Question

If the magnitude of the cross product of two vector is $\sqrt{3}$ times to the magnitude of their scalar product the angle between two vector will be :

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

Answer 1

The correct option is C $\frac{\pi }{3}$

$\text{Step 1: Simplify the given equation}$

Let the two vectors be $\overrightarrow{A}$ and $\overrightarrow{B}$

Magnitude of Cross product of vectors= $\sqrt{3}\times$ Magnitude of Dot product of vectors

$\Rightarrow |\overrightarrow{A}\times \overrightarrow{B}|=\sqrt{3}|\overrightarrow{A}.\overrightarrow{B}|$

Squaring on both the sides

$|\overrightarrow{A}\times \overrightarrow{B}{|}^2=3|\overrightarrow{A}.\overrightarrow{B}{|}^2$

$\Rightarrow (\overrightarrow{A}\times \overrightarrow{B})(\overrightarrow{A}\times \overrightarrow{B})=3(\overrightarrow{A}.\overrightarrow{B})(\overrightarrow{A}.\overrightarrow{B})$

$\Rightarrow \left(|\overrightarrow{A}||\overrightarrow{B}|\sin \theta \right)\left(|\overrightarrow{A}||\overrightarrow{B}|\sin \theta \right)$$=\left(|\overrightarrow{A}||\overrightarrow{B}|\cos \theta \right)\left(|\overrightarrow{A}|\overrightarrow{B}|\cos \theta \right)$ $....\left(1\right)$

$\text{Step 2: Angle between the Vectors}$

Eq $\left(1\right)$ $\Rightarrow {\sin }^2\theta =3{\cos }^2\theta$

$\Rightarrow ta{n}^2\theta =3$

$\Rightarrow \theta ={\tan }^{-1}\left(\sqrt{3}\right)$$=\frac{\pi }{3}$

Hence, Option C is correct.