The resultant of two vectors $A$ and $B$ is perpendicular to vector $A$ and its magnitude is equal to half the magnitude of vector $B$. The angle between $A$ and $B$ is

The correct option is B

$150°$

Find the angle between the given vectors

The magnitude of resultant of $A$ and $B$ is given by,

$R=\sqrt{A^2+B^2+2·A·B·\cos \left(\theta \right)}$

According to question,

Magnitude of resultant = half of Magnitude of $B$

$$\begin{gathered} \Rightarrow \sqrt{A^2+B^2+2·A·B·\cos \left(\theta \right)}=\frac{B}{2} \\ \Rightarrow A^2+B^2+2·A·B·\cos \left(\theta \right)={\left(\frac{B}{2}\right)}^2 \\ \Rightarrow A^2+B^2+2·A·B·\cos \left(\theta \right)=\frac{B^2}{4} \\ \Rightarrow A^2+\frac{3B^2}{4}+2·A·B·\cos \left(\theta \right)=0.................\left(i\right) \end{gathered}$$

Now,

$$\begin{gathered} \tan \left(90°\right)=\frac{B\sin \left(\theta \right)}{A+B\cos \left(\theta \right)} \\ \Rightarrow A+B\cos \left(\theta \right)=0 \\ \Rightarrow \cos \left(\theta \right)=-\frac{A}{B}....................\left(ii\right) \end{gathered}$$

Now using $\left(ii\right)$in $\left(i\right)$ we get

$$\begin{gathered} A^2+\frac{3B^2}{4}+2·A·B·\left(-\frac{A}{B}\right)=0 \\ \Rightarrow A^2+\frac{3B^2}{4}-2A^2=0 \\ \Rightarrow A^2=\frac{3B^2}{4} \\ \Rightarrow A=\frac{\sqrt{3}B}{2} \end{gathered}$$

Now,

$$\begin{gathered} \Rightarrow \cos \left(\theta \right)=-\frac{A}{B} \\ \Rightarrow \cos \left(\theta \right)=-\frac{\sqrt{3}B}{2B} \\ \Rightarrow \cos \left(\theta \right)=-\frac{\sqrt{3}}{2} \\ \Rightarrow \theta =150°\left[\because \cos \left(150°\right)=-\frac{\sqrt{3}}{2}\right] \end{gathered}$$

Hence, the correct option is $\left(\mathrm{B}\right)$, $150°$.