Rd Sharma XII Vol 2 2020 for Class 12 Science Maths Chapter 6 - Vector Or Cross Product

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Rd Sharma XII Vol 2 2020 Solutions for Class 12 Science Maths Chapter 6 Vector Or Cross Product are provided here with simple step-by-step explanations. These solutions for Vector Or Cross Product are extremely popular among Class 12 Science students for Maths Vector Or Cross 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 6 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 24.30:

Question 1:

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

Answer:

$Given : \vec{a} = \hat{i} + 3 \hat{j} - 2 \hat{k} \vec{b} = - \hat{i} + 0 \hat{j} + 3 \hat{k} ∴ \vec{a} \times \vec{b} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & - 2 \\ - 1 & 0 & 3 \end{matrix}| = (9 + 0) \hat{i} - (3 - 2) \hat{j} + (0 + 3) \hat{k} = 9 \hat{i} - \hat{j} + 3 \hat{k} \Rightarrow |\vec{a} \times \vec{b}| = \sqrt{9^{2} + {(- 1)}^{2} + 3^{2}} = \sqrt{91}$

Page No 24.30:

Question 2:

(i) If $\vec{a} = 3 \hat{i} + 4 \hat{j} and \vec{b} = \hat{i} + \hat{j} + \hat{k},$ find the value of $|\vec{a} \times \vec{b}| .$

(ii) If $\vec{a} = 2 \hat{i} + \hat{k}, \vec{b} = \hat{i} + \hat{j} + \hat{k},$ find the magnitude of $\vec{a} \times \vec{b} .$

Answer:

$(i) Given : \vec{a} = 3 \hat{i} + 4 \hat{j} + 0 \hat{k} \vec{b} = \hat{i} + \hat{j} + \hat{k} ∴ \vec{a} \times \vec{b} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 0 \\ 1 & 1 & 1 \end{matrix}| = (4 - 0) \hat{i} - (3 - 0) \hat{j} + (3 - 4) \hat{k} = 4 \hat{i} - 3 \hat{j} - \hat{k} \Rightarrow |\vec{a} \times \vec{b}| = \sqrt{16 + 9 + 1} = \sqrt{26}$

$(ii) Given : \vec{a} = 2 \hat{i} + 0 \hat{j} + \hat{k} \vec{b} = \hat{i} + \hat{j} + \hat{k} \vec{a} \times \vec{b} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 1 \\ 1 & 1 & 1 \end{matrix}| = (0 - 1) \hat{i} - (2 - 1) \hat{j} + (2 - 0) \hat{k} = - \hat{i} - \hat{j} + 2 \hat{k} \Rightarrow |\vec{a} \times \vec{b}| = \sqrt{{(- 1)}^{2} + {(- 1)}^{2} + 2^{2}} = \sqrt{6}$

Page No 24.30:

Question 3:

(i) Find a unit vector perpendicular to both the vectors $4 \hat{i} - \hat{j} + 3 \hat{k} and - 2 \hat{i} + \hat{j} - 2 \hat{k} .$

(ii) Find a unit vector perpendicular to the plane containing the vectors $\vec{a} = 2 \hat{i} + \hat{j} + \hat{k} and \vec{b} = \hat{i} + 2 \hat{j} + \hat{k} .$

Answer:

$(i) Given : \vec{a} = 4 \hat{i} - \hat{j} + 3 \hat{k} \vec{b} = - 2 \hat{i} + \hat{j} - 2 \hat{k} ∴ \vec{a} \times \vec{b} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & - 1 & 3 \\ - 2 & 1 & - 2 \end{matrix}| = (2 - 3) \hat{i} - (- 8 + 6) \hat{j} + (4 - 2) \hat{k} = - \hat{i} + 2 \hat{j} + 2 \hat{k} \Rightarrow |\vec{a} \times \vec{b}| = \sqrt{1 + 2^{2} + 2^{2}} = \sqrt{9} = 3 Unit vector perpendicular to \vec{a} and \vec{b} = \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} = \frac{- \hat{i} + 2 \hat{j} + 2 \hat{k}}{3}$

$(ii) Given : \vec{a} = 2 \hat{i} + \hat{j} + \hat{k} \vec{b} = \hat{i} + 2 \hat{j} + \hat{k} \overset{}{∴ \vec{a}} \times \overset{}{\vec{b}} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 1 \\ 1 & 2 & 1 \end{matrix}| = (1 - 2) \hat{i} - (2 - 1) \hat{j} + \overset{}{(4 - 1) \overset{⏜}{k}} = - \hat{i} - \hat{j} + 3 \hat{k} \Rightarrow |\vec{a} \times \vec{b}| = \sqrt{1 + 1 + 9} = \sqrt{11} Unit vector perpendicular to the plane containing vectors \vec{a} and \vec{b} = \pm \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} Unit vector perpendicular to the plane containing vectors \vec{a} and \vec{b} = \pm \frac{1}{\sqrt{11}} (- \hat{i} - \hat{j} + 3 \hat{k})$

Page No 24.30:

Question 4:

Find the magnitude of $\vec{a} = (3 \hat{k} + 4 \hat{j}) \times (\hat{i} + \hat{j} - \hat{k}) .$

Answer:

$\vec{a} = (0 \hat{i} + 4 \hat{j} + 3 \hat{k}) \times (\hat{i} + \hat{j} - \hat{k}) = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 4 & 3 \\ 1 & 1 & - 1 \end{matrix}| = \hat{i} (- 4 - 3) - \hat{j} (0 - 3) + \hat{k} (0 - 4) = - 7 \hat{i} + 3 \hat{j} - 4 \hat{k} \Rightarrow |\vec{a}| = \sqrt{{(- 7)}^{2} + 3^{2} + {(- 4)}^{2}} = \sqrt{74}$

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Question 5:

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

Answer:

$Given : \vec{a} = 4 \hat{i} + 3 \hat{j} + \hat{k} 2 \vec{b} = 2 \hat{i} + 0 \hat{j} - 4 \hat{k} 2 \vec{b} \times \vec{a} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & - 4 \\ 4 & 3 & 1 \end{matrix}| = (0 + 12) \hat{i} - (2 + 16) \hat{j} + (6 - 0) \hat{k} = 12 \hat{i} - 18 \hat{j} + 6 \hat{k} \Rightarrow |2 \vec{b} \times \vec{a}| = \sqrt{12^{2} + (- 18^{2}) + 6^{2}} = \sqrt{504}$

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Question 6:

$If \vec{a} = 3 \hat{i} - \hat{j} - 2 \hat{k} and \vec{b} = 2 \hat{i} + 3 \hat{j} + \hat{k}, find (\vec{a} + 2 \vec{b}) \times (2 \vec{a} - \vec{b}) .$

Answer:

$Given : \vec{a} = 3 \hat{i} - \hat{j} - 2 \hat{k} \vec{b} = 2 \hat{i} + 3 \hat{j} + \hat{k} ∴ \vec{a} + 2 \vec{b} = 3 \hat{i} - \hat{j} - 2 \hat{k} + 2 (2 \hat{i} + 3 \hat{j} + \hat{k}) = 7 \hat{i} + 5 \hat{j} + 0 \hat{k} ∴ 2 \vec{a} - \vec{b} = 2 (3 \hat{i} - \hat{j} - 2 \hat{k}) - (2 \hat{i} + 3 \hat{j} + \hat{k}) = 4 \hat{i} - 5 \hat{j} - 5 \hat{k} (\vec{a} + 2 \vec{b}) \times (2 \vec{a} - \vec{b}) = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & 5 & 0 \\ 4 & - 5 & - 5 \end{matrix}| = \hat{i} (- 25 + 0) - \hat{j} (- 35 + 0) + \hat{k} (- 35 - 20) = - 25 \hat{i} + 35 \hat{j} - 55 \hat{k}$

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Question 7:

(i) Find a vector of magnitude 49, which is perpendicular to both the vectors $2 \hat{i} + 3 \hat{j} + 6 \hat{k} and 3 \hat{i} - 6 \hat{j} + 2 \hat{k} .$

(ii) Find a vector whose length is 3 and which is perpendicular to the vector $\vec{a} = 3 \hat{i} + \hat{j} - 4 \hat{k} and \vec{b} = 6 \hat{i} + 5 \hat{j} - 2 \hat{k} .$

Answer:

$(i) Given : \vec{a} = 2 \hat{i} + 3 \hat{j} + 6 \hat{k} \vec{b} = 3 \hat{i} - 6 \hat{j} + 2 \hat{k} \vec{a} \times \vec{b} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 6 \\ 3 & - 6 & 2 \end{matrix}| = (6 + 36) \hat{i} - (4 - 18) \hat{j} + (- 12 - 9) \hat{k} = 42 \hat{i} + 14 \hat{j} - 21 \hat{k} \Rightarrow |\vec{a} \times \vec{b}| = \sqrt{42^{2} + 14^{2} + (- 21^{2})} = \sqrt{2401} = 49 Required vector = 49 \times \{\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}\} = 49 \times \frac{42 \hat{i} + 14 \hat{j} - 21 \hat{k}}{49} = 42 \hat{i} + 14 \hat{j} - 21 \hat{k}$

$(ii) Given : \vec{a} = 3 \hat{i} + \hat{j} - 4 \hat{k} \vec{b} = 6 \hat{i} + 5 \hat{j} - 2 \hat{k} ∴ \vec{a} \times \vec{b} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & - 4 \\ 6 & 5 & - 2 \end{matrix}| = (- 2 + 20) \hat{i} - (- 6 + 24) \hat{j} + (15 - 6) \hat{k} = 18 \hat{i} - 18 \hat{j} + 9 \hat{k} \Rightarrow |\vec{a} \times \vec{b}| = \sqrt{18^{2} + (- 18^{2}) + 9^{2}} = \sqrt{729} = 27 Required vector = 3 \times \{\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}\} = 3 \times \frac{18 \hat{i} - 18 \hat{j} + 9 \hat{k}}{27} = \frac{3 (2 \hat{i} - 2 \hat{j} + \hat{k})}{3} = 2 \hat{i} - 2 \hat{j} + \hat{k}$

Page No 24.30:

Question 8:

Find the area of the parallelogram determined by the vectors:
(i) $2 \hat{i} and 3 \hat{j}$

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

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

(iv) $\hat{i} - 3 \hat{j} + \hat{k} and \hat{i} + \hat{j} + \hat{k} .$

Answer:

$(i) Let : a = 2 \hat{i} + 0 \hat{j} + 0 \hat{k} \vec{b} = 0 \hat{i} + 3 \hat{j} + 0 \hat{k} ∴ \vec{a} \times \vec{b} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 0 \\ 0 & 3 & 0 \end{matrix}| = (0 - 0) \hat{i} - (0 - 0) \hat{j} + (6 - 0) \hat{k} = 0 \hat{i} + 0 \hat{j} + 6 \hat{k} Area of the parallelogram= |\vec{a} \times \vec{b}| = \sqrt{0 + 0 + 6^{2}} = 6 sq. units$

$(ii) Let : \vec{a} = 2 \hat{i} + \hat{j} + 3 \hat{k} \vec{b} = \hat{i} - \hat{j} + 0 \hat{k} ∴ \vec{a} \times \vec{b} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 1 & - 1 & 0 \end{matrix}| = (0 + 3) \hat{i} - (0 - 3) \hat{j} + (- 2 - 1) \hat{k} = 3 \hat{i} + 3 \hat{j} - 3 \hat{k} Area of the parallelogram = |\vec{a} \times \vec{b}| = \sqrt{3^{2} + 3^{2} + 3^{2}} = \sqrt{27} = 3 \sqrt{3} sq. units$

$(iii) Let : \vec{a} = 3 \hat{i} + \hat{j} - 2 \hat{k} \vec{b} = 1 \hat{i} - 3 \hat{j} + 4 \hat{k} \vec{a} \times \vec{b} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & - 2 \\ 1 & - 3 & 4 \end{matrix}| = \hat{i} (4 - 6) - \hat{j} (12 + 2) + \hat{k} (- 9 - 1) = - 2 \hat{i} - 14 \hat{j} - 10 \hat{k} Area of the parallelogram= |\vec{a} \times \vec{b}| = \sqrt{{(- 2)}^{2} + {(- 14)}^{2} + {(- 10)}^{2}} = \sqrt{300} = 10 \sqrt{3} sq. units$

$(iv) Let : \vec{a} = \hat{i} - 3 \hat{j} + \hat{k} \vec{b} = \hat{i} + \hat{j} + \hat{k} \vec{a} \times \vec{b} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 3 & 1 \\ 1 & 1 & 1 \end{matrix}| = (- 3 - 1) \hat{i} - (1 - 1) \hat{j} + (1 + 3) \hat{k} = - 4 \hat{i} + 0 \hat{j} + 4 \hat{k} Area of the parallelogram= |\vec{a} \times \vec{b}| = \sqrt{{(- 4)}^{2} + 0 + 4^{2}} = \sqrt{32} = 4 \sqrt{2} sq. units.$

Page No 24.30:

Question 9:

Find the area of the parallelogram whose diagonals are:
(i) $4 \hat{i} - \hat{j} - 3 \hat{k} and - 2 \hat{j} + \hat{j} - 2 \hat{k}$

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

(iii) $3 \hat{i} + 4 \hat{j} and \hat{i} + \hat{j} + \hat{k}$

(iv) $2 \hat{i} + 3 \hat{j} + 6 \hat{k} and 3 \hat{i} - 6 \hat{j} + 2 \hat{k}$

Answer:

$(i) Let : \vec{a} = 4 \hat{i} - \hat{j} - 3 \hat{k} \vec{b} = - 2 \hat{i} + \hat{j} - 2 \hat{k} ∴ \vec{a} \times \vec{b} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & - 1 & - 3 \\ - 2 & 1 & - 2 \end{matrix}| = (2 + 3) \hat{i} - (- 8 - 6) \hat{j} + (4 - 2) \hat{k} = 5 \hat{i} + 14 \hat{j} + 2 \hat{k} \Rightarrow |\vec{a} \times \vec{b}| = \sqrt{25 + 196 + 4} = \sqrt{225} = 15 Area of the parallelogram = \frac{1}{2} |\vec{a} \times \vec{b}| = \frac{15}{2} sq. units.$

$(ii) Let : \vec{a} = 2 \hat{i} + 0 \hat{j} + \hat{k} \vec{b} = \hat{i} + \hat{j} + \hat{k} \vec{a} \times \vec{b} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 1 \\ 1 & 1 & 1 \end{matrix}| = (0 - 1) \hat{i} - (2 - 1) \hat{j} + (2 - 0) \hat{k} = - \hat{i} - \hat{j} + 2 \hat{k} \Rightarrow |\vec{a} \times \vec{b}| = \sqrt{{(- 1)}^{2} + {(- 1)}^{2} + (2)} = \sqrt{6} Area of the parallelogram = \frac{1}{2} |\vec{a} \times \vec{b}| = \frac{\sqrt{6}}{2} sq. units$

$(iii) Let : \vec{a} = 3 \hat{i} + 4 \hat{j} + 0 \hat{k} \vec{b} = \hat{i} + \hat{j} + \hat{k} ∴ \vec{a} \times \vec{b} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 0 \\ 1 & 1 & 1 \end{matrix}| = (4 - 0) \hat{i} - (3 - 0) \hat{j} + (3 - 4) \hat{k} = 4 \hat{i} - 3 \hat{j} - \hat{k} \Rightarrow |\vec{a} \times \vec{b}| = \sqrt{4^{2} + {(- 3)}^{2} + {(- 1)}^{2}} = \sqrt{26} Area of the parallelogram = \frac{1}{2} |\vec{a} \times \vec{b}| = \frac{\sqrt{26}}{2} sq. units$

$(iv) Let : \vec{a} = 2 \hat{i} + 3 \hat{j} + 6 \hat{k} \vec{b} = 3 \hat{i} - 6 \hat{j} + 2 \hat{k} ∴ \vec{a} \times \vec{b} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 6 \\ 3 & - 6 & 2 \end{matrix}| = (6 + 36) \hat{i} - (4 - 18) \hat{j} + (- 12 - 9) \hat{k} = 42 \hat{i} + 14 \hat{j} - 21 \hat{k} \Rightarrow |\vec{a} \times \vec{b}| = \sqrt{42^{2} + 14^{2} + {(- 21)}^{2}} = \sqrt{2401} = 49 Area of the parallelogram = \frac{1}{2} |\vec{a} \times \vec{b}| = \frac{49}{2} sq. units$

Disclaimer: The answer given for (iii) and (iv) in the textbook is incorrect.

