}^{2} = [(\vec{a} \times \vec{b}) . (\vec{a} \times \vec{b})] + {(\vec{a} . \vec{b})}^{2} (By definition of scalar triple product) = {|(\vec{a} \times \vec{b})|}^{2} + {(\vec{a} . \vec{b})}^{2} = {(|\begin{matrix} \vec{a} \end{matrix}| |\begin{matrix} \vec{b} \end{matrix}| s inθ)}^{2} + {(|\begin{matrix} \vec{a} \end{matrix}| |\begin{matrix} \vec{b} \end{matrix}| c osθ)}^{2} = {(|\begin{matrix} \vec{a} \end{matrix}| |\begin{matrix} \vec{b} \end{matrix}|)}^{2} s {in}^{2} θ + {(|\begin{matrix} \vec{a} \end{matrix}| |\begin{matrix} \vec{b} \end{matrix}|)}^{2} \cos^{2} θ = {(|\begin{matrix} \vec{a} \end{matrix}| |\begin{matrix} \vec{b} \end{matrix}|)}^{2} (\sin^{2} θ + c {os}^{2} θ) = {(|\begin{matrix} \vec{a} \end{matrix}| |\begin{matrix} \vec{b} \end{matrix}|)}^{2} (∵ \sin^{2} θ + c {os}^{2} θ = 1) = {(3 \times 4)}^{2} (Given : |\begin{matrix} \vec{a} \end{matrix}| = 3 and |\begin{matrix} \vec{b} \end{matrix}| = 4) = 144$

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Question 9:

If $[3 \vec{a} + 7 \vec{b} \vec{c} \vec{d}] = λ [\vec{a} \vec{c} \vec{d}] + μ [\vec{b} \vec{c} \vec{d}],$ then find the value of λ + μ.

Answer:

$We have [3 \vec{a} + 7 \vec{b} \vec{c} \vec{d}] = λ [\vec{a} \vec{c} \vec{d}] + μ [\vec{b} \vec{c} \vec{d}] \Rightarrow [(3 \vec{a} + 7 \vec{b}) \times \vec{c}] . \vec{d} = λ [\vec{a} \vec{c} \vec{d}] + μ [\vec{b} \vec{c} \vec{d}] (By definition of scalar triple product) \Rightarrow [(3 \vec{a} \times \vec{c}) + (7 \vec{b} \times \vec{c})] . \vec{d} = λ [\vec{a} \vec{c} \vec{d}] + μ [\vec{b} \vec{c} \vec{d}] \Rightarrow (3 \vec{a} \times \vec{c}) . \vec{d} + (7 \vec{b} \times \vec{c}) . \vec{d} = λ [\vec{a} \vec{c} \vec{d}] + μ [\vec{b} \vec{c} \vec{d}] \Rightarrow [3 \vec{a} \vec{c} \vec{d}] + [7 \vec{b} \vec{c} \vec{d}] = λ [\vec{a} \vec{c} \vec{d}] + μ [\vec{b} \vec{c} \vec{d}] \Rightarrow 3 [\vec{a} \vec{c} \vec{d}] + 7 [\vec{b} \vec{c} \vec{d}] = λ [\vec{a} \vec{c} \vec{d}] + μ [\vec{b} \vec{c} \vec{d}] (∵ [λ \vec{a} \vec{b} \vec{c}] = λ [\vec{a} \vec{b} \vec{c}] for any scalar λ) Comparing both sides, we get λ = 3 μ = 7 ∴ λ + μ = 3 + 7 = 10$

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Question 10:

If $\vec{a,} \vec{b,} \vec{c}$ are non-coplanar vectors, then find the value of $\frac{\vec{a} \cdot (\vec{b} \times \vec{c})}{(\vec{c} \times \vec{a}) \cdot \vec{b}} + \frac{\vec{b} \cdot (\vec{a} \times \vec{c})}{\vec{c} \cdot (\vec{a} \times \vec{b})}$

Answer:

$We have \frac{\vec{a} . (\vec{b} \times \vec{c})}{(\vec{c} \times \vec{a}) . \vec{b}} + \frac{\vec{b} . (\vec{a} \times \vec{c})}{\vec{c} . (\vec{a} \times \vec{b})} = \frac{[\vec{a} \vec{b} \vec{c}]}{[\vec{c} \vec{a} \vec{b}]} + \frac{[\vec{b} \vec{a} \vec{c}]}{[\vec{c} \vec{a} \vec{b}]} (By definition of scalar triple product) = \frac{[\vec{a} \vec{b} \vec{c}]}{[\vec{a} \vec{b} \vec{c}]} + \frac{- [\vec{a} \vec{b} \vec{c}]}{[\vec{a} \vec{b} \vec{c}]} (Change of cyclic order of vectors changes the sign of the scalar triple product) = 1 - 1 = 0$

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Question 11:

Find $\vec{a} . (\vec{b} \times \vec{c})$ , if $\vec{a} = 2 \hat{i} + \hat{j} + 3 \hat{k}, \vec{b} = - \hat{i} + 2 \hat{j} + \hat{k}$ and $\vec{c} = 3 \hat{i} + \hat{j} + 2 \hat{k}$ . [CBSE 2014]

Answer:

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

Now,

$\vec{b} \times \vec{c} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ - 1 & 2 & 1 \\ 3 & 1 & 2 \end{matrix}| = 3 \hat{i} + 5 \hat{j} - 7 \hat{k}$

$∴ \vec{a} . (\vec{b} \times \vec{c}) = (2 \hat{i} + \hat{j} + 3 \hat{k}) . (3 \hat{i} + 5 \hat{j} - 7 \hat{k}) = 2 \times 3 + 1 \times 5 + 3 \times (- 7) = 6 + 5 - 21 = - 10$

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