NCERT Solutions Class 12 Maths Chapter-10 (Vector Algebra) Exercise 10.4

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-10 (Vector Algebra)Exercise 10.4 This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

Exercise 10.4

Q1. $Find | \vec{a} \times \vec{b} |_{, if} \vec{a} = \hat{i} - 7 \hat{j} + 7 \hat{k} and \vec{b} = 3 \hat{i} - 2 \hat{j} + 2 \hat{k}$

Answer. $\vec{a} = \hat{i} - 7 \hat{j} + 7 \hat{k} and \vec{b} = 3 \hat{i} - 2 \hat{j} + 2 \hat{k}$ $\vec{a} \times \vec{b} = | \begin{array}{rrr} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 7 & 7 \\ 3 & - 2 & 2 \end{array} |$ $\begin{matrix} = \hat{i} (- 14 + 14) - \hat{j} (2 - 21) + \hat{k} (- 2 + 21) = 19 \hat{j} + 19 \hat{k} \\ ∴ | \vec{a} \times \vec{b} | = \sqrt{(19)^{2} + (19)^{2}} = \sqrt{2 \times (19)^{2}} = 19 \sqrt{2} \end{matrix}$

Q2. $\begin{array}{l} Find a unit vector perpendicular to each of the vector \vec{a} + \vec{b} and \vec{a} - \vec{b}, where \\ \vec{a} = 3 \hat{i} + 2 \hat{j} + 2 \hat{k} and \vec{b} = \hat{i} + 2 \hat{j} - 2 \hat{k} \end{array}$

Answer. $\begin{array}{l} We have, \\ \vec{a} = 3 \hat{i} + 2 \hat{j} + 2 \hat{k} and \vec{b} = \hat{i} + 2 \hat{j} - 2 \hat{k} \\ ∴ \vec{a} + \vec{b} = 4 \hat{i} + 4 \hat{j}, \vec{a} - \vec{b} = 2 \hat{i} + 4 \hat{k} \end{array}$ $(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{array} | = \hat{i} (16) - \hat{j} (16) + \hat{k} (- 8) = 16 \hat{i} - 16 \hat{j} - 8 \hat{k}$ $\begin{aligned} ∴ (\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) & = \sqrt{16^{2} + (- 16)^{2} + (- 8)^{2}} \\ = \sqrt{2^{2} \times 8^{2} + 2^{2} \times 8^{2} + 8^{2}} \\ = 8 \sqrt{2^{2} + 2^{2} + 1} = 8 \sqrt{9} = 8 \times 3 = 24 \end{aligned}$ $\begin{array}{l} Hence, the unit vector perpendicular to each of the vectors \vec{a} + \vec{b} and \vec{a} - \vec{b} is given by the \\ relation, \\ = \pm \frac{(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})}{(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})} = \pm \frac{16 \hat{i} - 16 \hat{j} - 8 \hat{k}}{24} \\ = \pm \frac{2 \hat{i} - 2 \hat{j} - \hat{k}}{3} = \pm \frac{2}{3} \hat{i} \mp \frac{2}{3} \hat{j} \mp \frac{1}{3} \hat{k} \end{array}$

Q3. $\begin{array}{l} If a unit vector \vec{a} makes an angles 3 {with}^{i, \frac{π}{4}} with \hat{ȷ} and an acute angle θ with \hat{k}, then \\ find θ and hence, the compounds of \vec{a} . \end{array}$

Answer. $\begin{array}{l} Let unit vector \vec{a} have (a_{1}, a_{2}, a_{3}) components. \\ ◻ \bar{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k} \\ since {\vec{a}}_{is a unit vector,} | \vec{a} | = 1 \end{array}$ $\begin{array}{l} \cos \frac{π}{3} = \frac{a_{1}}{| \vec{a} |} \\ \Rightarrow \frac{1}{2} = a_{1} \\ \cos \frac{π}{4} = \frac{a_{2}}{| \vec{a} |} \\ \Rightarrow \frac{1}{\sqrt{2}} = a_{2} \\ Also, \cos θ = \frac{a_{3}}{| \vec{a} |} \end{array}$ $\begin{array}{l} Now \\ | a | = 1 \\ \Rightarrow \sqrt{a_{1}^{2} + a_{2}^{2} + a_{3}^{2}} = 1 \\ \Rightarrow {(\frac{1}{2})}^{2} + {(\frac{1}{\sqrt{2}})}^{2} + \cos^{2} θ = 1 \\ \Rightarrow \frac{3}{4} + \cos^{2} θ = 1 \\ \Rightarrow \frac{3}{4} + \cos^{2} θ = 1 \end{array}$ $\begin{array}{l} \Rightarrow \frac{3}{4} + \cos^{2} θ = 1 \\ \Rightarrow \cos^{2} θ = 1 - \frac{3}{4} = \frac{1}{4} \\ \Rightarrow \cos θ = \frac{1}{2} \Rightarrow θ = \frac{π}{3} \\ ∴ a_{3} = \cos \frac{π}{3} = \frac{1}{2} \end{array}$ $θ = \frac{π}{3} and the components of \bar{a} are (\frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{1}{2})$

