Question

If three unit vector $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ ae such that $\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})=\frac{\overrightarrow{b}}{2}$ then vector $\overrightarrow{a}$ makes angles with $\overrightarrow{b}\&\overrightarrow{c}$

• A. ${60}^{\circ },{90}^{\circ }$
• B. ${45}^{\circ },{45}^{\circ }$
• C. ${30}^{\circ },{60}^{\circ }$
• D. ${90}^{\circ },{60}^{\circ }$

Answer 1

Answer: (A)

${60}^{\circ },{90}^{\circ }$

Given $\overrightarrow{a}\times \overrightarrow{b}\times \overrightarrow{c}=\frac{\overrightarrow{b}}{2}$ and $a$ & $b$ & $c$ are unit vectors

$(\overrightarrow{a}\cdot \overrightarrow{c})\overrightarrow{b}-(\overrightarrow{a}\cdot \overrightarrow{b})c=\frac{\overrightarrow{b}}{2}(\because \overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})=(\overrightarrow{a}.\overrightarrow{c})\overrightarrow{b}-(\overrightarrow{a}.\overrightarrow{b}\left)\overrightarrow{c}\right)$

$\cos \alpha \overrightarrow{b}-\cos \beta \overrightarrow{c}=\frac{\overrightarrow{b}}{2}+0.\overrightarrow{c}$

where $\alpha$ is the angle between $a$ & $c$ and $\beta$ is the angle between $a$ & $b$ comparing, we get

$\cos \alpha =\frac{1}{2},\cos \beta =0$

$\alpha ={60}^o,\beta ={90}^o$