Vectors 4 Question 4
4. Let $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$ be three non-zero vectors such that no two of them are collinear and $(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \times \overrightarrow{\mathbf{c}}=\frac{1}{3}|\overrightarrow{\mathbf{b}}||\overrightarrow{\mathbf{c}}| \overrightarrow{\mathbf{a}}$. If $\theta$ is the angle between vectors $\overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$, then a value of $\sin \theta$ is
(a) $\frac{2 \sqrt{2}}{3}$
(b) $\frac{-\sqrt{2}}{3}$
(c) $\frac{2}{3}$
(d) $\frac{-2 \sqrt{3}}{3}$
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Answer:
Correct Answer: 4. (a)
Solution:
- Given,
$$ (\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \times \overrightarrow{\mathbf{c}}=\frac{1}{3}|\overrightarrow{\mathbf{b}}||\overrightarrow{\mathbf{c}}| \overrightarrow{\mathbf{a}} $$
$$ \begin{array}{ll} \Rightarrow & -\overrightarrow{\mathbf{c}} \times(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}})=\frac{1}{3}|\overrightarrow{\mathbf{b}}||\overrightarrow{\mathbf{c}}| \overrightarrow{\mathbf{a}} \\ \Rightarrow & -(\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{b}}) \cdot \overrightarrow{\mathbf{a}}+(\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}) \overrightarrow{\mathbf{b}}=\frac{1}{3}|\overrightarrow{\mathbf{b}}||\overrightarrow{\mathbf{c}}| \overrightarrow{\mathbf{a}} \\ \Rightarrow & {\left[\frac{1}{3}|\overrightarrow{\mathbf{b}}||\overrightarrow{\mathbf{c}}|+(\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{b}})\right] a=(\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}) \overrightarrow{\mathbf{b}}} \end{array} $$
Since, $a$ and $b$ are not collinear.
$$ \begin{aligned} & \therefore \overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{b}}+\frac{1}{3}|\overrightarrow{\mathbf{b}}||\overrightarrow{\mathbf{c}}|=0 \text { and } \overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}=0 \\ & \Rightarrow|\overrightarrow{\mathbf{b}}||\overrightarrow{\mathbf{c}}| \cos \theta+\frac{1}{3}|\overrightarrow{\mathbf{b}}||\overrightarrow{\mathbf{c}}|=0 \Rightarrow|\overrightarrow{\mathbf{b}}||\overrightarrow{\mathbf{c}}|\left(\cos \theta+\frac{1}{3}\right)=0 \\ & \Rightarrow \quad \cos \theta+\frac{1}{3}=0 \quad[\because|\mathbf{b}| \neq 0,|\mathbf{c}| \neq 0] \\ & \Rightarrow \quad \cos \theta=-\frac{1}{3} \Rightarrow \sin \theta=\frac{\sqrt{8}}{3}=\frac{2 \sqrt{2}}{3} \end{aligned} $$
