Question

The magnitude of the cross product of two vectors is $\sqrt{3}$ times the dot product. The angle between the vector is

• A. $\frac{\mathrm{\pi }}{6}$
• B. $\frac{\mathrm{\pi }}{3}$
• C. $\frac{\mathrm{\pi }}{2}$
• D. $\frac{\mathrm{\pi }}{4}$

Answer 1

The correct option is B

$\frac{\mathrm{\pi }}{3}$

Explanation for the correct option

Let $\overrightarrow{A},\overrightarrow{B}$ be the two vector

Given, $\left|\overrightarrow{A}\times \overrightarrow{B}\right|=\sqrt{3}\left|\overrightarrow{A}.\overrightarrow{B}\right|$

We know that, $\left|\stackrel{\rightharpoonup }{A}\times \overrightarrow{B}\right|=\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\sin \left(\theta \right)$ and $\left|\overrightarrow{A}.\overrightarrow{B}\right|=\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\cos \left(\theta \right)$

$$\begin{gathered} \Rightarrow \left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\sin \left(\theta \right)=\sqrt{3}\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\cos \left(\theta \right) \\ \Rightarrow \frac{\sin \left(\theta \right)}{\cos \left(\theta \right)}=\sqrt{3} \\ \Rightarrow \tan \left(\theta \right)=\sqrt{3} \\ \Rightarrow \theta ={\tan }^{-1}\left(\sqrt{3}\right) \\ \Rightarrow \theta =\frac{\mathrm{\pi }}{3} \end{gathered}$$

Hence, $\mathrm{Option}\left(\mathrm{B}\right)$ is the correct answer.