Q4. $\begin{array}{l} Show that \\ (\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) = 2 (\vec{a} \times \vec{b}) \end{array}$

Answer. $\begin{array}{l} (\vec{a} - \vec{b}) \times (\vec{a} + \hat{b}) \\ = (\vec{a} - \vec{b}) \times \vec{a} + (\vec{a} - \vec{b}) \times \vec{b} \\ = \vec{a} \times \vec{a} - \vec{b} \times \vec{a} + \vec{a} \times \vec{b} - \vec{b} \times \vec{b} \\ = \vec{0} + \vec{a} \times \vec{b} + \vec{a} \times \vec{b} - \vec{0} \\ = 2 \vec{a} \times \vec{b} \end{array}$

Q5. Find $λ and μ$ if $(2 \hat{i} + 6 \hat{j} + 27 \hat{k}) \times (\hat{i} + λ \hat{j} + μ \hat{k}) = \vec{0}$

Answer. $\begin{array}{l} (2 \hat{i} + 6 \hat{j} + 27 \hat{k}) \times (\hat{i} + λ \hat{j} + μ \hat{k}) = \vec{0} \\ \Rightarrow | \begin{array}{ccc} \hat{j} & \hat{k} \\ 2 & 6 & 27 \\ 1 & λ & μ \end{array} | = 0 \hat{i} + 0 \hat{j} + 0 \hat{k} \\ \Rightarrow \hat{i} (6 μ - 27 λ) - \hat{j} (2 μ - 27) + \hat{k} (2 λ - 6) = 0 \hat{i} + 0 \hat{j} + 0 \hat{k} \end{array}$ $\begin{array}{l} On comparing the corresponding components, we have: \\ 6 μ - 27 λ = 0 \\ 2 μ - 27 = 0 \\ 2 λ - 6 = 0 \\ Now, \\ 2 λ - 6 = 0 \Rightarrow λ = \frac{27}{2} \\ 2 μ - 27 = 0 \Rightarrow μ = \frac{27}{2} \\ Hence, \end{array}$

Q6. Given that $\vec{a} \cdot \vec{b} = 0 and \vec{a} \times \vec{b} = \vec{0} . What can you conclude about the vectors \bar{a} and \vec{b}$

Answer. $\begin{array}{l} \vec{a} \cdot \vec{b} = 0 \\ Then, \\ (i) Either | \vec{a} | = 0 or | \vec{b} | = 0 or \vec{a} ⊥ \vec{b} (in case \vec{a} and \vec{b} are non-zero) \\ \vec{a} \times \vec{b} = 0 \end{array}$ (ii) Either $| \bar{a} | = 0 or | = 0, or \vec{a} ‖ \vec{b} (in case \vec{a} and \vec{b} are non-zero)$ $\begin{array}{l} But, \vec{a} and \vec{b} cannot be perpendicular and parallel simultaneously. \\ Hence, | \vec{a} | = 0 or | \bar{b} | = 0 \end{array}$

Q7. $\begin{array}{l} Let the vectors \vec{a}, \vec{b}, \vec{c} given as a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}, b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}, c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k} . Then show \\ that = \vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c} \end{array}$

