Question

Let $\stackrel{\rightharpoonup }{a}$, $\stackrel{\rightharpoonup }{b}$and $\stackrel{\rightharpoonup }{c}$ be a non-zero vectors such that $\left(\stackrel{\rightharpoonup }{a}\times \stackrel{\rightharpoonup }{b}\right)\times \stackrel{\rightharpoonup }{c}=-\frac{1}{4}\left|\stackrel{\rightharpoonup }{b}\right|\left|\stackrel{\rightharpoonup }{c}\right|\stackrel{\rightharpoonup }{a}$.If $\theta$is the acute angle between the vectors $\stackrel{\rightharpoonup }{b}$ and $\stackrel{\rightharpoonup }{c}$, then the angle between $\stackrel{\rightharpoonup }{a}$and $\stackrel{\rightharpoonup }{c}$ is?

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

Answer 1

The correct option is D

$\frac{\mathrm{\pi }}{2}$

Determine the angle between $\stackrel{\rightharpoonup }{a}$and $\stackrel{\rightharpoonup }{c}$.

Let $\stackrel{\rightharpoonup }{a}$, $\stackrel{\rightharpoonup }{b}$and $\stackrel{\rightharpoonup }{c}$ be a non-zero vector.

Given $\left(\stackrel{\rightharpoonup }{a}\times \stackrel{\rightharpoonup }{b}\right)\times \stackrel{\rightharpoonup }{c}=-\frac{1}{4}\left|\stackrel{\rightharpoonup }{b}\right|\left|\stackrel{\rightharpoonup }{c}\right|\stackrel{\rightharpoonup }{a}$

Using vector triple cross product formula $\left(\stackrel{\rightharpoonup }{a}\times \stackrel{\rightharpoonup }{b}\right)\mathbf{\times }\stackrel{\mathbf{\rightharpoonup }}{\mathbf{c}}\mathbf{=}\left(\stackrel{\rightharpoonup }{a.}\stackrel{\rightharpoonup }{c}\right)\stackrel{\mathbf{\rightharpoonup }}{\mathbf{b}}\mathbf{-}\left(\stackrel{\rightharpoonup }{b.}\stackrel{\rightharpoonup }{c}\right)\stackrel{\mathbf{\rightharpoonup }}{\mathbf{a}}$

$\left(\stackrel{\rightharpoonup }{a.}\stackrel{\rightharpoonup }{c}\right)\stackrel{\rightharpoonup }{b}-\left(\stackrel{\rightharpoonup }{b.}\stackrel{\rightharpoonup }{c}\right)\stackrel{\rightharpoonup }{a}=-\frac{1}{4}\left|\stackrel{\rightharpoonup }{b}\right|\left|\stackrel{\rightharpoonup }{c}\right|\stackrel{\rightharpoonup }{a}+0.\stackrel{\rightharpoonup }{b}$

Equating the coefficients of $\stackrel{\rightharpoonup }{a}$, (since $\theta$is the acute angle value be positive )

$$\begin{gathered} \stackrel{\rightharpoonup }{b.}\stackrel{\rightharpoonup }{c}=0 \\ \Rightarrow \left|\stackrel{\rightharpoonup }{b}\right|\left|\stackrel{\rightharpoonup }{c}\right|\cos \left(\theta \right)=0 \\ \Rightarrow \cos \left(\theta \right)=0 \\ \Rightarrow \theta ={\cos }^{-1}\left(0\right) \\ \Rightarrow \theta =\frac{\mathrm{\pi }}{2} \end{gathered}$$

Hence, option (D) is the correct answer.