Page No 24.30:

Question 10:

If $\vec{a} = 2 \hat{i} + 5 \hat{j} - 7 \hat{k}, \vec{b} = - 3 \hat{i} + 4 \hat{j} + \hat{k} and \vec{c} = \hat{i} - 2 \hat{j} - 3 \hat{k},$ compute $(\vec{a} \times \vec{b}) \times \vec{c} and \vec{a} \times (\vec{b} \times \vec{c})$ and verify that these are not equal.

Answer:

$Given : \vec{a} = 2 \hat{i} + 5 \hat{j} - 7 \hat{k} \vec{b} = - 3 \hat{i} + 4 \hat{j} + \hat{k} \vec{c} = \hat{i} - 2 \hat{j} - 3 \hat{k} ∴ \vec{a} \times \vec{b} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 5 & - 7 \\ - 3 & 4 & 1 \end{matrix}| = (5 + 28) \hat{i} - (2 - 21) \hat{j} + (8 + 15) \hat{k} = 33 \hat{i} + 19 \hat{j} + 23 \hat{k} \Rightarrow (\vec{a} \times \vec{b}) \times \vec{c} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 33 & 19 & 23 \\ 1 & - 2 & - 3 \end{matrix}| = (- 57 + 46) \hat{i} - (- 99 - 23) \hat{j} + (- 66 - 19) \hat{k} \Rightarrow (\vec{a} \times \vec{b}) \times \vec{c} = - 11 \hat{i} + 122 \hat{j} - 85 \hat{k} . . . (1) ∴ \vec{b} \times \vec{c} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ - 3 & 4 & 1 \\ 1 & - 2 & - 3 \end{matrix}| = (- 12 + 2) \hat{i} - (9 - 1) \hat{j} + (6 - 4) \hat{k} = - 10 \hat{i} - 8 \hat{j} + 2 \hat{k} \Rightarrow \vec{a} \times (\vec{b} \times \vec{c}) = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 5 & - 7 \\ - 10 & - 8 & 2 \end{matrix}| = (10 - 56) \hat{i} - (4 - 70) \hat{j} + (- 16 + 50) \hat{k} \Rightarrow \vec{a} \times (\vec{b} \times \vec{c}) = - 46 \hat{i} + 66 \hat{j} + 34 \hat{k} . . . (2) From (1) and (2), we get (\vec{a} \times \vec{b}) \times \vec{c} \neq \vec{a} \times (\vec{b} \times \vec{c})$

Page No 24.30:

Question 11:

$If |\vec{a}| = 2, |\vec{b}| = 5 and |\vec{a} \times \vec{b}| = 8, find \vec{a} \cdot \vec{b} .$

Answer:

$We know {(\vec{a} . \vec{b})}^{2} + {|\vec{a} \times \vec{b}|}^{2} = {|\vec{a}|}^{2} {|\vec{b}|}^{2} \Rightarrow {(\vec{a} . \vec{b})}^{2} + 8^{2} = 2^{2} \times 5^{2} (∵ |\vec{a} \times \vec{b}| = 8, |\vec{a}| = 2 and |\vec{b}| = 5) \Rightarrow {(\vec{a} . \vec{b})}^{2} + 64 = 100 \Rightarrow {(\vec{a} . \vec{b})}^{2} = 36 \Rightarrow (\vec{a} . \vec{b}) = 6$

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Question 12:

Given $\vec{a} = \frac{1}{7} (2 \hat{i} + 3 \hat{j} + 6 \hat{k}), \vec{b} = \frac{1}{7} (3 \hat{i} - 6 \hat{j} + 2 \hat{k}), \vec{c} = \frac{1}{7} (6 \hat{i} + 2 \hat{j} - 3 \hat{k}), \hat{i}, \hat{j}, \hat{k}$ being a right handed orthogonal system of unit vectors in space, show that $\vec{a}, \vec{b}, \vec{c}$ is also another system.

Answer:

$Given : \vec{a} = \frac{1}{7} (2 \hat{i} + 3 \hat{j} + 6 \hat{k}) \vec{b} = \frac{1}{7} (3 \hat{i} - 6 \hat{j} + 2 \hat{k}) \vec{c} = \frac{1}{7} (6 \hat{i} + 2 \hat{j} - 3 \hat{k}) \vec{a} \times \vec{b} = (\frac{1}{7}) (\frac{1}{7}) |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 6 \\ 3 & - 6 & 2 \end{matrix}| = \frac{1}{49} (42 \hat{i} + 14 \hat{j} - 21 \hat{k}) = \frac{1}{49} [7 (6 \hat{i} + 2 \hat{j} - 3 \hat{k})] = \frac{1}{7} (6 \hat{i} + 2 \hat{j} - 3 \hat{k}) = \vec{c} \vec{b} \times \vec{c} = (\frac{1}{7}) (\frac{1}{7}) |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & - 6 & 2 \\ 6 & 2 & - 3 \end{matrix}| = \frac{1}{49} (14 \hat{i} + 21 \hat{j} + 42 \hat{k}) = \frac{1}{49} [7 (2 \hat{i} + 3 \hat{j} + 6 \hat{k})] = \frac{1}{7} (2 \hat{i} + 3 \hat{j} + 6 \hat{k}) = \vec{a} \vec{c} \times \vec{a} = (\frac{1}{7}) (\frac{1}{7}) |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 6 & 2 & - 3 \\ 2 & 3 & 6 \end{matrix}| = \frac{1}{49} (21 \hat{i} - 42 \hat{j} + 14 \hat{k}) = \frac{1}{49} [7 (3 \hat{i} - 6 \hat{j} + 2 \hat{k})] = \frac{1}{7} (3 \hat{i} - 6 \hat{j} + 2 \hat{k}) = \vec{b} |\vec{a}| = \frac{1}{7} \sqrt{4 + 9 + 36} = \frac{7}{7} = 1 |\vec{b}| = \frac{1}{7} \sqrt{9 + 36 + 4} = \frac{7}{7} = 1 |\vec{c}| = \frac{1}{7} \sqrt{36 + 4 + 9} = \frac{7}{7} = 1 Thus, \vec{a}, \vec{b} and \vec{c} form a right handed orthogonal system of unit vectors.$

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Question 13:

$If |\vec{a}| = 13, |\vec{b}| = 5 and \vec{a} . \vec{b} = 60, then find |\vec{a} \times \vec{b}| .$

Answer:

$We know {(\vec{a} . \vec{b})}^{2} + {|\vec{a} \times \vec{b}|}^{2} = {|\vec{a}|}^{2} {|\vec{b}|}^{2} \Rightarrow {(60)}^{2} + {|\vec{a} \times \vec{b}|}^{2} = {(13)}^{2} \times 5^{2} (∵ \vec{a} . \vec{b} = 60, |\vec{a}| = 13 and |\vec{b}| = 5) \Rightarrow 3600 + {|\vec{a} \times \vec{b}|}^{2} = 4225 \Rightarrow {|\vec{a} \times \vec{b}|}^{2} = 625 \Rightarrow |\vec{a} \times \vec{b}| = 25$

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Question 14:

Find the angle between two vectors $\vec{a} and \vec{b}$ , if $|\vec{a} \times \vec{b}| = \vec{a} \cdot \vec{b} .$

Answer:

$Let θ be the angle between \vec{a} and \vec{b} . Given: |\vec{a} \times \vec{b}| = \vec{a} . \vec{b} \Rightarrow |\vec{a}| |\vec{b}| \sin θ = |\vec{a}| |\vec{b}| \cos θ \Rightarrow \sin θ = \cos θ \Rightarrow \tan θ = 1 \Rightarrow θ = \frac{π}{4}$

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Question 15:

If $\vec{a} \times \vec{b} = \vec{b} \times \vec{c} \neq 0,$ then show that $\vec{a} + \vec{c} = m \vec{b},$ where m is any scalar.

Answer:

$\vec{a} \times \vec{b} = \vec{b} \times \vec{c} \Rightarrow \vec{a} \times \vec{b} = - \vec{c} \times \vec{b} \Rightarrow \vec{a} \times \vec{b} + \vec{c} \times \vec{b} = 0 \Rightarrow (\vec{a} + \vec{c}) \times \vec{b} = 0 (Using right distributive property) Thus, \vec{a} + \vec{c} is parallel to \vec{b} . \Leftrightarrow \vec{a} + \vec{c} = m \vec{b}, for some scalar m.$

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Question 16:

If $|\vec{a}| = 2, |\vec{b}| = 7 and \vec{a} \times \vec{b} = 3 \hat{i} + 2 \hat{j} + 6 \hat{k},$ find the angle between $\vec{a} and \vec{b} .$

Answer:

$Let θ be the angle between \vec{a} and \vec{b} . \vec{a} \times \vec{b} = 3 \hat{i} + 2 \hat{j} + 6 \hat{k} (Given) \Rightarrow |\vec{a} \times \vec{b}| = \sqrt{9 + 4 + 36} = 7 We know |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin θ \Rightarrow 7 = (2) (7) \sin θ \Rightarrow \sin θ = \frac{1}{2} \Rightarrow θ = \frac{π}{6}$

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Question 17:

What inference can you draw if $\vec{a} \times \vec{b} = \vec{0} and \vec{a} \cdot \vec{b} = 0 .$

Answer:

$Given: |\vec{a} \times \vec{b}| = \vec{0} \Rightarrow \vec{a} = 0 \vec{b} =0 ∴ \vec{a} ∥ \vec{b} Also, \vec{a} . \vec{b} = 0 \Rightarrow |\vec{a}| |\vec{b}| \cos θ = 0 \Rightarrow \vec{a} = \vec{0} or \vec{b} = \vec{0} or, \vec{a} ⊥ \vec{b} But \vec{a} cannot be both perpendicular as well as parallel to \vec{b} . ∴ |\vec{a}| =0 |\vec{b}| =0$

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Question 18:

If $\vec{a,} \vec{b,} \vec{c}$ are three unit vectors such that $\vec{a} \times \vec{b} = \vec{c}, \vec{b} \times \vec{c} = \vec{a,} \vec{c} \times \vec{a} = \vec{b} .$ Show that $\vec{a,} \vec{b,} \vec{c}$ form an orthonormal right handed triad of unit vectors.

Answer:

$Given: \vec{a} \times \vec{b} = \vec{c} \vec{b} \times \vec{c} = \vec{a} \vec{c} \times \vec{a} = \vec{b} . . . (1) Now, |\vec{a} \times \vec{b}| = |\vec{c}| = 1 (∵ \vec{c} is a unit vector) |\vec{b} \times \vec{c}| = |\vec{a}| = 1 (∵ \vec{a} is a unit vector) |\vec{c} \times \vec{a}| = |\vec{b}| = 1 (∵ \vec{b} is a unit vector) ∴ |\vec{a} \times \vec{b}| = |\vec{b} \times \vec{c}| = |\vec{c} \times \vec{a}| = 1 . . . (2) From (1) and (2), we know \vec{a}, \vec{b} and \vec{c} form an orthonormal right handed triad of unit vectors.$

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Question 19:

Find a unit vector perpendicular to the plane ABC, where the coordinates of A, B and C are A (3, −1, 2), B (1, −1, −3) and C (4, −3, 1).

Answer:

$The vector \vec{A B} \times \vec{A C} is perpendicular to the vectors \vec{A B} and \vec{A C} . ∴ Required unit vector = \frac{\vec{A B} \times \vec{A C}}{|\vec{A B} \times \vec{A C}|} Now, \vec{A B} = Position vector of B - Position vector of A = (\hat{i} - \hat{j} - 3 \hat{k}) - (3 \hat{i} - \hat{j} + 2 \hat{k}) = - 2 \hat{i} + 0 \hat{j} - 5 k \vec{A C} = Position vector of C - Position vector of A = (4 \hat{i} - 3 \hat{j} + \hat{k}) - (3 \hat{i} - \hat{j} + 2 \hat{k}) = \hat{i} - 2 \hat{j} - \hat{k} ∴ \vec{A B} \times \vec{A C} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ - 2 & 0 & - 5 \\ 1 & - 2 & - 1 \end{matrix}| = (0 - 10) i - (2 + 5) j + (4 - 0) \hat{k} = - 10 \hat{i} - 7 \hat{j} + 4 \hat{k} |\vec{A B} \times \vec{B C}| = \sqrt{{(- 10)}^{2} + {(- 7)}^{2} + 4^{2}} = \sqrt{165} Unit vector perpendicular to the plane ABC = \frac{\vec{A B} \times \vec{A C}}{|\vec{A B} \times \vec{A C}|} = \frac{- 10 \hat{i} - 7 \hat{j} + 4 \hat{k}}{\sqrt{165}}$

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Question 20:

If a, b, c are the lengths of sides, BC, CA and AB of a triangle ABC, prove that $\vec{B C} + \vec{C A} + \vec{A B} = \vec{0}$ and deduce that $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} .$

Answer:

$We have \vec{B C} = \vec{a} \vec{C A} = \vec{b} \vec{A B} = \vec{c} |\vec{a}| = a |\vec{b}| =b (∵ Length is always positive) \vec{c} =c Now, \vec{B C} + \vec{C A} + \vec{A B} = \vec{0} (Given) \Rightarrow \vec{a} + \vec{b} + \vec{c} = \vec{0} \Rightarrow \vec{a} \times (\vec{a} + \vec{b} + \vec{c}) = \vec{a} \times \vec{0} \Rightarrow \vec{a} \times \vec{a} + \vec{a} \times \vec{b} + \vec{a} \times \vec{c} = \vec{0} \Rightarrow \vec{0} + \vec{a} \times \vec{b} - \vec{c} \times \vec{a} = \vec{0} \Rightarrow \vec{a} \times \vec{b} = \vec{c} \times \vec{a} \Rightarrow |\vec{a}| |\vec{b}| \sin C = |\vec{c}| |\vec{a}| \sin B \Rightarrow a b \sin C = c a \sin B Dividing both sides by abc, we get \Rightarrow \frac{\sin C}{c} = \frac{\sin B}{b} . . . (1) Again, \vec{B C} + \vec{C A} + \vec{A B} = \vec{0} \Rightarrow \vec{a} + \vec{b} + \vec{c} = \vec{0} \Rightarrow \vec{b} \times (\vec{a} + \vec{b} + \vec{c}) = \vec{b} \times \vec{0} \Rightarrow \vec{b} \times \vec{a} + \vec{b} \times \vec{b} + \vec{b} \times \vec{c} = \vec{0} \Rightarrow - \vec{a} \times \vec{b} + \vec{0} + \vec{b} \times \vec{c} = \vec{0} \Rightarrow \vec{a} \times \vec{b} = \vec{b} \times \vec{c} \Rightarrow |\vec{a}| |\vec{b}| \sin C = |\vec{b}| |\vec{c}| \sin A \Rightarrow a b \sin C = b c \sin A Dividing both sides by abc, we get \Rightarrow \frac{\sin C}{c} = \frac{\sin A}{a} . . . (2) From (1) and (2), we get \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} \Rightarrow \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

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Question 21:

If $\vec{a} = \hat{i} - 2 \hat{j} + 3 \hat{k}, and \vec{b} = 2 \hat{i} + 3 \hat{j} - 5 \hat{k},$ then find $\vec{a} \times \vec{b} .$ Verify that $\vec{a} and \vec{a} \times \vec{b}$ are perpendicular to each other.

