Q99 / 170 · Manipal (MET) 2015 · Mathematics
Let a,b and c be non-zero vectors such that no two are collinear and $(a \times b) \times c=\frac{1}{3}|b||c| a$. If $\theta$ is the acute angle between the vectors $b$ and $c$, then $\sin \theta$ equals
- Correct answerA$\frac{2 \sqrt{2}}{3}$
- B$\frac{\sqrt{2}}{3}$
- C$\frac{2}{3}$
- D$\frac{1}{3}$
Solution
We have,
$(a \times b) \times c=\frac{1}{3}|b||c| a$
$\begin{aligned} \Rightarrow & (a \cdot c) b-(b \cdot c) a=\frac{1}{3}|b||c| a \\ \Rightarrow \quad & (a \cdot c) b-\left\{(b \cdot c)+\frac{1}{3}|b||c|\right\} a=0 \\ \Rightarrow \quad & a \cdot c=0 \text { and } b \cdot c+\frac{1}{3}|b||c|=0 \\ & {[\because a, b \text { are non-collinear }] }\end{aligned}$
$\begin{aligned} & \Rightarrow \quad|\mathbf{b}||\mathbf{c}| \cos \theta+\frac{1}{3}|\mathbf{b}||\mathbf{c}|=0 \\ & \Rightarrow \quad \cos \theta=-\frac{1}{3} \\ & \Rightarrow \quad \sin \theta=\sqrt{1-\cos ^2 \theta}=\sqrt{1-\left(-\frac{1}{3}\right)^2}\end{aligned}$
$=\sqrt{1-\frac{1}{9}}=\sqrt{\frac{8}{9}}=\frac{2 \sqrt{2}}{3}$
Hence, the correct answer is option: 'A'.