Vectors 4 Question 9
9. If $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{c}}$ are non-coplanar unit vectors such that $\overrightarrow{\mathbf{a}} \times(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})=\frac{(\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}})}{\sqrt{2}}$, then the angle between $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ is
(1995, 2M)
(a) $\frac{3 \pi}{4}$
(b) $\frac{\pi}{4}$
(c) $\frac{\pi}{2}$
(d) $\pi$
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Answer:
Correct Answer: 9. (a)
Solution:
- Since, $\quad \overrightarrow{\mathbf{a}} \times(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})=\frac{\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}}{\sqrt{2}}$
$\Rightarrow \quad(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{c}}) \overrightarrow{\mathbf{b}}-(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}) \overrightarrow{\mathbf{c}}=\frac{1}{\sqrt{2}} \overrightarrow{\mathbf{b}}+\frac{1}{\sqrt{2}} \overrightarrow{\mathbf{c}}$
On equating the coefficient of $\overrightarrow{\mathbf{c}}$, we get
$$ \begin{aligned} \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} & =-\frac{1}{\sqrt{2}} \\ \Rightarrow \quad|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}| \cos \theta & =-\frac{1}{\sqrt{2}} \\ \therefore \quad \cos \theta & =-\frac{1}{\sqrt{2}} \Rightarrow \theta=\frac{3 \pi}{4} \end{aligned} $$