Answer:

$Given : \vec{a} = \overset{⏜}{i} - 2 \overset{⏜}{j} + 3 \overset{⏜}{k} \vec{b} = 2 \overset{⏜}{i} + 3 \overset{⏜}{j} - 5 \overset{⏜}{k}$

$\vec{a} \times \vec{b} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 2 & 3 \\ 2 & 3 & - 5 \end{matrix}| = \hat{i} + 11 \hat{j} + 7 \hat{k} Now, \vec{a} . (\vec{a} \times \vec{b}) = 1 - 22 + 21 = 0 Thus, \vec{a} is perpendicular to \vec{a} \times \vec{b} .$

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Question 22:

If $\vec{p} and \vec{q}$ are unit vectors forming an angle of 30°; find the area of the parallelogram having $\vec{a} = \vec{p} + 2 \vec{q} and \vec{b} = 2 \vec{p} + \vec{q}$ as its diagonals.

Answer:

$Given : \vec{a} = \hat{p} + 2 \overset{⏜}{q} \vec{b} = 2 \overset{⏜}{p} + \hat{q} \vec{a} \times \vec{b} = (\vec{p} + 2 \vec{q}) \times (2 \vec{p} + \vec{q}) = 2 \vec{p} \times \vec{p} + \vec{p} \times \vec{q} + 4 \vec{q} \times \vec{p} + 2 \vec{q} \times \vec{q} = 2 (0) + \vec{p} \times \vec{q} - 4 \vec{p} \times \vec{q} + 2 (0) = - 3 \vec{p} \times \vec{q} Area of the parallelogram = \frac{1}{2} |\vec{a} \times \vec{b}| = \frac{1}{2} |- 3 (\vec{p} \times \vec{q})| = \frac{3}{2} |\vec{p}| |\vec{q}| \sin 30^{o} = \frac{3}{2} (1) (1) (\frac{1}{2}) (∵ \vec{p} and \vec{q} are unit vectors) = \frac{3}{4} sq. units$

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Question 23:

For any two vectors $\vec{a} and \vec{b}$ , prove that ${|\vec{a} \times \vec{b}|}^{2} = |\begin{matrix} \vec{a} . \vec{a} & \vec{a} . \vec{b} \\ \vec{b} . \vec{a} & \vec{b} . \vec{b} \end{matrix}|$ .

Answer:

$RHS= |\begin{matrix} \vec{a} . \vec{a} & \vec{a} . \vec{b} \\ \vec{b} . \vec{a} & \vec{b} . \vec{b} \end{matrix}| = |\begin{matrix} {|\vec{a}|}^{2} & |\vec{a}| |\vec{b}| \cos θ \\ |\vec{a}| |\vec{b}| \cos θ & {|\vec{b}|}^{2} \end{matrix}| = {|\vec{a}|}^{2} {|\vec{b}|}^{2} - {|\vec{a}|}^{2} {|\vec{b}|}^{2} \cos^{2} θ = {|\vec{a}|}^{2} {|\vec{b}|}^{2} (1 - \cos^{2} θ) = {|\vec{a}|}^{2} {|\vec{b}|}^{2} \sin^{2} θ = {(|\vec{a}| |\vec{b}| \sin θ)}^{2} = {|\vec{a} \times \vec{b}|}^{2} = LHS Hence proved .$

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Question 24:

Define $\vec{a} \times \vec{b}$ and prove that $|\vec{a} \times \vec{b}| = (\vec{a} . \vec{b})$ tan θ, where θ is the angle between $\vec{a} and \vec{b}$ .

Answer:

$If \vec{a} and \vec{b} are two non-zero non-parallel vectors, then the vector product denoted by \vec{a} × \vec{b} is defined as \vec{a} \times \vec{b} = |\vec{a}| |\vec{b}| \sin θ \hat{η} . Here, θ is the angle between \vec{a} and \vec{b} and \hat{η} is the unit vector perpendicular to the plane of \vec{a} and \vec{b} such that \vec{a}, b and \hat{η} form a right handed system. LHS= |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin θ = |\vec{a}| |\vec{b}| \sin θ \times \frac{\cos θ}{\cos θ} = |\vec{a}| |\vec{b}| \cos θ \frac{\sin θ}{\cos θ} = |\vec{a}| |\vec{b}| \cos θ \tan θ = (\vec{a} . \vec{b}) \tan θ = RHS Hence proved .$

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Question 25:

$If |\vec{a}| = \sqrt{26}, |\vec{b}| = 7 and |\vec{a} \times \vec{b}| = 35, find \vec{a} . \vec{b} .$

Answer:

$We know {(\vec{a} . \vec{b})}^{2} + {|\vec{a} \times \vec{b}|}^{2} = {|\vec{a}|}^{2} {|\vec{b}|}^{2} \Rightarrow {(\vec{a} . \vec{b})}^{2} + 35^{2} = {(\sqrt{26})}^{2} \times 7^{2} (∵ |\vec{a} \times \vec{b}| = 35, |\vec{a}| = \sqrt{26} and |\vec{b}| = 7) \Rightarrow {(\vec{a} . \vec{b})}^{2} + 1225 = 1274 \Rightarrow {(\vec{a} . \vec{b})}^{2} = 49 \Rightarrow (\vec{a} . \vec{b}) = 7$

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Question 26:

Find the area of the triangle formed by O, A, B when $\vec{O A} = \hat{i} + 2 \hat{j} + 3 \hat{k}, \vec{O B} = - 3 \hat{i} - 2 \hat{j} + \hat{k} .$

Answer:

$Given : \vec{O A} = \hat{i} + 2 \hat{j} + 3 \hat{k} \vec{O B} = - 3 \hat{i} - 2 \hat{j} + \hat{k} \vec{O A} \times \vec{O B} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ - 3 & - 2 & 1 \end{matrix}| = 8 \hat{i} - 10 \hat{j} + 4 \hat{k} |\vec{O A} \times \vec{O B}| = \sqrt{64 + 100 + 16} = \sqrt{180} = 6 \sqrt{5} Area of the triangle= \frac{1}{2} |\vec{O A} \times \vec{O B}| = \frac{1}{2} (6 \sqrt{5}) =3 \sqrt{5} sq. units$

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Question 27:

(i) Let $\vec{a} = \hat{i} + 4 \hat{j} + 2 \hat{k}, \vec{b} = 3 \hat{i} - 2 \hat{j} + 7 \hat{k} and \vec{c} = 2 \hat{i} - \hat{j} + 4 \hat{k} .$ Find a vector $\vec{d}$ which is perpendicular to both $\vec{a} and \vec{b}$ $and \vec{c} \cdot \vec{d} = 15 .$

(ii) Let $\vec{a} = 4 \hat{i} + 5 \hat{j} - \hat{k}, \vec{b} = \hat{i} - 4 \hat{j} + 5 \hat{k} and \vec{c} = 3 \hat{i} + \hat{j} - \hat{k}$ . Find a vector $\vec{d}$ which is perpendicular to both $\vec{c} and \vec{b} and \vec{d} . \vec{a} = 21$ .

Answer:

(i)
$Given : \vec{a} = \hat{i} + 4 \hat{j} + 2 \hat{k} \vec{b} = 3 \hat{i} - 2 \hat{j} + 7 \hat{k} \vec{c} = 2 \hat{i} - \hat{j} + 4 \hat{k} Since d is perpendicular to both a and b, it is parallel to \vec{a} \times \vec{b} . Suppose d = λ (\vec{a} \times \vec{b}) for some scalar λ . d = λ |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 4 & 2 \\ 3 & - 2 & 7 \end{matrix}| = λ [(28 + 4) \hat{i} - (7 - 6) \hat{j} + (- 2 - 12) \hat{k}] = λ [32 \hat{i} - \hat{j} - 14 \hat{k}] \vec{c .} \vec{d} = 15 (Given) \Rightarrow (2 \hat{i} - \hat{j} + 4 \hat{k}) . λ (32 \hat{i} - \hat{j} - 14 \hat{k}) = 15 \Rightarrow λ (64 + 1 - 56) = 15 \Rightarrow λ = \frac{5}{3} ∴ \vec{d} = \frac{5}{3} (32 \hat{i} - \hat{j} - 14 \hat{k}) \Rightarrow \vec{d} = \frac{1}{3} (160 \hat{i} - 5 \hat{j} - 70 \hat{k})$

Disclaimer: The question should contain
$"which is perpendicular to both \vec{a} and \vec{b} " instead of "which is perpendicular to both \vec{a} and \vec{d} "$

(ii)

$Let \vec{d} = x \hat{i} + y \hat{j} + z \hat{k} Since \vec{d} is perpendicular to both \vec{c} and \vec{b}, so \vec{d .} \vec{c} = 0 and \vec{d .} \vec{b} = 0 3 x + y - z = 0 . . . . . (1) x - 4 y + 5 z = 0 . . . . . (2) \vec{d .} \vec{a} = 21 4 x + 5 y - z = 21 . . . . . (3) Solving (1), (2) and (3) x = (- \frac{1}{3}), y = \frac{16}{3}, z = \frac{13}{3} \vec{d} = \frac{1}{3} (- \hat{i} + 16 \hat{j} + 13 \hat{k}) \vec{d} = - \frac{1}{3} (\hat{i} - 16 \hat{j} - 13 \hat{k})$

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Question 28:

Find a unit vector perpendicular to each of the vectors $\vec{a} + \vec{b} and \vec{a} - \vec{b}, where \vec{a} = 3 \hat{i} + 2 \hat{j} + 2 \hat{k} and \vec{b} = \hat{i} + 2 \hat{j} - 2 \hat{k} .$

Answer:

$Given : \vec{a} = 3 \hat{i} + 2 \hat{j} + 2 \hat{k} \vec{b} = \hat{i} + 2 \hat{j} - 2 \hat{k} ∴ \vec{a} + \vec{b} = 4 \hat{i} + 4 \hat{j} + 0 \hat{k} \vec{a} - \vec{b} = 2 \hat{i} + 0 \hat{j} + 4 \hat{k} (\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{matrix}| = 16 \hat{i} - 16 \hat{j} - 8 \hat{k} ∴ |(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})| = \sqrt{256 + 256 + 64} = \sqrt{576} = 24 Unit vector perpendicular to both \vec{a} + \vec{b} and \vec{a} - \vec{b} = \frac{(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})}{|(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})|} = \frac{16 \hat{i} - 16 \hat{j} - 8 \hat{k}}{24} = \frac{8 (2 \hat{i} - 2 \hat{j} - \hat{k})}{24} = \frac{1}{3} (2 \hat{i} - 2 \hat{j} - \hat{k})$

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Question 29:

Using vectors find the area of the triangle with vertices, A (2, 3, 5), B (3, 5, 8) and C (2, 7, 8).

Answer:

$Let \vec{a}, \vec{b} and \overset{}{\vec{c}} be the position vector s of A, B and C, respectively . Then, \vec{a} = 2 \hat{i} + 3 \hat{j} + 5 \hat{k} \vec{b} = 3 \hat{i} + 5 \hat{j} + 8 \hat{k} \vec{c} = 2 \hat{i} + 7 \hat{j} + 8 \hat{k} Now, \vec{A B} = \vec{b} - \vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k} \vec{A C} = \vec{c} - \vec{a} = 0 \hat{i} + 4 \hat{j} + 3 \hat{k} ∴ \vec{A B} \times \vec{A C} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 0 & 4 & 3 \end{matrix}| = - 6 \hat{i} - 3 \hat{j} + 4 \hat{k} \Rightarrow |\vec{A B} \times \vec{A C}| = \sqrt{36 + 9 + 16} = \sqrt{61} Area of triangle ABC = \frac{1}{2} |\vec{A B} \times \vec{A C}| = \frac{\sqrt{61}}{2} sq. units$

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Question 30:

If $\vec{a} = 2 \hat{i} - 3 \hat{j} + \hat{k}, \vec{b} = - \hat{i} + \hat{k}, \vec{c} = 2 \hat{j} - \hat{k}$ are three vectors, find the area of the parallelogram having diagonals $(\vec{a} + \vec{b})$ and $(\vec{b} + \vec{c})$ . [CBSE 2014]

Answer:

It is given that $\vec{a} = 2 \hat{i} - 3 \hat{j} + \hat{k}, \vec{b} = - \hat{i} + \hat{k}, \vec{c} = 2 \hat{j} - \hat{k}$ .

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

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

We know that the area of parallelogram is $\frac{1}{2} |\vec{d_{1}} \times \vec{d_{2}}|$ , where $\vec{d_{1}}$ and $\vec{d_{2}}$ are the diagonal vectors.

Now,

$(\vec{a} + \vec{b}) \times (\vec{b} + \vec{c}) = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 3 & 2 \\ - 1 & 2 & 0 \end{matrix}| = - 4 \hat{i} - 2 \hat{j} - \hat{k}$

∴ Area of the parallelogram having diagonals $(\vec{a} + \vec{b})$ and $(\vec{b} + \vec{c})$

$= \frac{1}{2} |(\vec{a} + \vec{b}) \times (\vec{b} + \vec{c})| = \frac{1}{2} |- 4 \hat{i} - 2 \hat{j} - \hat{k}| = \frac{1}{2} \sqrt{{(- 4)}^{2} + {(- 2)}^{2} + {(- 1)}^{2}} = \frac{\sqrt{21}}{2} square units$

Thus, the required area of the parallelogram is $\frac{\sqrt{21}}{2}$ square units.

