Question

If $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are non-coplanar unit vectors such that $\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})=\frac{(\overrightarrow{b}+\overrightarrow{c})}{\sqrt{2}}$, then the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is

• A. $\frac{3\pi }{4}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{2}$
• D. $\pi$

Answer 1

Answer: (A)

Since, $\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})=\frac{\overrightarrow{b}+\overrightarrow{c}}{\sqrt{2}}$
$\Rightarrow (\vec{a}\cdot \overrightarrow{c})\overrightarrow{b}-(\overrightarrow{a}\cdot \overrightarrow{b})\overrightarrow{c}=\frac{1}{\sqrt{2}}\overrightarrow{b}+\frac{1}{\sqrt{2}}\overrightarrow{c}$
On equating the coefficient of $\overrightarrow{c}$, we get
$\overrightarrow{a}\cdot \overrightarrow{b}=-\frac{1}{\sqrt{2}}\Rightarrow |\overrightarrow{a}||\overrightarrow{b}|\cos \theta =-\frac{1}{\sqrt{2}}$
$\therefore \cos \theta =-\frac{1}{\sqrt{2}}\Rightarrow \theta =\frac{3\pi }{4}$