Question

If magnitude of scalar and vector products of two vectors are $48\sqrt{3}$ and $144$ respectively, then the angle between the two vectors is :

• A. ${30}^{\circ }$
• B. ${60}^{\circ }$
• C. ${90}^{\circ }$
• D. ${180}^{\circ }$

Answer 1

The correct option is B ${60}^{\circ }$

Let $\overrightarrow{a}$ and $\overrightarrow{b}$ be two vectors.
Given:
$|\overrightarrow{a}\times \overrightarrow{b}|=144$
$\Rightarrow |a||b|\sin \theta =144$
and $|\overrightarrow{a}.\overrightarrow{b}|=48\sqrt{3}$
$\Rightarrow |a||b|\cos \theta =48\sqrt{3}$
So, $\tan \theta =\frac{144}{48\sqrt{3}}$
$\Rightarrow \tan \theta =\sqrt{3}\Rightarrow \theta ={60}^{\circ }$