Question

If $a,b$ and $c$are unit vectors such that$ax(bxc)=\frac{1}{2}b$, then the angle between $a$and $c$ is?

• A. $\pi /6$
• B. $\pi /4$
• C. $\pi /2$
• D. $\pi /3$

Answer 1

The correct option is D

$\pi /3$

Find the angle between $a$and $c$:

Given that $a,b$ and $c$are unit vectors such that$ax(bxc)=\frac{1}{2}b$

$\Rightarrow (a.c)b–(a.b)c=\frac{1}{2}b$

If we compare both the sides,

$a.c=1/2$

The angle between $a$and $c$ is given by

$$\begin{gathered} \Rightarrow \cos \theta =\frac{(a.c)}{\left|a\right|\left|c\right|} \\ \Rightarrow \cos \theta =\frac{{\displaystyle \frac{1}{2}}}{1\times 1}\mathbf{}\mathbf{}\left[sinceaandcareunitvectors\Rightarrow \left|a\right|=1,\left|c\right|=1\right] \\ \Rightarrow \cos \theta =\frac{1}{2} \\ \Rightarrow \cos \theta =\cos (\pi /3) \end{gathered}$$

Therefore, $\theta =\pi /3$

Hence, the correct option is D.