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Question 31:

The two adjacent sides of a parallelogram are $2 \hat{i} - 4 \hat{j} + 5 \hat{k} and \hat{i} - 2 \hat{j} - 3 \hat{k} .$ Find the unit vector parallel to one of its diagonals. Also, find its area.

Answer:

$Suppose □ A B C D is the given parallelogram and A C is its diagonal . Let : \vec{A B} = 2 \hat{i} - 4 \hat{j} + 5 \hat{k} \vec{B C} = \hat{i} - 2 \hat{j} - 3 \hat{k} ∴ Diagonal \vec{A C} = \vec{A B} + \vec{B C} = 3 \hat{i} - 6 \hat{j} + 2 \hat{k} \Rightarrow |\vec{A C}| = \sqrt{9 + 36 + 4} = 7 Unit vector parallel to \vec{A C} = \frac{\vec{A C}}{|\vec{A C}|} = \frac{3 \hat{i} - 6 \hat{j} + 2 \hat{k}}{7} Now, \vec{A B} \times \vec{B C} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & - 4 & 5 \\ 1 & - 2 & - 3 \end{matrix}| = 22 \hat{i} + 11 \hat{j} + 0 \hat{k} \Rightarrow |\vec{A B} \times \vec{A C}| = \sqrt{484 + 121} = \sqrt{605} = 11 \sqrt{5} Area of triangle ABC = \frac{1}{2} |\vec{A B} \times \vec{A C}| = \frac{11 \sqrt{5}}{2} sq . units$

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Question 32:

If either $\vec{a} = \vec{0} or \vec{b} = \vec{0}, then \vec{a} \times \vec{b} = \vec{0} .$ Is the converse true? Justify your answer with an example.

Answer:

$If \vec{a} = \vec{0} or \vec{b} =0, then |\vec{a}| |\vec{b}| \sin θ \hat{n} = \vec{0 .} \Rightarrow \vec{a} \times \vec{b} = \vec{0} But the converse is not true as whenever \vec{a} × \vec{b} = \vec{0}, we cannot be sure that either \vec{a} = \vec{0} or \vec{b} = \vec{0} . For example: \vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k} \vec{b} = \hat{i} + 2 \hat{j} + 3 \hat{k} Here, \vec{a} ≠0 \vec{b} ≠0 But \vec{a} \times \vec{b} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 1 & 2 & 3 \end{matrix}| = 0 \hat{i} + 0 \hat{j} + 0 \hat{k} = \vec{0}$

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Question 33:

If $\vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}, \vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k} and \vec{c} = c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k},$ then verify that $\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c} .$

Answer:

$Given : \vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k} \vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k} \vec{c} = c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k} \vec{b} + \vec{c} = (b_{1} + c_{1}) \hat{i} + (b_{2} + c_{2}) \hat{j} + (b_{3} + c_{3}) \hat{k} ∴ \vec{a} \times (\vec{b} + c) = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ b_{1} + c_{1} & b_{2} + c_{2} & b_{3} + c_{3} \end{matrix}| = (a_{2} b_{3} + a_{2} c_{3} - a_{3} b_{2} - a_{3} c_{2}) \hat{i} - (a_{1} b_{3} + a_{1} c_{3} - a_{3} b_{1} - a_{3} c_{1}) \hat{j} + (a_{1} b_{2} + a_{1} c_{2} - a_{2} b_{1} - a_{2} c_{1}) \hat{k} . . . (1) Now, \vec{a} \times \vec{b} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \end{matrix}| = (a_{2} b_{3} - a_{3} b_{2}) \hat{i} - (a_{1} b_{3} - a_{3} b_{1}) \hat{j} + (a_{1} b_{2} - a_{2} b_{1}) \hat{k} \vec{a} \times \vec{c} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ c_{1} & c_{2} & c_{3} \end{matrix}| = (a_{2} c_{3} - a_{3} c_{2}) \hat{i} - (a_{1} c_{3} - a_{3} c_{1}) \hat{j} + (a_{1} c_{2} - a_{2} c_{1}) \hat{k} \vec{a} \times \vec{b} + \vec{b} \times \vec{c} = (a_{2} b_{3} + a_{2} c_{3} - a_{3} b_{2} - a_{3} c_{2}) \hat{i} - (a_{1} b_{3} + a_{1} c_{3} - a_{3} b_{1} - a_{3} c_{1}) \hat{j} + (a_{1} b_{2} + a_{1} c_{2} - a_{2} b_{1} - a_{2} c_{1}) \hat{k} . . . (2) From (1) and (2), we get \vec{a} \times (\vec{b} + c) = \vec{a} \times \vec{b} + \vec{b} \times \vec{c}$

Page No 24.32:

Question 34:

Using vectors, find the area of the triangle with vertices:
(i) A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5)
(ii) A(1, 2, 3), B(2, −1, 4) and C(4, 5, −1) [CBSE 2011, NCERT EXEMPLAR]

Answer:

(i) The vertices of the triangle are A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5).

Position vector of A = $\hat{i} + \hat{j} + 2 \hat{k}$

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

Position vector of C = $\hat{i} + 5 \hat{j} + 5 \hat{k}$

$\vec{AB} = (2 \hat{i} + 3 \hat{j} + 5 \hat{k}) - (\hat{i} + \hat{j} + 2 \hat{k}) = \hat{i} + 2 \hat{j} + 3 \hat{k}$

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

Now,

$\vec{AB} \times \vec{AC} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 0 & 4 & 3 \end{matrix}| = - 6 \hat{i} - 3 \hat{j} + 4 \hat{k}$

∴ Area of ∆ABC = $\frac{1}{2} |\vec{AB} \times \vec{AC}|$

$= \frac{1}{2} |- 6 \hat{i} - 3 \hat{j} + 4 \hat{k}| = \frac{1}{2} \sqrt{{(- 6)}^{2} + {(- 3)}^{2} + 4^{2}} = \frac{1}{2} \sqrt{36 + 9 + 16} = \frac{\sqrt{61}}{2} square units$

(ii) The vertices of the triangle are A(1, 2, 3), B(2, −1, 4) and C(4, 5, −1).

Position vector of A = $\hat{i} + 2 \hat{j} + 3 \hat{k}$

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

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

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

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

Now,

$\vec{AB} \times \vec{AC} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 3 & 1 \\ 3 & 3 & - 4 \end{matrix}| = 9 \hat{i} + 7 \hat{j} + 12 \hat{k}$

∴ Area of ∆ABC = $\frac{1}{2} |\vec{AB} \times \vec{AC}|$

$= \frac{1}{2} |9 \hat{i} + 7 \hat{j} + 12 \hat{k}| = \frac{1}{2} \sqrt{9^{2} + 7^{2} + 12^{2}} = \frac{1}{2} \sqrt{81 + 49 + 144} = \frac{\sqrt{274}}{2} square units$

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Question 35:

Find all vectors of magnitude $10 \sqrt{3}$ that are perpendicular to the plane of $\hat{i} + 2 \hat{j} + \hat{k}$ and $- \hat{i} + 3 \hat{j} + 4 \hat{k}$ . [NCERT EXEMPLAR]

Answer:

Let $\vec{a} = \hat{i} + 2 \hat{j} + \hat{k}$ and $\vec{b} = - \hat{i} + 3 \hat{j} + 4 \hat{k}$ .

Unit vectors perpendicular to both $\vec{a}$ and $\vec{b}$ = $\pm \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}$

Now,
$\vec{a} \times \vec{b} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ - 1 & 3 & 4 \end{matrix}| = 5 \hat{i} - 5 \hat{j} + 5 \hat{k} ∴ |\vec{a} \times \vec{b}| = |5 \hat{i} - 5 \hat{j} + 5 \hat{k}| = \sqrt{5^{2} + {(- 5)}^{2} + 5^{2}} = \sqrt{75} = 5 \sqrt{3}$

Unit vectors perpendicular to both $\vec{a}$ and $\vec{b}$ = $\pm \frac{5 \hat{i} - 5 \hat{j} + 5 \hat{k}}{5 \sqrt{3}} = \pm \frac{\hat{i} - \hat{j} + \hat{k}}{\sqrt{3}}$

∴ Required vectors = $10 \sqrt{3} (\pm \frac{\hat{i} - \hat{j} + \hat{k}}{\sqrt{3}}) = \pm 10 (\hat{i} - \hat{j} + \hat{k})$

Thus, the vectors of magnitude $10 \sqrt{3}$ that are perpendicular to the plane of $\hat{i} + 2 \hat{j} + \hat{k}$ and $- \hat{i} + 3 \hat{j} + 4 \hat{k}$ are $\pm 10 (\hat{i} - \hat{j} + \hat{k})$ .

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Question 36:

The two adjacent sides of a parallelogram are $2 \hat{i} - 4 \hat{j} - 5 \hat{k} and 2 \hat{i} + 2 \hat{j} + 3 \hat{k}$ . Find the two unit vectors parallel to its diagonals. Using the diagonal vectors, find the area of the parallelogram.

Answer:

The two adjacent sides of a parallelogram are $2 \hat{i} - 4 \hat{j} - 5 \hat{k} and 2 \hat{i} + 2 \hat{j} + 3 \hat{k}$ .
Suppose $\vec{a} = 2 \hat{i} - 4 \hat{j} - 5 \hat{k} and \vec{b} = 2 \hat{i} + 2 \hat{j} + 3 \hat{k}$

Then any one diagonal of a parallelogram is $\vec{P} = \vec{a} + \vec{b}$ .
$\vec{P} = \vec{a} + \vec{b} = 2 \hat{i} - 4 \hat{j} - 5 \hat{k} + 2 \hat{i} + 2 \hat{j} + 3 \hat{k} = 4 \hat{i} - 2 \hat{j} - 2 \hat{k}$
Therefore, unit vector along the diagonal is $\frac{\vec{P}}{|\vec{P}|} = \frac{4 \hat{i} - 2 \hat{j} - 2 \hat{k}}{\sqrt{16 + 4 + 4}} = \frac{2 \hat{i} - \hat{j} - \hat{k}}{\sqrt{6}}$ .
Another diagonal of a parallelogram is $\vec{P'} = \vec{b} - \vec{a}$ .
$\vec{P}' = \vec{b} - \vec{a} = 2 \hat{i} + 2 \hat{j} + 3 \hat{k} - 2 \hat{i} + 4 \hat{j} + 5 \hat{k} = 6 \hat{j} + 8 \hat{k}$
Therefore, unit vector along the diagonal is $\frac{\vec{P'}}{|\vec{P'}|} = \frac{6 \hat{j} + 8 \hat{k}}{\sqrt{36 + 64}} = \frac{6 \hat{j} + 8 \hat{k}}{10} = \frac{3 \hat{j} + 4 \hat{k}}{5}$ .
Now,
$\vec{P} \times \vec{P'} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & - 2 & - 2 \\ 0 & 6 & 8 \end{matrix}| = \hat{i} (- 16 + 12) - \hat{j} (32 - 0) + \hat{k} (24 - 0) = - 4 \hat{i} - 32 \hat{j} + 24 \hat{k}$
Area of parallelogram = $\frac{|\vec{P} \times \vec{P'}|}{2} = \frac{\sqrt{16 + 1024 + 576}}{2} = \frac{\sqrt{1616}}{2} = \frac{4 \sqrt{101}}{2} = 2 \sqrt{101}$ square units

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Question 37:

If ${|\vec{a} \times \vec{b}|}^{2} + {|\vec{a} \cdot \vec{b}|}^{2} = 400$ and $|\vec{a}| = 5,$ then write the value of $|\vec{b}| .$

Answer:

${|\vec{a} \times \vec{b}|}^{2} + {|\vec{a} \cdot \vec{b}|}^{2} = 400 \Rightarrow {\{|\vec{a}| |\vec{b}| \sin θ\}}^{2} + {\{|\vec{a}| |\vec{b}| \cos θ\}}^{2} = 400 \Rightarrow {|\vec{a}|}^{2} {|\vec{b}|}^{2} \sin^{2} θ + {|\vec{a}|}^{2} {|\vec{b}|}^{2} \cos^{2} θ = 400 \Rightarrow {|\vec{a}|}^{2} {|\vec{b}|}^{2} = 400 \Rightarrow 25 \times {|\vec{b}|}^{2} = 400 \Rightarrow {|\vec{b}|}^{2} = 16 \Rightarrow |\vec{b}| = 4$

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Question 38:

If θ is the angle between two vectors $\hat{i} - 2 \hat{j} + 3 \hat{k} and 3 \hat{i} - 2 \hat{j} + \hat{k}, find \sin θ$ .

