Question

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

• A. $\frac{8\pi }{2}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{2}$
• D. $\pi$

Answer 1

The correct option is B $\frac{\pi }{4}$

Given $\overline{a}\times (\overline{b}\times \overline{c})=\frac{\overline{b}+\overline{c}}{\sqrt{2}}$
$\Rightarrow (\overline{a}\cdot \overline{c})\overline{b}-(\overline{a}\cdot \overline{b})\overline{c}=\frac{\overline{b}+\overline{c}}{\sqrt{2}}$
$\Rightarrow (\overline{a}\cdot \overline{c}-\frac{1}{\sqrt{2}})\overline{b}-(\overline{a}\cdot \overline{b}+\frac{1}{\sqrt{2}})\overline{c}=\overline{0}$
$\Rightarrow \overline{a}\cdot \overline{c}=\frac{1}{\sqrt{2}}$ and $\overline{a}\cdot \overline{b}=-\frac{1}{\sqrt{2}}$
($\because \overline{a},\overline{b},\overline{c}$ are non-coplanar)
$\cos \theta =\frac{\overline{a}\cdot \overline{b}}{\left|\overline{a}\right|\left|\overline{b}\right|}=\frac{-1/\sqrt{2}}{1.1}$
$=\frac{-1}{\sqrt{2}}$
$\therefore \theta =\pi -\frac{\pi }{4}$
$=\frac{3\pi }{4}$
$\therefore$ angle between $\overline{a}$ and $\overline{b}$ is $\frac{\pi }{4}$