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 of the correct option.

Compute the angle:

Let two vectors are $\overrightarrow{a}$ and $\overrightarrow{b}$

Given: The magnitude of the cross product of two vectors is $\sqrt{3}$ times the dot product.

$\left|\overrightarrow{a}\times \overrightarrow{b}\right|=\sqrt{3}\left|\overrightarrow{a}\overrightarrow{.b}\right|$

$\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)$

$\Rightarrow$ $\tan \left(\theta \right)=\sqrt{3}$

Since $\tan \left(\frac{\mathrm{\pi }}{3}\right)=\sqrt{3}$,

$\Rightarrow$ $\theta =\frac{\mathrm{\pi }}{3}$

Therefore, the angle between the vector is $\frac{\mathrm{\pi }}{3}$.

Hence, option $\left(B\right)$ is the correct answer.