Answer. $\begin{array}{l} we have, \\ \vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}, \vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}, \vec{c} = c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k} \\ (\vec{b} + \vec{c}) = (b_{1} + c_{1}) \hat{i} + (b_{2} + c_{2}) \hat{j} + (b_{3} + c_{3}) \hat{k} \end{array}$ $Now, \vec{a} \times (\vec{b} + \vec{c}) | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ b_{1} + c_{1} & b_{2} + c_{2} & b_{3} + c_{3} \end{array} |$ $\begin{array}{l} = i [a_{2} (b_{3} + c_{3}) - a_{3} (b_{2} + c_{2})] - \hat{j} [a_{1} (b_{3} + c_{3}) - a_{3} (b_{1} + c_{1})] + \hat{k} [a_{1} (b_{2} + c_{2}) - a_{2} (b_{1} + c_{1})] \\ = \hat{i} [a_{2} b_{3} + a_{2} c_{3} - a_{3} b_{2} - a_{3} c_{2}] + \hat{j} [- a_{1} b_{3} - a_{1} c_{3} + a_{3} c_{1}] + \hat{k} [a_{1} b_{2} + a_{1} c_{2} - a_{2} b_{1} - a_{2} c_{1}] \dots (1) \end{array}$ $\begin{aligned} \vec{a} \times \vec{b} & = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \end{array} | \\ = \hat{i} [a_{2} b_{3} - a_{3} b_{2}] + \hat{j} [b_{1} a_{3} - a_{1} b_{3}] + \hat{k} [a_{1} b_{2} - a_{2} b_{1}] \end{aligned}$ $\begin{array}{l} (\vec{a} \times \vec{b}) + (\vec{a} \times \vec{c}) = \hat{i} [a_{2} b_{3} + a_{2} c_{3} - a_{3} b_{2} - a_{3} c_{2}] + \hat{j} [b a_{3} + a_{3} c_{1} - a_{1} b_{3} - a_{1} c_{3}] \\ + \hat{k} [a_{1} b_{2} + a_{1} c_{2} - a_{2} b_{1} - a_{2} c_{1}] \\ Now, from (1) and (4), we have: \\ \vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c} \\ Hence, the given result is proved. \end{array}$

Q8. $\begin{array}{l} If either \vec{a} = {\vec{0}}_{or} \vec{b} = \vec{0}, then \vec{a} \times \vec{b} = \vec{0} . Is the converse true? Justify your answer with an \\ example. \end{array}$

Answer. $\begin{array}{l} Take any parallel non-zero vectors so that \bar{a} \times \vec{b} = \bar{0} \\ Let \vec{a} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}, \vec{b} = 4 \hat{i} + 6 \hat{j} + 8 \hat{k} \\ Then, \end{array}$ $\vec{a} \times \vec{b} = | \begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 4 & 6 & 8 \end{array} | = \hat{i} (24 - 24) - \hat{j} (16 - 16) + \hat{k} (12 - 12) = 0 \hat{i} + 0 \hat{ȷ} + 0 \hat{k} = \vec{0}$ $\begin{array}{l} It can now be observed that: \\ | \vec{a} | = \sqrt{2^{2} + 3^{2} + 4^{2}} = \sqrt{29} \\ ∴ \vec{a} \neq \vec{0} \\ | \vec{b} | = \sqrt{4^{2} + 6^{2} + 8^{2}} = \sqrt{116} \\ ∴ \vec{b} \neq \vec{0} \end{array}$ Hence, the converse of the given statement need not be true.

Q9. $\begin{array}{l} Find the area of the triangle with vertices A (1, 1, 2), B (2, 3, 5) and \\ C (1, 5, 5) . \end{array}$

Answer. $\begin{array}{l} The vertices of triangle A B C are given as A (1, 1, 2), B (2, 3, 5), and \\ C (1, 5, 5) . \\ The adjacent sides \vec{A B} and \vec{B C} of Δ A B C are given as: \\ \vec{A B} = (2 - 1) \hat{i} + (3 - 1) \hat{j} + (5 - 2) \hat{k} = \hat{i} + 2 \hat{j} + 3 \hat{k} \\ \vec{B C} = (1 - 2) \hat{i} + (5 - 3) \hat{j} + (5 - 5) \hat{k} = - \hat{i} + 2 \hat{j} \end{array}$ $\vec{A B} \times \vec{B C} = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ - 1 & 2 & 0 \end{array} | = \hat{i} (- 6) - \hat{j} (3) + \hat{k} (2 + 2) = - 6 \hat{i} - 3 \hat{j} + 4 \hat{k}$ $\begin{matrix} ∴ | \vec{A B} \times \vec{B C} | = \sqrt{(- 6)^{2} + (- 3)^{2} + 4^{2}} = \sqrt{36 + 9 + 16} = \sqrt{61} \\ Hence, the area of Δ A B C square units. \end{matrix}$

Q10. $\begin{array}{l} Find the area of the parallelogram whose adjacent sides are determined by the vector \\ \vec{a} = \hat{i} - \hat{j} + 3 \hat{k} and \vec{b} = 2 \hat{i} - 7 \hat{j} + \hat{k} \end{array}$

