Ncert Exercises & Solutions

If $|\vec{A}|=2$ and $|\vec{B}|=4$, then match the relations in column-I with the angle $\theta$ between $\vec{A}$ and $\vec{B}$ in column-II.

	Column-I		Column-II
(a)	$\vec{A}.\vec{B}=0$	(i)	$\theta=0^{\circ}$
(b)	$\vec{A}.\vec{B}=8$	(ii)	$\theta=90^{\circ}$
(c)	$\vec{A}.\vec{B}=4$	(iii)	$\theta=180^{\circ}$
(d)	$\vec{A}.\vec{B}=-8$	(iv)	$\theta=60^{\circ}$

Choose the correct answer from the options given below:

1.	(a)–(iii), (b)-(ii), (c)-(i), (d)-(iv)
2.	(a)–(ii), (b)-(i), (c)-(iv), (d)-(iii)
3.	(a)–(ii), (b)-(iv), (c)-(iii), (d)-(i)
4.	(a)–(iii), (b)-(i), (c)-(ii), (d)-(iv)

Given |A| = 2 and |B| = 4

(a)

$\begin{array}{l} A B = A B \cos θ = 0 \Rightarrow 2 \times 4 \cos θ = 0 \\ \Rightarrow \cos θ = 0 = \cos 90^{\circ} \Rightarrow θ = 90^{\circ} \\ ∴ Option (a) matches with option (ii). \end{array}$

(b)

$\begin{array}{l} AB = ABcos θ = 8 \\ \Rightarrow \cos θ = 1 = \cos \infty \Rightarrow 2 \times 4 \cos θ = 8 \\ ∴ Option (b) matches with option (i). \end{array}$

(c)

$\begin{array}{l} AB = ABcos θ = 4 \Rightarrow 2 \times 4 \cos θ = 4 \\ \Rightarrow \cos θ = \frac{1}{2} = \cos 60^{\circ} \Rightarrow θ = 60^{\circ} \\ ∴ Option (c) matches with option (iv). \end{array}$

(d)

$\begin{array}{l} AB = ABcos θ = - 8 \Rightarrow 2 \times 4 \cos θ = - 8 \\ \Rightarrow \cos θ = - 1 = \cos 180^{\circ} \Rightarrow θ = 180 ° \end{array}$
$∴ Option (d) matches with option (iii) .$

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