Vectors 4 Question 2
2. Let $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ be three unit vectors, out of which vectors $\mathbf{b}$ and $\mathbf{c}$ are non-parallel. If $\alpha$ and $\beta$ are the angles which vector $\mathbf{a}$ makes with vectors $\mathbf{b}$ and $\mathbf{c}$ respectively and $\mathbf{a} \times(\mathbf{b} \times \mathbf{c})=\frac{1}{2} \mathbf{b}$, then $|\alpha-\beta|$ is equal to
(2019 Main, 12 Jan II)
(a) $30^{\circ}$
(b) $45^{\circ}$
(c) $90^{\circ}$
(d) $60^{\circ}$
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Answer:
Correct Answer: 2. (a)
Solution:
- Given, $\mathbf{a} \times(\mathbf{b} \times \mathbf{c})=\frac{1}{2} \mathbf{b} \Rightarrow(\mathbf{a} \cdot \mathbf{c}) \mathbf{b}-(\mathbf{a} \cdot \mathbf{b}) \mathbf{c}=\frac{1}{2} \mathbf{b}$
$$ [\because \mathbf{a} \times(\mathbf{b} \times \mathbf{c})=(\mathbf{a} \cdot \mathbf{c}) \mathbf{b}-(\mathbf{a} \cdot \mathbf{b}) \mathbf{c}] $$
On comparing both sides, we get
$$ \mathbf{a} \cdot \mathbf{c}=\frac{1}{2} $$
and $\quad \mathbf{a} \cdot \mathbf{b}=0$
$\because \mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ are unit vectors, and angle between $\mathbf{a}$ and $\mathbf{b}$ is $\alpha$ and angle between $\mathbf{a}$ and $\mathbf{c}$ is $\beta$, so
$$ \begin{aligned} & |\mathbf{a}||\mathbf{c}| \cos \beta=\frac{1}{2} \\ & \text { [from Eq. (i)] } \\ & \Rightarrow \quad \cos \beta=\frac{1}{2} \quad[\because|\mathbf{a}|=1=|\mathbf{c}|] \\ & \Rightarrow \quad \beta=\frac{\pi}{3} \quad \text {…(iii) }\left[\because \cos \frac{\pi}{3}=\frac{1}{2}\right] \end{aligned} $$
and $|\mathbf{a}||\mathbf{b}| \cos \alpha=0 \quad$ [from Eq. (ii)]
$\Rightarrow \quad \alpha=\frac{\pi}{2}$
From Eqs. (iii) and (iv), we get
$$ |\alpha-\beta|=\frac{\pi}{2}-\frac{\pi}{3}=\frac{\pi}{6}=30^{\circ} $$
