Ncert Exercises & Solutions

If $\left| \vec{A}\right|$ = $2$ and $\left| \vec{B}\right|$ = $4,$ 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) $\left\| \vec{A}\times \vec{B}\right\|$ $=0$	(p) $\theta=30^\circ$
(B) $\left\| \vec{A}\times \vec{B}\right\|$ $=8$	(q) $\theta=45^\circ$
(C) $\left\| \vec{A}\times \vec{B}\right\|$ $=4$	(r) $\theta=90^\circ$
(D) $\left\| \vec{A}\times \vec{B}\right\|$ $=4\sqrt2$	(s) $\theta=0^\circ$

1.	A(s), B(r), C(q), D(p)
2.	A(s), B(p), C(r), D(q)
3.	A(s), B(p), C(q), D(r)
4.	A(s), B(r), C(p), D(q)

Hint: Recall vector product.

Step 1: Use

$| A \times B | = ABsinθ$ and calculate

$θ$ in each part.

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

$\begin{array}{l} (a) | A \times B | = ABsinθ = 0 \Rightarrow 2 \times 4 \times sinθ = 0 \\ \Rightarrow sinθ = 0 = \sin 0^{\circ} \Rightarrow θ = 0^{\circ} \\ ∴ Option (a) matches with option (iv). \end{array}$

$\begin{array}{l} (b) | A \times B | = ABsin θ = 8 \Rightarrow 2 \times 4 \sin θ = 8 \\ \Rightarrow \sin θ = 1 = \sin 90^{\circ} \Rightarrow θ = 90^{\circ} \\ ∴ Option (b) matches with option (iii). \end{array}$

$\begin{array}{l} (c) | A \times B | = ABsin θ = 4 \Rightarrow 2 \times 4 \sin θ = 4 \\ \Rightarrow \sin θ = \frac{1}{2} = \sin 30^{\circ} \Rightarrow θ = 30^{\circ} \\ ∴ Option (c) matches with option (i). \end{array}$

$\begin{array}{l} (d) | A \times B | = ABsin θ = 4 \sqrt{2} \Rightarrow 2 \times 4 \sin θ = 4 \sqrt{2} \\ \Rightarrow \sin θ = \frac{1}{\sqrt{2}} = \sin 45^{\circ} \Rightarrow θ = 45^{\circ} \\ ∴ Option (d) matches with option (ii). \end{array}$

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