Question

The resultant of 2 vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ is perpendicular to the $\overrightarrow{A}$ and its magnitude is equal to half the magnitude of vector B. Find the angle between vector $\overrightarrow{A}$ and $\overrightarrow{B}$

Answer 1

Let the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$ be $\Theta$ .

$\overrightarrow{A}+\overrightarrow{B}$ is perpendicular to $\overrightarrow{A}$ .

Therefore , ($\overrightarrow{A}+\overrightarrow{B}$) . ($\overrightarrow{A}$) = 0

${\left|\overrightarrow{A}\right|}^2$ + $\overrightarrow{A}.\overrightarrow{B}$ = 0

$\overrightarrow{A}.\overrightarrow{B}$ = -${\left|\overrightarrow{A}\right|}^2$ ....(i)

And $|\overrightarrow{A}+\overrightarrow{B}|$ = $\frac{\left|\overrightarrow{B}\right|}{2}$

${\left|\overrightarrow{A}\right|}^2$ + ${\left|\overrightarrow{B}\right|}^2$ + $2$ $\overrightarrow{A}.\overrightarrow{B}$ = $\frac{{\left|\overrightarrow{B}\right|}^2}{4}$ ...(ii)

From (i) and (ii) , we get :

cos$\Theta$ = -$\frac{\left|\overrightarrow{A}\right|}{\left|\overrightarrow{B}\right|}$

= -$\frac{\sqrt{3}}{2}$

Hence, $\Theta$ = ${150}^{\circ }$.