Question

Two vectors have magnitudes $3$ units and $6$ units respectively. What should be the angle between these two vectors if the resultant vector is perpendicular to the smaller one?

• A. ${60}^{\circ }$
• B. ${120}^{\circ }$
• C. ${150}^{\circ }$
• D. ${180}^{\circ }$

Answer 1

The correct option is B ${120}^{\circ }$

Let the angle between the vectors be $\theta$ and as per given question, the angle between the vector of magnitude $3$ units and the resultant is $\alpha ={90}^{\circ }$.
From parallelogram law we have
$\tan \alpha =\frac{6\sin \theta }{3+6\cos \theta }=\tan {90}^{\circ }$
$\Rightarrow 3+6\cos \theta =0\Rightarrow \cos \theta =\frac{-3}{6}=\frac{-1}{2}$
$\Rightarrow \theta ={\cos }^{-1}\left(\frac{-1}{2}\right)={120}^{\circ }$