Answer:

Let $\vec{a} = \hat{i} - 2 \hat{j} + 3 \hat{k}$ and $\vec{b} = 3 \hat{i} - 2 \hat{j} + \hat{k}$ . If θ is the angle between them. Then,
$\cos θ = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$
Now,
$\vec{a} \cdot \vec{b} = (\hat{i} - 2 \hat{j} + 3 \hat{k}) \cdot (3 \hat{i} - 2 \hat{j} + \hat{k}) = 3 + 4 + 3 = 10$

$|\vec{a}| = \sqrt{1 + 4 + 9} = \sqrt{14} and |\vec{b}| = \sqrt{9 + 4 + 1} = \sqrt{14}$
$∴ \cos θ = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} \Rightarrow \cos θ = \frac{10}{\sqrt{14} \sqrt{14}} = \frac{10}{14} = \frac{5}{7}$
Now,
$\sin θ = \sqrt{1 - \cos^{2} θ} = \sqrt{1 - {(\frac{5}{7})}^{2}} = \sqrt{\frac{24}{49}} = \frac{2 \sqrt{6}}{7}$

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Question 1:

If $\vec{a}$ is any vector, then ${(\vec{a} \times \hat{i})}^{2} + {(\vec{a} \times \hat{j})}^{2} + {(\vec{a} \times \hat{k})}^{2} =$
(a) ${\vec{a}}^{2}$

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

(c) $3 {\vec{a}}^{2}$

(d) $4 {\vec{a}}^{2}$

Answer:

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

$Let \vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k} \vec{a} \times \hat{i} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ 1 & 0 & 0 \end{matrix}| = a_{3} \hat{j} - a_{2} \hat{k} \Rightarrow {(\vec{a} \times \hat{i})}^{2} = {(a_{3} \hat{j} - a_{2} \hat{k})}^{2} = {a_{3}}^{2} {|\hat{j}|}^{2} + {a_{2}}^{2} {|\hat{k}|}^{2} - 2 a_{3} a_{2} (\hat{j} . \hat{k}) = {a_{3}}^{2} + {a_{2}}^{2} (∵ \hat{j} . \hat{k} =0) . . . (1) ∴ \vec{a} \times \hat{j} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ 0 & 1 & 0 \end{matrix}| = - a_{3} \hat{i} + a_{1} \hat{k} \Rightarrow {(\vec{a} \times \hat{j})}^{2} = {(- a_{3} \hat{i} + a_{1} \hat{k})}^{2} = {a_{3}}^{2} {|\hat{i}|}^{2} + {a_{1}}^{2} {|\hat{k}|}^{2} - 2 a_{3} a_{2} (\hat{i} . \hat{k}) = {a_{3}}^{2} + {a_{1}}^{2} (∵ \hat{i} . \hat{k} =0) . . . (2) ∴ \vec{a} \times \hat{k} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ 0 & 0 & 1 \end{matrix}| = a_{2} \hat{i} - a_{1} \hat{j} \Rightarrow {(\vec{a} \times k)}^{2} = {(a_{2} \hat{i} - a_{1} \hat{j})}^{2} = {a_{2}}^{2} {|\hat{i}|}^{2} + {a_{1}}^{2} {|j|}^{2} + 2 a_{1} a_{2} (\hat{i} . \hat{j}) = {a_{2}}^{2} + {a_{1}}^{2} (∵ \hat{i} . \hat{j} =0) . . . (3) Adding (1), (2) and (3), we get {(\vec{a} \times \hat{i})}^{2} + {(\vec{a} \times \hat{j})}^{2} + {(\vec{a} \times k)}^{2} = {a_{3}}^{2} + {a_{2}}^{2} + {a_{3}}^{2} + {a_{1}}^{2} + {a_{2}}^{2} + {a_{1}}^{2} = 2 ({a_{1}}^{2} + {a_{2}}^{2} + {a_{3}}^{2}) = 2 {\vec{a}}^{2} (∵ |\vec{a}| = \sqrt{{a_{1}}^{2} + {a_{2}}^{2} + {a_{3}}^{2}})$

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Question 2:

If $\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c}$ and $\vec{a} \times \vec{b} = \vec{a} \times \vec{c,} \vec{a} \neq 0,$ then
(a) $\vec{b} = \vec{c}$

(b) $\vec{b} = \vec{0}$

(c) $\vec{b} + \vec{c} = \vec{0}$

(d) none of these

Answer:

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

$\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} \overset{}{\Rightarrow \vec{a}} \cdot \vec{b} - \vec{a} \cdot \vec{c} = 0 \Rightarrow \vec{a .} (\vec{b} - \vec{c}) = 0 Let θ be the angle between \vec{a} and (\vec{b} - \vec{c}) |\vec{a}| |(\vec{b} - \vec{c})| \cos θ . . . (1)$

$and \vec{a} \times \vec{b} = \vec{a} \times \vec{c} \Rightarrow \vec{a} \times \vec{b} - \vec{a} \times \vec{c} = 0 \Rightarrow \vec{a} \times (\vec{b} - \vec{c}) = 0 Then, |\vec{a}| |(\vec{b} - \vec{c})| \sin θ = 0 . . . (2) Here, it is given that \vec{a} \neq 0 Therefore, for eq (1) and eq (2) to be 0 We have, |(\vec{b} - \vec{c})| \cos θ = 0 For |(\vec{b} - \vec{c})| \cos θ = 0, one of |(\vec{b} - \vec{c})| or \cos θ must be 0 Case 1 : Let \cos θ = 0 \Rightarrow θ = 90 ° \Rightarrow \sin θ = 1 & if |(\vec{b} - \vec{c})| \sin θ = 0 and \sin θ = 1 Then |(\vec{b} - \vec{c})| = 0 \Rightarrow \vec{b} = \vec{c} Case 2 : Let |(\vec{b} - \vec{c})| = 0 \Rightarrow \vec{b} = \vec{c} Hence, \vec{b} = \vec{c}$

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Question 3:

The vector $\vec{b} = 3 \hat{i} + 4 \hat{k}$ is to be written as the sum of a vector $\vec{α}$ parallel to $\vec{a} = \hat{i} + \hat{j}$ and a vector $\vec{β}$ perpendicular to $\vec{a}$ . Then $\vec{α} =$
(a) $\frac{3}{2} (\hat{i} + \hat{j})$

(b) $\frac{2}{3} (\hat{i} + \hat{j})$

(c) $\frac{1}{2} (\hat{i} + \hat{j})$

(d) $\frac{1}{3} (\hat{i} + \hat{j})$

Answer:

(a) $\frac{3}{2} (\hat{i} + \hat{j})$

$Let: \vec{α} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k} \vec{β} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k} Now, \vec{b} =3 \hat{i} +4 \hat{k} = \vec{α} + \vec{β} (Given) \Rightarrow 3 \hat{i} + 0 \hat{j} + 4 \hat{k} = (a_{1} + b_{1}) \hat{i} + (a_{2} + b_{2}) \hat{j} + (a_{3} + b_{3}) \hat{k} \Rightarrow a_{1} + b_{1} = 3; a_{2} + b_{2} = 0; a_{3} + b_{3} = 4 \Rightarrow a_{1} + b_{1} = 3; a_{2} = - b_{2}; a_{3} + b_{3} = 4 . . . (1) \vec{a} = \hat{i} + \hat{j} (Given) Also, \vec{α} is parallel to \vec{a} . \Rightarrow \vec{α} \times \vec{a} = \vec{0} \Rightarrow |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ 1 & 1 & 0 \end{matrix}| = \vec{0} \Rightarrow - a_{3} \hat{i} + a_{3} \hat{j} + (a_{1} - a_{2}) \hat{k} = 0 \hat{i} + 0 \hat{j} + 0 \hat{k} \Rightarrow a_{3} = 0; a_{1} - a_{2} = 0 \Rightarrow a_{3} = 0; a_{1} = a_{2} . . . (2) Since \vec{β} is perpendicular to \vec{a}, we get \Rightarrow \vec{β} . \vec{a} = 0 \Rightarrow (b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}) . (\hat{i} + \hat{j}) = 0 \Rightarrow b_{1} + b_{2} = 0 \Rightarrow b_{1} = - b_{2} . . . (3) Solving (1), (2) and (3), we get a_{1} = \frac{3}{2}; a_{2} = \frac{3}{2}; a_{3} = 0 ∴ \vec{α} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k} = \frac{3}{2} \hat{i} + \frac{3}{2} \hat{j} + 0 \hat{k} = \frac{3}{2} (\hat{i} + \hat{j})$

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Question 4:

The unit vector perpendicular to the plane passing through points $P (\hat{i} - \hat{j} + 2 \hat{k}), Q (2 \hat{i} - \hat{k}) and R (2 \hat{j} + \hat{k})$ is
(a) $2 \hat{i} + \hat{j} + \hat{k}$

(b) $\sqrt{6} (2 \hat{i} + \hat{j} + \hat{k})$

(c) $\frac{1}{\sqrt{6}} (2 \hat{i} + \hat{j} + \hat{k})$

(d) $\frac{1}{6} (2 \hat{i} + \hat{j} + \hat{k})$

Answer:

(c) $\frac{1}{\sqrt{6}} (2 \hat{i} + \hat{j} + \hat{k})$

$The vector \vec{P Q} \times \vec{P R} is perpendicular to the vectors \vec{P Q} and \vec{P R} . ∴ Required unit vector= = \frac{\vec{P Q} \times \vec{P R}}{|\vec{P Q} \times \vec{P R}|} Now, \vec{P Q} = P . V . of Q - P . V . of P = \hat{i} + \hat{j} - 3 \hat{k} \vec{P R} = P . V . of R - P . V . of P = - \hat{i} + 3 \hat{j} - \hat{k} ∴ \vec{P Q} \times \vec{P R} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & - 3 \\ - 1 & 3 & - 1 \end{matrix}| = 8 \hat{i} + 4 \hat{j} + 4 \hat{k} = 4 (2 \hat{i} + \hat{j} + \hat{k}) \Rightarrow |\vec{P Q} \times \vec{P R}| = \sqrt{64 + 16 + 16} = \sqrt{96} = 4 \sqrt{6} Required unit vector = \frac{\vec{P Q} \times \vec{P R}}{|\vec{P Q} \times \vec{P R}|} = \frac{4 (2 \hat{i} + \hat{j} + \hat{k})}{4 \sqrt{6}} = \frac{1}{\sqrt{6}} (2 \hat{i} + \hat{j} + \hat{k})$

Page No 24.34:

Question 5:

If $\vec{a,} \vec{b}$ represent the diagonals of a rhombus, then
(a) $\vec{a} \times \vec{b} = \vec{0}$

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

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

(d) $\vec{a} \times \vec{b} = \vec{a}$

Answer:

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

$We know that the diagonals in a rhombus (\vec{a} and \vec{b}) are perpendicular. Therefore, their dot product is zero. \Rightarrow \vec{a} . \vec{b} = 0$

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Question 6:

Vectors $\vec{a} and \vec{b}$ are inclined at angle θ = 120°. If $|\vec{a}| = 1, |\vec{b}| = 2,$ then ${[(\vec{a} + 3 \vec{b}) \times (3 \vec{a} - \vec{b})]}^{2}$ is equal to
(a) 300
(b) 325
(c) 275
(d) 225

Answer:

(a) 300

$(\vec{a} + 3 \vec{b}) \times (3 \vec{a} - \vec{b}) = 3 (\vec{a} \times \vec{a}) - \vec{a} \times \vec{b} + 9 (\vec{b} \times \vec{a}) - 3 (\vec{b} \times \vec{b}) = 3 (0) - \vec{a} \times \vec{b} - 9 (\vec{a} \times \vec{b}) - 3 (0) = - 10 (\vec{a} \times \vec{b}) Now, |(\vec{a} \times 3 \vec{b}) \times (3 \vec{a} - \vec{b})|^{2} = {|- 10 (\vec{a} \times \vec{b})|}^{2} = 100 {|(\vec{a} \times \vec{b})|}^{2} = 100 {|\vec{a}|}^{2} {|\vec{b}|}^{2} \sin^{2} 120 = 100 {(1)}^{2} {(2)}^{2} {(\frac{\sqrt{3}}{2})}^{2} = 400 \times \frac{3}{4} = 300$

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Question 7:

If $\vec{a} = \hat{i} + \hat{j} - \hat{k}, \vec{b} = - \hat{i} + 2 \hat{j} + 2 \hat{k} and \vec{c} = - \hat{i} + 2 \hat{j} - \hat{k},$ then a unit vector normal to the vectors $\vec{a} + \vec{b} and \vec{b} - \vec{c}$ is
(a) $\hat{i}$

(b) $\hat{j}$

(c) $\hat{k}$

(d) none of these

Answer:

(a) $\hat{i}$

$\vec{a} + \vec{b} = 0 \hat{i} + 3 \hat{j} + \hat{k} \vec{b} - \vec{c} = 0 \hat{i} - 0 \hat{j} + 3 \hat{k} (\vec{a} + \vec{b}) \times (\vec{b} - \vec{c}) = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 3 & 1 \\ 0 & 0 & 3 \end{matrix}| = 9 \hat{i} |(\vec{a} + \vec{b}) \times (\vec{b} - \vec{c})| = 9 |\hat{i}| = 9 (1) = 9 Unit vector perpendicular to both \vec{a} + \vec{b} and \vec{b} - \vec{c} = \frac{(\vec{a} + \vec{b}) \times (\vec{b} - \vec{c})}{|(\vec{a} + \vec{b}) \times (\vec{b} - \vec{c})|} = \frac{9 \hat{i}}{9} = \hat{i}$

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Question 8:

A unit vector perpendicular to both $\hat{i} + \hat{j} and \hat{j} + \hat{k}$ is

(a) $\hat{i} - \hat{j} + \hat{k}$

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

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

(d) $\frac{1}{\sqrt{3}} (\hat{i} - \hat{j} + \hat{k})$

Answer:

(d) $\frac{1}{\sqrt{3}} (\hat{i} - \hat{j} + \hat{k})$

$Let: \vec{a} = \hat{i} + \hat{j} + 0 \hat{k} \vec{b} = 0 \hat{i} + \hat{j} + \hat{k} ∴ \vec{a} \times \vec{b} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{matrix}| = \hat{i} - \hat{j} + \hat{k} \Rightarrow |\vec{a} \times \vec{b}| = \sqrt{1 + 1 + 1} = \sqrt{3} Unit vector perpendicular to \vec{a} and \vec{b} = \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} = \frac{\hat{i} - \hat{j} + \hat{k}}{\sqrt{3}}$

Disclaimer: The answer given for this question in the textbook is incorrect.

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Question 9:

If $\vec{a} = 2 \hat{i} - 3 \hat{j} - \hat{k} and \vec{b} = \hat{i} + 4 \hat{j} - 2 \hat{k}, then \vec{a} \times \vec{b}$ is
(a) $10 \hat{i} + 2 \hat{j} + 11 \hat{k}$

(b) $10 \hat{i} + 3 \hat{j} + 11 \hat{k}$

(c) $10 \hat{i} - 3 \hat{j} + 11 \hat{k}$

(d) $10 \hat{i} - 2 \hat{j} - 10 \hat{k}$

Answer:

(b) $10 \hat{i} + 3 \hat{j} + 11 \hat{k}$

$\vec{a} \times \vec{b} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & - 3 & - 1 \\ 1 & 4 & - 2 \end{matrix}| = 10 \hat{i} + 3 \hat{j} + 11 \hat{k}$

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Question 10:

If $\hat{i}, \hat{j}, \hat{k}$ are unit vectors, then
(a) $\hat{i} \cdot \hat{j} = 1$

(b) $\hat{i} \cdot \hat{i} = 1$

(c) $\hat{i} \times \hat{j} = 1$

(d) $\hat{i} \times (\hat{j} \times \hat{k}) = 1$

Answer:

(b) $\hat{i} \cdot \hat{i} = 1$

$Let us check each option one by one. (a) We know \hat{i} . \hat{j} = 0 \neq 1 (b) We know \hat{i} . \hat{i} = {|\hat{i}|}^{2} = 1^{2} = 1 (c) \hat{i} \times \hat{j} = \hat{k} \neq 1 (d) \hat{i} \times (\hat{j} \times \hat{k}) = \hat{i} \times \hat{i} = 0 \neq 1$

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Question 11:

If θ is the angle between the vectors $2 \hat{i} - 2 \hat{j} + 4 \hat{k} and 3 \hat{i} + \hat{j} + 2 \hat{k},$ then sin θ =
(a) $\frac{2}{3}$

(b) $\frac{2}{\sqrt{7}}$

(c) $\frac{\sqrt{2}}{7}$

(d) $\sqrt{\frac{2}{7}}$

Answer:

