Question:
The angle between vector $(\overrightarrow{\mathrm{A}})$ and $(\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}})$ is :
Correct Option: , 3
Solution:
Angle between $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}, \theta=60^{\circ}$
Angle betwenn $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}$
$\tan \alpha=\frac{B \sin \theta}{A-B \cos \theta}$
$=\frac{B \sqrt{\frac{3}{2}}}{A-B \times \frac{1}{2} 2}$
$\tan \alpha=\frac{\sqrt{3} B}{2 A-B}$
Ans 3