Answer. $\begin{array}{l} The area of the parallelogram whose adjacent sides are \bar{a} and {\vec{b}}_{is} | \bar{a} \times \vec{b} | \\ Adjacent sides are given as: \\ \vec{a} = \hat{i} - \hat{j} + 3 \hat{k} and \vec{b} = 2 \hat{i} - 7 \hat{j} + \hat{k} \end{array}$ $∴ \vec{a} \times \vec{b} = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 1 & 3 \\ 2 & - 7 & 1 \end{array} | = \hat{i} (- 1 + 21) - \hat{j} (1 - 6) + \hat{k} (- 7 + 2) = 20 \hat{i} + 5 \hat{j} - 5 \hat{k}$ $| \vec{a} \times \vec{b} | = \sqrt{20^{2} + 5^{2} + 5^{2}} = \sqrt{400 + 25 + 25} = 15 \sqrt{2}$ $15 \sqrt{2}$

Q11. $\begin{array}{l} Let the vectors \vec{a} and \vec{b} be such that | \vec{a} | = 3 and | \vec{b} | = \frac{\sqrt{2}}{3}, then \vec{a} \times \vec{b} is a unit vector, if \\ the anale between \vec{a} and \vec{b} is \end{array}$ $(A) \frac{π}{6} (B) \frac{π}{4} (C) \frac{π}{3} (D) \frac{π}{2}$

Answer. It is given that $\begin{array}{l} | \vec{a} | = 3 and | \vec{b} | = \frac{\sqrt{2}}{3} \\ we know that \end{array} \vec{a} \times \vec{b} = | \vec{a} | | \vec{b} | \sin θ \hat{n}$ $\begin{array}{l} Now, \vec{a} \times \vec{b} is a unit vector if | \vec{a} \times \vec{b} | = 1 \\ | \vec{a} \times \vec{b} | = 1 \\ \Rightarrow | \vec{a} | | \vec{b} | \sin θ \hat{n} | = 1 \\ \Rightarrow | \vec{a} | | \vec{b} | \sin θ | = 1 \\ \Rightarrow 3 \times \frac{\sqrt{2}}{3} \times \sin θ = 1 \\ \Rightarrow \sin θ = \frac{1}{\sqrt{2}} \\ \Rightarrow θ = \frac{π}{4} \end{array}$ $\begin{array}{l} Hence, \vec{a} \times \vec{b} is a unit vector if the angle between \vec{a} and \vec{b} is \frac{π}{4} \\ The correct answer is B . \end{array}$

Q12. $\begin{array}{l} Area of a rectangle having vertices A, B, C, and D with position vectors \\ - \hat{i} + \frac{1}{2} \hat{j} + 4 \hat{k}, \hat{i} + \frac{1}{2} \hat{j} + 4 \hat{k}, \hat{i} - \frac{1}{2} \hat{j} + 4 \hat{k} \end{array}$ respectively $\begin{array}{l} (A) \frac{1}{2} (B) 1 \\ (C) 2 (D) 4 \end{array}$

\begin{array}{l}  \end{array}

Answer. $\vec{O A} = - \hat{i} + \frac{1}{2} \hat{j} + 4 \hat{k}, \vec{O B} = \hat{i} + \frac{1}{2} \hat{j} + 4 \hat{k}, \vec{O C} = \hat{i} - \frac{1}{2} \hat{j} + 4 \hat{k}, \vec{O D} = - \hat{i} - \frac{1}{2} \hat{j} + 4 \hat{k}$ $\begin{array}{l} \vec{A B} = (1 + 1) \hat{i} + (\frac{1}{2} - \frac{1}{2}) \hat{j} + (4 - 4) \hat{k} = 2 \hat{i} \\ \vec{B C} = (1 - 1) \hat{i} + (- \frac{1}{2} - \frac{1}{2}) \hat{j} + (4 - 4) \hat{k} = - \hat{j} \end{array}$ $∴ \vec{A B} \times \vec{B C} = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 0 \\ 0 & - 1 & 0 \end{array} | = \hat{k} (- 2) = - 2 \hat{k}$ $\begin{array}{l} | \vec{A B} \times \vec{A C} | = \sqrt{(- 2)^{2}} = 2 \\ Now, it is known that the area of a parallelogram whose adjacent sides are \\ \vec{a} and {\vec{b}}_{is} | \vec{a} \times \vec{b} | \end{array}$ $\begin{array}{l} Hence, the area of the given rectangle is | \vec{A B} \times \vec{B C} | = 2 square units. \\ The correct answer is C . \end{array}$