(b) $\frac{2}{\sqrt{7}}$

$Let: \vec{a} =2 \hat{i} -2 \hat{j} +4 \hat{k} \vec{b} =3 \hat{i} + \hat{j} +2 \hat{k} |\vec{a}| = \sqrt{2^{2} + {(- 2)}^{2} + 4^{2}} = \sqrt{4 + 4 + 16} = \sqrt{24} = 2 \sqrt{6} |\vec{b}| = \sqrt{3^{2} + 1^{2} + 2^{2}} = \sqrt{9 + 1 + 4} = \sqrt{14} \vec{a} \times \vec{b} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & - 2 & 4 \\ 3 & 1 & 2 \end{matrix}| = - 8 \hat{i} + 8 \hat{j} + 8 \hat{k} |\vec{a} \times \vec{b}| = \sqrt{64 + 64 + 64} = \sqrt{192} = 8 \sqrt{3} Let θ be the angle between \vec{a} and \vec{b .} |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin θ \Rightarrow 8 \sqrt{3} = (2 \sqrt{6}) (\sqrt{14}) \sin θ \Rightarrow \sin θ = \frac{8 \sqrt{3}}{4 \sqrt{21}} = \frac{2}{\sqrt{7}} \Rightarrow θ = \sin^{- 1} (\frac{2}{\sqrt{7}})$

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Question 12:

If $|\vec{a} \times \vec{b}| = 4, |\vec{a} \cdot \vec{b}| = 2, then {|\vec{a}|}^{2} {|\vec{b}|}^{2} =$

(a) 6
(b) 2
(c) 20
(d) 8

Answer:

(c) 20

$We know {(\vec{a} . \vec{b})}^{2} + {|\vec{a} \times \vec{b}|}^{2} = {|\vec{a}|}^{2} {|\vec{b}|}^{2} . . . (1) |\vec{a} . \vec{b}| = 2 (Given) \Rightarrow {|\vec{a} . \vec{b}|}^{2} = {(\vec{a} . \vec{b})}^{2} From (1), we get {(2)}^{2} + {(4)}^{2} = {|\vec{a}|}^{2} {|\vec{b}|}^{2} \Rightarrow {|\vec{a}|}^{2} {|\vec{b}|}^{2} = 20$

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Question 13:

The value of ${(\vec{a} \times \vec{b})}^{2}$ is

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

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

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

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

Answer:

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

${(\vec{a} . \vec{b})}^{2} + {|\vec{a} \times \vec{b}|}^{2} = {(|\vec{a}| |\vec{b}| \cos θ)}^{2} + {(|\vec{a}| |\vec{b}| \sin θ)}^{2} = {|\vec{a}|}^{2} {|\vec{b}|}^{2} (\cos^{2} θ + \sin^{2} θ) = {|\vec{a}|}^{2} {|\vec{b}|}^{2} (1) = {|\vec{a}|}^{2} {|\vec{b}|}^{2} ∴ {|\vec{a} \times \vec{b}|}^{2} = {|\vec{a}|}^{2} {|\vec{b}|}^{2} - {(\vec{a} . \vec{b})}^{2} Thus, the value of {(\vec{a} \times \vec{b})}^{2} is {|\vec{a}|}^{2} {|\vec{b}|}^{2} - {(\vec{a} . \vec{b})}^{2} .$

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Question 14:

The value of $\hat{i} \cdot (\hat{j} \times \hat{k}) + \hat{j} \cdot (\hat{i} \times \hat{k}) + \hat{k} \cdot (\hat{i} \times \hat{j}),$ is
(a) 0
(b) −1
(c) 1
(d) 3

Answer:

(c) 1

$\hat{i} . (\hat{j} \times \hat{k}) + \hat{j} . (\hat{i} \times \hat{k}) + \hat{k} . (\hat{i} \times \hat{j}) = \hat{i} . \hat{i} + \hat{j} . (- \hat{j}) + \hat{k} . \hat{k} = {|\hat{i}|}^{2} - {|\hat{j}|}^{2} + {|\hat{k}|}^{2} = 1 - 1 + 1 = 1$

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Question 15:

If θ is the angle between any two vectors $\vec{a} and \vec{b}$ , then $|\vec{a} \cdot \vec{b}| = |\vec{a} \times \vec{b}|$ when θ is equal to

(a) 0
(b) π/4
(c) π/2
(d) π

Answer:

(b) π/4

$Let θ be the angle between \vec{a} and \vec{b} . We know |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin θ \vec{a} . \vec{b} = |\vec{a}| |\vec{b}| \cos θ \Rightarrow |\vec{a} . \vec{b}| = ||\vec{a}| |\vec{b}| \cos θ| = |\overset{' \to}{a}| |\vec{b}| |\cos θ| Given: |\vec{a} . \vec{b}| = |\vec{a} \times \vec{b}| \Rightarrow |\vec{a}| |\vec{b}| |\cos θ| = |\vec{a}| |\vec{b}| \sin θ \Rightarrow |\cos θ| = \sin θ \Rightarrow θ = \frac{π}{4}$

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Question 16:

If $|\vec{a}| = 10, |\vec{b}| = 2 and \vec{a} \cdot \vec{b} = 12,$ then the value of $|\vec{a} \times \vec{b}|$ is
(a) 5
(b) 10
(c) 14
(d) 16

Answer:

$Given |\vec{a}| = 10, |\vec{b}| = 2 \vec{a} . \vec{b} = 12$

$\begin{array}{rcl} Since {|\vec{a} \times \vec{b}|}^{2} & = & {|a|}^{2} {|b|}^{2} - {(\vec{a} . \vec{b})}^{2} By lagrange' s identity \\ = & {(10)}^{2} {(2)}^{2} - {(12)}^{2} \\ = & 100 \times 4 - 144 \\ = & 400 - 144 \end{array}$
$i . e {|\vec{a} \times \vec{b}|}^{2} = 256 i . e |\vec{a} \times \vec{b}| = 16$
Hence, the correct answer is option D.

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Question 17:

The number of vectors of unit length perpendicular to the vectors $\vec{a} = 2 \hat{i} + \hat{j} + 2 \hat{k} and \vec{b} = \hat{j} + \hat{k} is$
(a) one
(b) two
(c) three
(d) infinite

Answer:

$Unit vector perpendicular to \vec{a} and \vec{b} is \pm \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} ∴ for \vec{a} = 2 \hat{i} + \hat{j} + 2 \hat{k} and \vec{b} = \hat{j} + \hat{k} \vec{a} \times \vec{b} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 2 \\ 0 & 1 & 1 \end{matrix}| \vec{a} \times \vec{b} = \hat{i} (1 - 2) - \hat{j} (2 - 0) + \hat{k} (2 - 0)$

$\begin{array}{rcl} ∴ |\vec{a} \times \vec{b}| & = & \sqrt{{(- 1)}^{2} + {(2)}^{2} + {(2)}^{2}} \\ = & \sqrt{9} = 3 \end{array}$
$∴ unit vector perpendicular to both \vec{a} and \vec{b} is \pm \frac{- \hat{i} - 2 \hat{j} + 2 \hat{k}}{3} = \pm (- \frac{\hat{i}}{3} - \frac{2 \hat{j}}{3} + \frac{2 \hat{k}}{3})$

i.e. number of vectors of unit length perpendicular to the vectors $\vec{a} and \vec{b}$ is two.

Hence, the correct answer is option B.

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Question 18:

$If |\vec{a}| = 8, |\vec{b}| = 3 and |\vec{a} \times \vec{b}| = 12,$ then the value of $\vec{a} \cdot \vec{b}$ is

$(a) 6 \sqrt{3} (b) 8 \sqrt{3} (c) 12 \sqrt{3} (d) none of these$

Answer:

$Given |\vec{a}| = 8 |\vec{b}| = 3 and |\vec{a} \times \vec{b}| = 12 using Lagrange' s identity, {|\vec{a} \times \vec{b}|}^{2} = {|a|}^{2} {|b|}^{2} - {(\vec{a} . \vec{b})}^{2} i . e . {(12)}^{2} = {(8)}^{2} {(3)}^{2} - {(\vec{a} . \vec{b})}^{2} i . e . {(\vec{a} . \vec{b})}^{2} = 64 \times 9 - 144 i . e . {(\vec{a} . \vec{b})}^{2} = 432 i . e . \vec{a} . \vec{b} = 12 \sqrt{3}$
Hence, the correct answer is option C.

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Question 19:

The vectors from origin O to the points A and B are $\vec{a} = 2 \hat{i} - 3 \hat{j} + 2 \hat{k} and \vec{b} = 2 \hat{i} + 3 \hat{j} + \hat{k}$ respectively, then area of triangle OAB is
(a) 340
$(b) \sqrt{25} (c) \sqrt{229} (d) \frac{1}{2} \sqrt{229}$

Answer:

$\begin{array}{rcl} Area of Δ O A B & = & \frac{1}{2} |\vec{a} \times \vec{b}| \\ Since |\vec{a} \times \vec{b}| & = & |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & - 3 & 2 \\ 2 & 3 & 1 \end{matrix}| \\ = & \hat{i} (- 3 - 6) - \hat{j} (2 - 4) + \hat{k} (6 + 6) \end{array}$

$\vec{a} \times \vec{b} = \hat{i} (- 9) + 2 \hat{j} + 12 \hat{k} i . e . |\vec{a} \times \vec{b}| = \sqrt{81 + 4 + 144} |\vec{a} \times \vec{b}| = \sqrt{229} ∴ Area of Δ O A B = \frac{1}{2} \sqrt{229}$
Hence, the correct answer is option D.

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Question 1:

The value of the expression ${|\vec{a} \times \vec{b}|}^{2} + {(\vec{a} \cdot \vec{b})}^{2}$ is ______________.

Answer:

By lagrange's identity,

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

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Question 2:

$If {|\vec{a} \times \vec{b}|}^{2} + {|\vec{a} \cdot \vec{b}|}^{2} = 144 and |\vec{a}| = 4, then |\vec{b}|$ is equal to ____________.

Answer:

$If given {|\vec{a} \times \vec{b}|}^{2} + {|\vec{a} . \vec{b}|}^{2} = 144 and |\vec{a}| = 4 By lagrange' s identity, {|\vec{a} \times \vec{b}|}^{2} + {|\vec{a} . \vec{b}|}^{2} = {|a|}^{2} {|b|}^{2} i . e 144 = {(4)}^{2} {|b|}^{2} i . e . {|b|}^{2} = \frac{144}{16} = 9 i . e . |b| = 3$

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Question 3:

If $\vec{a} and \vec{b}$ are unit vectors such that $\vec{a} \times \vec{b}$ is also a unit vector, then the angle between $\vec{a} and \vec{b}$ is ___________.

Answer:

$Given for \vec{a}, \vec{b} \vec{a}, \vec{b} and \vec{a} \times \vec{b} are unit vectors . i . e |\vec{a}| = 1 = |\vec{b}| = (\vec{a} \times \vec{b}) ∴ By lagrange' s identity, {|\vec{a} \times \vec{b}|}^{2} + {|\vec{a} . \vec{b}|}^{2} = {|a|}^{2} {|b|}^{2} i . e {|\vec{a} . \vec{b}|}^{2} = 1 - 1 = 0 i . e {|\vec{a} . \vec{b}|}^{2} = 0 i . e |\vec{a} . \vec{b}| = 0 i . e . \vec{a} . \vec{b} = 0 i . e . |\vec{a}| |\vec{b}| \cos θ = 0; where θ is angle between \vec{a} and \vec{b} i . e . 1 \times 1 \cos θ = 0 i . e . \cos θ = 0 i . e . θ = \frac{π}{2} i . e . angle between \vec{a} and \vec{b} is \frac{π}{2}$

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Question 4:

For any two vectors $\vec{a} and \vec{b}, (2 \vec{a} + 3 \vec{b}) \times (5 \vec{a} + 7 \overset{\leftrightarrow}{b})$ = _________________.

Answer:

$For any two vectors, \vec{a} and \vec{b} (2 \vec{a} + 3 \vec{b}) \times (5 \vec{a} + 7 \vec{b}) = 2 \vec{a} \times 5 \vec{a} + 3 \vec{b} \times 5 \vec{a} + 2 \vec{a} \times 7 \vec{b} + 3 \vec{b} \times 7 \vec{b} = 10 \vec{a} \times \vec{a} + 15 \vec{b} \times \vec{a} + 14 \vec{a} \times \vec{b} + 21 \vec{b} \times \vec{b} Since \vec{a} \times \vec{a} = 0 = \vec{b} \times \vec{b} and \vec{a} \times \vec{b} = - \vec{b} \times \vec{a} = 0 + 15 \vec{b} \times \vec{a} - 14 \vec{b} \times \vec{a} + 0 (2 \vec{a} + 3 \vec{b}) \times (5 \vec{a} + 7 \vec{b}) = \vec{b} \times \vec{a}$

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Question 5:

The number of vectors of unit length perpendicular to vectors $\vec{a} = \hat{i} + \hat{j} and \vec{b} = \hat{j} + \hat{k} is_________________.$

Answer:

The unit vector perpendicular to $\vec{a} and \vec{b} is$ given by $\pm \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}$

$\begin{array}{rcl} for \vec{a} & = & \hat{i} + \hat{j} and \vec{b} = \hat{j} + \hat{k} \\ \vec{a} \times \vec{b} & = & |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{matrix}| \\ = & \hat{i} (1 - 0) - \hat{j} (1 - 0) + \hat{k} (1 - 0) \end{array}$
$\vec{a} \times \vec{b} = \hat{i} - \hat{j} + \hat{k} i . e . |\vec{a} \times \vec{b}| = \sqrt{1 + 1 + 1} = \sqrt{3} ∴ number of unit vectors perpendicular to \vec{a} and \vec{b} is two i . e \pm \frac{(\hat{i} - \hat{j} + \hat{k})}{\sqrt{3}}$

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Question 6:

$If \vec{a} = \hat{i} - \hat{j} and \vec{b} = \hat{j} + \hat{k}, then {|\vec{a} \times \vec{b}|}^{2} + {|\vec{a} . \vec{b}|}^{2} =________________.$

Answer:

$For \vec{a} = \hat{i} - \hat{j} \vec{b} = \hat{j} + \hat{k} |\vec{a}| = \sqrt{1^{2} + 1^{2}} = \sqrt{2} |\vec{b}| = \sqrt{1^{2} + 1^{2}} = \sqrt{2} By lagrange' s identity,$

$\begin{array}{rcl} {|\vec{a} \times \vec{b}|}^{2} + {|\vec{a} . \vec{b}|}^{2} & = & {|a|}^{2} {|b|}^{2} \\ = & {(\sqrt{2})}^{2} {(\sqrt{2})}^{2} \\ = & 2 \times 2 \end{array}$

$i . e {|\vec{a} \times \vec{b}|}^{2} + {|\vec{a} . \vec{b}|}^{2} = 4$

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Question 7:

For any non-zero vector $\vec{r}, {|\vec{r} \times \hat{i}|}^{2} + {|\vec{r} \times \hat{j}|}^{2} + {|\vec{r} \times \hat{k}|}^{2} = λ {|\vec{r}|}^{2}, then λ =___________________.$

Answer:

$Let \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} Then \vec{r} \times \hat{i} = (x \hat{i} + y \hat{j} + z \hat{k}) \times \hat{i} = x (\hat{i} \times \hat{i}) + y (\hat{j} \times \hat{i}) + z (\hat{k} \times \hat{i}) i . e \vec{r} \times \hat{i} = 0 - y \hat{k} + z \hat{j} . . . (1) (∵ \hat{j} \times \hat{i} = - \hat{k} and \hat{k} \times \hat{i} = \hat{j}) Similarly, \vec{r} \times \hat{k} and \vec{r} \times \hat{j} are calculated i . e . \vec{r} \times \hat{j} = (x \hat{i} + y \hat{j} + z \hat{k}) \times \hat{j} = x (\hat{i} \times \hat{j}) + y (\hat{j} \times \hat{j}) + z (\hat{k} \times \hat{j}) = x \hat{k} + y (0) - z \hat{i} i . e . \vec{r} \times \hat{j} = x \hat{k} - z \hat{i} . . . (2) and \vec{r} \times \hat{k} = (x \hat{i} + y \hat{j} + z \hat{k}) \times \hat{k} = x (\hat{i} \times \hat{k}) + y (\hat{j} \times \hat{k}) + z (\hat{k} \times \hat{k}) = - x \hat{j} + y \hat{i} + z (0) i . e . \vec{r} \times \hat{k} = - x \hat{j} + y \hat{i} . . . (3) ∴ From (1), (2) and (3) {|\vec{r} \times \hat{i}|}^{2} + {|\vec{r} \times \hat{j}|}^{2} + {|\vec{r} \times \hat{k}|}^{2} = {|- y \hat{k} + z \hat{j}|}^{2} + {|x \hat{k} - z \hat{i}|}^{2} + {|- x \hat{j} + y \hat{i}|}^{2} = y^{2} + z^{2} + x^{2} + z^{2} + x^{2} + y^{2} = 2 (x^{2} + y^{2} + z^{2}) = 2 {|r|}^{2} = λ {|r|}^{2} (given) ∴ λ = 2$

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Question 8:

If $\vec{a}, \vec{b} and \vec{c}$ are the position vectors of the vertices A, B and C respectively of a $∆ A B C such that |\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}| = 20,$ then area of $∆ A B C$ is ____________.

Answer:

$For \vec{a}, \vec{b} and \vec{c} position vectors of A, B and C vertices of a Δ A B C, given |\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}| = 20 Area of Δ = \frac{1}{2} |\vec{A B} \times \vec{A C}| Since \vec{A B} = \vec{b} - \vec{a} and \vec{A C} = \vec{c} - \vec{a} i . e . \vec{A B} \times \vec{A C} = (\vec{b} - \vec{a}) \times (\vec{c} - \vec{a}) = \vec{b} \times \vec{c} - \vec{b} \times \vec{a} - \vec{a} \times \vec{c} + \vec{a} \times \vec{a} = \vec{b} \times \vec{c} + \vec{a} \times \vec{b} + \vec{c} \times \vec{a} + 0 \{\begin{cases} ∴ \vec{a} \times \vec{b} = - \vec{b} \times \vec{a} \\ \vec{a} \times \vec{c} = - \vec{c} \times \vec{a} \\ \vec{a} \times \vec{a} = 0 \end{cases} = \vec{b} \times \vec{c} + \vec{a} \times \vec{b} + \vec{c} \times \vec{a} i . e . \frac{1}{2} |\vec{A B} \times \vec{A C}| = \frac{1}{2} |\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}| = \frac{1}{2} \times 20 (given) i . e . Area of Δ A B C = 10$

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Question 9:

If $\vec{a}, \vec{b}$ are two vectors such that $|\vec{a}| = 1, |\vec{b}| = 2 and |\vec{a} \times \overset{⇀}{b}| = \sqrt{2}$ then the angle between $\vec{a} and \vec{b}$ is _____________.

Answer:

$For given vectors \vec{a} and \vec{b} |\vec{a}| = 1, |\vec{b}| = 2 and |\vec{a} \times \vec{b}| = \sqrt{2} ∴ By lagrange' s identity, {|\vec{a} \times \vec{b}|}^{2} + {|\vec{a} . \vec{b}|}^{2} = {|\vec{a}|}^{2} {|\vec{b}|}^{2} i . e . 2 + {|\vec{a} . \vec{b}|}^{2} = 1 \times 4 i . e . {|\vec{a} . \vec{b}|}^{2} = 2 i . e . \vec{a} . \vec{b} = \sqrt{2} ∴ |\vec{a}| |\vec{b}| \cos θ = \sqrt{2} (where θ is angle between a and b) i . e . 1 \times 2 \cos θ = \sqrt{2} i . e . \cos θ = \frac{1}{\sqrt{2}} i . e . θ = \frac{π}{4} or \frac{3 π}{4} i . e angle between \vec{a} and \vec{b} is \frac{π}{4} or \frac{3 π}{4}$

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Question 10:

For any two-collinear vectors $\vec{a} and \vec{b}$ , the value of $\vec{a} . (\vec{a} \times \vec{b})$ is ____________.

Answer:

$For non ‐ collinear vector \vec{a} and \vec{b} . \vec{a} . (\vec{a} \times \vec{b}) Let \vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k} \vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k} then \vec{a} \times \vec{b} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \end{matrix}| \vec{a} \times \vec{b} = \hat{i} (a_{2} b_{3} - a_{3} b_{2}) - \hat{j} (a_{1} b_{3} - a_{3} b_{1}) + \hat{k} (a_{1} b_{2} - a_{2} b_{1}) then \vec{a} \cdot (\vec{a} \times \vec{b}) = (a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}) \cdot [\hat{i} (a_{2} b_{3} - a_{3} b_{2}) - \hat{j} (a_{1} b_{3} - a_{3} b_{1}) + \hat{k} (a_{1} b_{2} - a_{2} b_{1})] = a_{1} (a_{2} b_{3} - a_{3} b_{2}) + a_{2} (a_{3} b_{1} - a_{1} b_{3}) + a_{3} (a_{1} b_{2} - a_{2} b_{1}) = a_{1} a_{2} b_{3} - a_{1} a_{3} b_{2} + a_{2} a_{3} b_{1} - a_{2} a_{1} b_{3} + a_{3} a_{1} b_{2} - a_{3} a_{2} b_{1} i . e . \vec{a} . (\vec{a} \times \vec{b}) = 0$

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Question 11:

If $\vec{a} and \vec{b}$ are two non-zero non-collinear vectors such that $|\vec{a} \times \vec{b}| = 1 and |\vec{a} . \vec{b}| = \sqrt{3}$ , then the angle between $\vec{a} and \vec{b}$ is _____________.

Answer:

$For non ‐ zero non collinear vectors \vec{a} and \vec{b} Given |\vec{a} \times \vec{b}| = 1 |\vec{a} \cdot \vec{b}| = \sqrt{3} Since by lagrange' s identity, {|\vec{a} \times \vec{b}|}^{2} + {|\vec{a} . \vec{b}|}^{2} = {|\vec{a}|}^{2} {|\vec{b}|}^{2} i . e . 1 + 3 = {|\vec{a}|}^{2} {|\vec{b}|}^{2} i . e . 4 = {|\vec{a}|}^{2} {|\vec{b}|}^{2} i . e . |\vec{a}| |\vec{b}| = 2 ∴ |\vec{a} . \vec{b}| = |\vec{a}| |\vec{b}| \cos θ where θ is angle between \vec{a} and \vec{b} \sqrt{3} = 2 \cos θ i . e . \cos θ = \frac{\sqrt{3}}{2} i . e . θ = \frac{π}{6}$

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Question 12:

If three points with position vectors $\vec{a}, \vec{b} and \vec{c}$ are collinear, then $\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} =_______________.$

Answer:

If $\vec{a}, \vec{b} and \vec{c}$ are collinear,

Let points A, B, C be collinear, where position vectors are $\vec{a}, \vec{b} and \vec{c}$ respectively, since $\vec{A B} and \vec{B C}$ are parallel vectors,

$i . e \vec{A B} \times \vec{B C} = 0 i . e (\vec{b} - \vec{a}) \times (\vec{c} - \vec{b}) = 0 i . e \vec{b} \times \vec{c} + \vec{b} \times (- \vec{b}) + (- \vec{a}) \times \vec{c} + \vec{a} \times \vec{b} = 0 i . e \vec{b} \times \vec{c} + 0 + \vec{c} \times \vec{a} + \vec{a} \times \vec{b} = 0 i . e \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} = 0$

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Question 1:

Define vector product of two vectors.

Answer:

$If \vec{a} and \vec{b} are two non-zero non-parallel vectors, then the vector product denoted by \vec{a} × \vec{b} is defined as \vec{a} \times \vec{b} = |\vec{a}| |\vec{b}| \sin θ \hat{η} . Here, θ is the angle between \vec{a} and \vec{b} and \hat{η} is the unit vector perpendicular to the plane of \vec{a} and \vec{b} such that \vec{a}, \vec{b} and \hat{η} form a right handed system.$

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Question 2:

Write the value $(\hat{i} \times \hat{j}) \cdot \hat{k} + \hat{i} \cdot \hat{j} .$

Answer:

$(\hat{i} \times \hat{j}) . \hat{k} + \hat{i} . \hat{j} = \hat{k} . \hat{k} + 0 = {|\hat{k}|}^{2} + 0 = 1^{2} + 0 (∵ |k| = 1) = 1$

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Question 3:

Write the value of $\hat{i} . (\hat{j} \times \hat{k}) + \hat{j} . (\hat{k} \times \hat{i}) + \hat{k} . (\hat{j} \times \hat{i}) .$

Answer:

$\hat{i} . (\hat{j} \times \hat{k}) + \hat{j} . (\hat{k} \times \hat{i}) + \hat{k} . (\hat{j} \times \hat{i}) = \hat{i} . \hat{i} + \hat{j} . j + \hat{k} . (- \hat{k}) = {|\hat{i}|}^{2} + {|\hat{j}|}^{2} - {|\hat{k}|}^{2} = 1 + 1 - 1 (∵ |\hat{i}| = 1, |\hat{j}| = 1 and |\hat{k}| = 1) = 1$

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Question 4:

Write the value of $\hat{i} . (\hat{j} \times \hat{k}) + \hat{j} . (\hat{k} \times \hat{i}) + \hat{k} . (\hat{i} \times \hat{j}) .$

Answer:

$\hat{i} . (\hat{j} \times \hat{k}) + \hat{j} . (\hat{k} \times \hat{i}) + \hat{k} . (\hat{i} \times \hat{j}) = \hat{i} . \hat{i} + \hat{j} . j + \hat{k} . \hat{k} = {|\hat{i}|}^{2} + {|\hat{j}|}^{2} + {|\hat{k}|}^{2} = 1 + 1 + 1 = 3$

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Question 5:

Write the value of $\hat{i} \times (\hat{j} + \hat{k}) + \hat{j} \times (\hat{k} + \hat{i}) + \hat{k} \times (\hat{i} + \hat{j}) .$

Answer:

$\hat{i} \times (\hat{j} + \hat{k}) + \hat{j} \times (\hat{k} + \hat{i}) + \hat{k} \times (\hat{i} + \hat{j}) = (\hat{i} \times \hat{j}) + (\hat{i} \times \hat{k}) + (\hat{j} \times \hat{k}) + (\hat{j} \times \hat{i}) + (\hat{k} \times \hat{i}) + (\hat{k} \times \hat{j}) = \hat{k} - \hat{j} + \hat{i} - \hat{k} + \hat{j} - \hat{i} = \vec{0}$

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Question 6:

Write the expression for the area of the parallelogram having $\vec{a} and \vec{b}$ as its diagonals.

Answer:

$Given: \vec{a} and \vec{b} are diagonals of a parallelogram. Area of the parallelogram = \frac{1}{2} |\vec{a} \times \vec{b}|$

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Question 7:

For any two vectors $\vec{a} and \vec{b}$ write the value of ${(\vec{a} . \vec{b})}^{2} + {|\vec{a} \times \vec{b}|}^{2}$ in terms of their magnitudes.

Answer:

${(\vec{a} . \vec{b})}^{2} + {|\vec{a} \times \vec{b}|}^{2} = {(|\vec{a}| |\vec{b}| \cos θ)}^{2} + {(|\vec{a}| |\vec{b}| \sin θ)}^{2} = {|\vec{a}|}^{2} {|\vec{b}|}^{2} (\cos^{2} θ + \sin^{2} θ) = {|\vec{a}|}^{2} {|\vec{b}|}^{2} (1) = {|\vec{a}|}^{2} {|\vec{b}|}^{2}$

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Question 8:

If $\vec{a} and \vec{b}$ are two vectors of magnitudes 3 and $\frac{\sqrt{2}}{3}$ respectively such that $\vec{a} \times \vec{b}$ is a unit vector. Write the angle between $\vec{a} and \vec{b} .$

Answer:

$Let θ be the angle between \vec{a} and \vec{b} . It is given that \vec{a} \times \vec{b} is a unit vector. \Rightarrow |\vec{a} \times \vec{b}| = 1 We know |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin θ \Rightarrow 1 = (3) (\frac{\sqrt{2}}{3}) \sin θ \Rightarrow \sin θ = \frac{1}{\sqrt{2}} \Rightarrow θ = 45^{o}, 135^{o}$

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Question 9:

$If |\vec{a}| = 10, |\vec{b}| = 2 and |\vec{a} \times \vec{b}| = 16, find \vec{a} . \vec{b} .$

Answer:

$We know {(\vec{a} . \vec{b})}^{2} + {|\vec{a} \times \vec{b}|}^{2} = {|\vec{a}|}^{2} {|\vec{b}|}^{2} \Rightarrow {(\vec{a} . \vec{b})}^{2} + {(16)}^{2} = {(10)}^{2} \times 2^{2} (∵ |\vec{a} \times \vec{b}| = 16, |\vec{a}| = 10 and |\vec{b}| = 2) \Rightarrow {(\vec{a} . \vec{b})}^{2} + 256 = 400 \Rightarrow {(\vec{a} . \vec{b})}^{2} = 144 \Rightarrow (\vec{a} . \vec{b}) = \pm 12$

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Question 10:

For any two vectors $\vec{a}$ and $\vec{b}$ , find $\vec{a} . (\vec{b} \times \vec{a}) .$

Answer:

$Let: \vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k} \vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k} \vec{b} \times \vec{a} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ b_{1} & b_{2} & b_{3} \\ a_{1} & a_{2} & a_{3} \end{matrix}| = \hat{i} (b_{2} a_{3} - b_{3} a_{2}) - \hat{j} (b_{1} a_{3} - b_{3} a_{1}) + \hat{k} (b_{1} a_{2} - b_{2} a_{1}) Now, \vec{a} . (\vec{b} \times \vec{a}) = (a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}) . [\hat{i} (b_{2} a_{3} - b_{3} a_{2}) - \hat{j} (b_{1} a_{3} - b_{3} a_{1}) + \hat{k} (b_{1} a_{2} - b_{2} a_{1})] = a_{1} (b_{2} a_{3} - b_{3} a_{2}) - a_{2} (b_{1} a_{3} - b_{3} a_{1}) + a_{3} (b_{1} a_{2} - b_{2} a_{1}) = a_{1} b_{2} a_{3} - a_{1} b_{3} a_{2} - a_{2} b_{1} a_{3} + a_{2} b_{3} a_{1} + a_{3} b_{1} a_{2} - a_{3} b_{2} a_{1} = 0$

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Question 11:

If $\vec{a} and \vec{b}$ are two vectors such that $|\vec{a} \times \vec{b}| = \sqrt{3} and \vec{a} . \vec{b} = 1,$ find the angle between.

Answer:

$|\vec{a} \times \vec{b}| = \sqrt{3} \Rightarrow |\vec{a}| |\vec{b}| \sin θ = \sqrt{3} . . . (1) \vec{a} . \vec{b} = 1 \Rightarrow |\vec{a}| |\vec{b}| \cos θ = 1 . . . (2) Dividing (1) by (2), we get \frac{|\vec{a}| |\vec{b}| \sin θ}{|\vec{a}| |\vec{b}| \cos θ} = \sqrt{3} \Rightarrow \tan θ = \sqrt{3} \Rightarrow θ = 60^{o}$

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Question 12:

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

Answer:

$\vec{a} \times (\vec{b} + \vec{c}) + \vec{b} \times (\vec{c} + \vec{a}) + \vec{c} \times (\vec{a} + \vec{b}) = (\vec{a} \times \vec{b}) + (\vec{a} \times \vec{c}) + (\vec{b} \times \vec{c}) + (\vec{b} \times \vec{a}) + (\vec{c} \times \vec{a}) + (\vec{c} \times \vec{b}) = (\vec{a} \times \vec{b}) + (\vec{a} \times \vec{c}) + (\vec{b} \times \vec{c}) - (\vec{a} \times \vec{b}) - (\vec{a} \times \vec{c}) - (\vec{b} \times \vec{c}) = \vec{0}$

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Question 13:

For any two vectors $\vec{a} and \vec{b}, find (\vec{a} \times \vec{b}) . \vec{b} .$

Answer:

$Let: \vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k} \vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k} \vec{a} \times \vec{b} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \end{matrix}| = \hat{i} (a_{2} b_{3} - a_{3} b_{2}) - \hat{j} (a_{1} b_{3} - a_{3} b_{1}) + \hat{k} (a_{1} b_{2} - a_{2} b_{1}) (\vec{a} \times \vec{b}) . \vec{b} = [\hat{i} (a_{2} b_{3} - a_{3} b_{2}) - \hat{j} (a_{1} b_{3} - a_{3} b_{1}) + \hat{k} (a_{1} b_{2} - a_{2} b_{1})] . (b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}) = b_{1} (a_{2} b_{3} - a_{3} b_{2}) - b_{2} (a_{1} b_{3} - a_{3} b_{1}) + b_{3} (a_{1} b_{2} - a_{2} b_{1}) = a_{2} b_{1} b_{3} - a_{3} b_{1} b_{2} - a_{1} b_{2} b_{3} + a_{3} b_{1} b_{2} + a_{1} b_{2} b_{3} - a_{2} b_{1} b_{3} = 0$

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Question 14:

Write the value of $\hat{i} \times (\hat{j} \times \hat{k}) .$

Answer:

$\hat{i} \times (\hat{j} \times \hat{k}) = \hat{i} \times \hat{i} = \vec{0}$

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Question 15:

If $\vec{a} = 3 \hat{i} - \hat{j} + 2 \hat{k}$ and $\vec{b} = 2 \hat{i} + \hat{j} - \hat{k},$ then find $(\vec{a} \times \vec{b}) \vec{a} .$

Answer:

$Since \vec{a} × \vec{b} is a vector, (\vec{a} \times \vec{b}) \vec{a} without any dot or cross product in between is meaningless.$

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Question 16:

Write a unit vector perpendicular to $\hat{i} + \hat{j} and \hat{j} + \hat{k} .$

Answer:

$Let \vec{a} = \hat{i} + \hat{j} + 0 \hat{k}; \vec{b} = 0 \hat{i} + \hat{j} + \hat{k} \vec{a} \times \vec{b} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{matrix}| = \hat{i} - \hat{j} + \hat{k} \Rightarrow |\vec{a} \times \vec{b}| = \sqrt{1 + 1 + 1} = \sqrt{3} Unit vector perpendicular to \vec{a} and \vec{b} is, \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} = \frac{1}{\sqrt{3}} (\hat{i} - \hat{j} + \hat{k})$

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Question 17:

If ${|\vec{a} \times \vec{b}|}^{2} + {(\vec{a} . \vec{b})}^{2} = 144$ and $|\vec{a}| = 4,$ find $|\vec{b}|$ .

Answer:

$We know {|\vec{a} \times \vec{b}|}^{2} + {(\vec{a} . \vec{b})}^{2} = {|\vec{a}|}^{2} {|\vec{b}|}^{2} \Rightarrow 144 = 4^{2} {|\vec{b}|}^{2} \Rightarrow 144 = 16 {|\vec{b}|}^{2} \Rightarrow {|\vec{b}|}^{2} = 9 \Rightarrow |\vec{b}| = 3$

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Question 18:

If $\vec{r} = x \hat{i} + y \hat{j} + z \hat{k},$ then write the value of ${|\vec{r} \times \hat{i}|}^{2} .$

Answer:

$Given : \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} Now, \vec{i} = \hat{i} + 0 \hat{j} + 0 \hat{k} \vec{r} \times \vec{i} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ 1 & 0 & 0 \end{matrix}| = 0 \hat{i} + z \hat{j} - y \hat{k} \Rightarrow |\vec{r} \times \vec{i}| = \sqrt{z^{2} + y^{2}} \Rightarrow {|\vec{r} \times \vec{i}|}^{2} = z^{2} + y^{2}$

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Question 19:

If $\vec{a} and \vec{b}$ are unit vectors such that $\vec{a} \times \vec{b}$ is also a unit vector, find the angle between $\vec{a} and \vec{b}$ .

Answer:

$Let θ be the angle between \vec{a} and \vec{b} . Given: |\vec{a} \times \vec{b}| = 1 |\vec{a}| = 1 |\vec{b}| = 1 We know |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin θ \Rightarrow 1 = (1) (1) \sin θ \Rightarrow \sin θ = 1 \Rightarrow θ = \frac{π}{2}$

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Question 20:

If $\vec{a} and \vec{b}$ are two vectors such that $|\vec{a} . \vec{b}| = |\vec{a} \times \vec{b}|,$ write the angle between $\vec{a} and \vec{b} .$

Answer:

$Let θ be the angle between \vec{a} and \vec{b} . We know |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| |\sin θ| |\vec{a} . \vec{b}| = |\vec{a}| |\vec{b}| |\cos θ| Now, |\vec{a} \times \vec{b}| = |\vec{a} . \vec{b}| (Given) \Rightarrow |\vec{a}| |\vec{b}| |\sin θ| = |\vec{a}| |\vec{b}| |\cos θ| \Rightarrow |\sin θ| = |\cos θ| \Rightarrow θ = \frac{π}{4}$

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Question 21:

If $\vec{a} and \vec{b}$ are unit vectors, then write the value of ${|\vec{a} \times \vec{b}|}^{2} + {(\vec{a} . \vec{b})}^{2} .$

Answer:

$It is given that \vec{a} and \vec{b} are unit vectors. \Rightarrow |\vec{a}| = |\vec{b}| = 1 . . . (1) Now, {(\vec{a} . \vec{b})}^{2} + {|\vec{a} \times \vec{b}|}^{2} = {(|\vec{a}| |\vec{b}| \cos θ)}^{2} + {(|\vec{a}| |\vec{b}| \sin θ)}^{2} = {|\vec{a}|}^{2} {|\vec{b}|}^{2} (\cos^{2} θ + \sin^{2} θ) = {|\vec{a}|}^{2} {|\vec{b}|}^{2} (1) = {|\vec{a}|}^{2} {|\vec{b}|}^{2} = 1^{2} 1^{2} [From (1)] = 1$

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Question 22:

If $\vec{a}$ is a unit vector such that $\vec{a} \times \hat{i} = \hat{j}, find \vec{a} . \hat{i} .$

Answer:

$We know \hat{k} \times \hat{i} = \hat{j} . . . (1) Given : \vec{a} \times \hat{i} = \hat{j} . . . (2) Comparing (1) and (2), we get \vec{a} = \hat{k} Now, \vec{a} . \hat{i} = \hat{k} . \hat{i} = 0$

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Question 23:

If $\vec{c}$ is a unit vector perpendicular to the vectors $\vec{a} and \vec{b},$ write another unit vector perpendicular to $\vec{a} and \vec{b} .$

Answer:

$\vec{c} is a unit vector perpendicular to both \vec{a} and \vec{b} . ⇒ \vec{c} = \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} ⇒- \vec{c} = \frac{\vec{b} \times \vec{a}}{|\vec{a} \times \vec{b}|} Therefore, - \vec{c} is perpendicular to \vec{b} and \vec{a .} Thus, - \vec{c} is another unit vector perpendicular to \vec{a} and \vec{b .}$

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Question 24:

Find the angle between two vectors $\vec{a} and \vec{b}$ with magnitudes 1 and 2 respectively and when $|\vec{a} \times \vec{b}| = \sqrt{3} .$

Answer:

$Let θ be the angle between \vec{a} and \vec{b} . We know |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin θ \Rightarrow \sqrt{3} = (1) (2) \sin θ \Rightarrow \sin θ = \frac{\sqrt{3}}{2} \Rightarrow θ = \frac{π}{3}$

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Question 25:

Vectors $\vec{a} and \vec{b}$ are such that $|\vec{a}| = \sqrt{3}, |\vec{b}| = \frac{2}{3} and (\vec{a} \times \vec{b})$ is a unit vector. Write the angle between $\vec{a} and \vec{b}$ .

Answer:

$Given: \vec{a} \times \vec{b} is a unit vector. \Rightarrow |\vec{a} \times \vec{b}| = 1 . . . (1) Let θ be the angle between \vec{a} and \vec{b} . We know |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin θ From (1), we get 1 = (\sqrt{3}) (\frac{2}{3}) \sin θ \Rightarrow \sin θ = \frac{\sqrt{3}}{2} \Rightarrow θ = \frac{π}{3}$

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Question 26:

Find λ, if $(2 \hat{i} + 6 \hat{j} + 14 \hat{k}) \times (\hat{i} - λ \hat{j} + 7 \hat{k}) = \vec{0} .$

Answer:

$Given: |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 6 & 14 \\ 1 & - λ & 7 \end{matrix}| = \vec{0} \Rightarrow \hat{i} (42 + 14 λ) - 0 \hat{j} + \hat{k} (- 2 λ - 6) = 0 \hat{i} + 0 \hat{j} + 0 \hat{k} \Rightarrow 42 + 14 λ = 0; - 2 λ - 6 = 0 \Rightarrow λ = - 3 (This satisfies the above equations)$

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Question 27:

Write the value of the area of the parallelogram determined by the vectors $2 \hat{i} and 3 \hat{j} .$

Answer:

$Let: \vec{a} =2 \hat{i} \vec{b} =3 \hat{j} \vec{a} \times \vec{b} = 6 (\hat{i} \times \hat{j}) = 6 \hat{k} Area of the parallelogram = |\vec{a} \times \vec{b}| = 6 |\hat{k}| = 6 (1) = 6 sq. units$

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Question 28:

Write the value of $(\hat{i} \times \hat{j}) \cdot \hat{k} + (\hat{j} + \hat{k}) \cdot \hat{j}$

Answer:

$(\hat{i} \times \hat{j}) . \hat{k} + (\hat{j} + \hat{k}) . \hat{j} = \hat{k} . \hat{k} + \hat{j} . \hat{j} + \hat{k} . \hat{j} (∵ \hat{i} \times \hat{j} = \hat{k}) = {|\hat{k}|}^{2} + {\hat{|j|}}^{2} + 0 (∵ \hat{k} . \hat{j} = 0) = 1^{2} + 1^{2} = 2$

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Question 29:

Find a vector of magnitude $\sqrt{171}$ which is perpendicular to both of the vectors $\vec{a} = \hat{i} + 2 \hat{j} - 3 \hat{k}$ and $\vec{b} = 3 \hat{i} - \hat{j} + 2 \hat{k}$ .

Answer:

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

Unit vectors perpendicular to both $\vec{a}$ and $\vec{b}$ = $\pm \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}$

Now,
$\vec{a} \times \vec{b} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & - 3 \\ 3 & - 1 & 2 \end{matrix}| = \hat{i} - 11 \hat{j} - 7 \hat{k} ∴ |\vec{a} \times \vec{b}| = |\hat{i} - 11 \hat{j} - 7 \hat{k}| = \sqrt{1^{2} + {(- 11)}^{2} + {(- 7)}^{2}} = \sqrt{1 + 121 + 49} = \sqrt{171}$

Unit vectors perpendicular to both $\vec{a}$ and $\vec{b}$ = $\pm \frac{\hat{i} - 11 \hat{j} - 7 \hat{k}}{\sqrt{171}}$

∴ Required vectors = $\sqrt{171} (\pm \frac{\hat{i} - 11 \hat{j} - 7 \hat{k}}{\sqrt{171}}) = \pm (\hat{i} - 11 \hat{j} - 7 \hat{k})$

Thus, the vectors of magnitude $\sqrt{171}$ which are perpendicular to both the given vectors are $\pm (\hat{i} - 11 \hat{j} - 7 \hat{k})$ .

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Question 30:

Write the number of vectors of unit length perpendicular to both the vectors $\vec{a} = 2 \hat{i} + \hat{j} + 2 \hat{k} and \vec{b} = \hat{j} + \hat{k}$ .

Answer:

Unit vectors perpendicular to $\vec{a}$ and $\vec{b}$ are $\pm (\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|})$ .
$\vec{a} \times \vec{b} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 2 \\ 0 & 1 & 1 \end{matrix}| = - \hat{i} - 2 \hat{j} + 2 \hat{k}$

∴ Unit vectors perpendicular to $\vec{a}$ and $\vec{b}$ are $\pm \frac{- \hat{i} - 2 \hat{j} + 2 \hat{k}}{\sqrt{{(- 1)}^{2} + {(- 2)}^{2} + {(2)}^{2}}} = \pm (- \frac{1}{3} \hat{i} - \frac{2}{3} \hat{j} + \frac{2}{3} \hat{k})$

Thus, there are two unit vectors perpendicular to the given vectors.

Page No 24.37:

Question 31:

Write the angle between the vectors $\vec{a} \times \vec{b}$ and $\vec{b} \times \vec{a}$ .

Answer:

$\vec{b} \times \vec{a}$ = $- \vec{a} \times \vec{b}$

So, $\vec{a} \times \vec{b}$ and $\vec{b} \times \vec{a}$ are vectors of same magnitude but opposite in directions.

Thus, the angle between the vectors $\vec{a} \times \vec{b}$ and $\vec{b} \times \vec{a}$ is 180º.

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