Two vectors $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ are at an angle of $60^{\circ}$ with each other. Their resultant makes an angle of $45^{\circ}$ with $\overrightarrow{\mathrm{a}}$. If $|\vec{b}|=2$ units, then $|\vec{a}|$ is

• A

  $\sqrt{3}$
• Correct answer
  B

  $\sqrt{3}-1$
• C

  $\sqrt{3}+1$
• D

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

Solution

Angle between the vectors is, $\theta=60^{\circ}$. Angle between resultant and $\vec{a}$ is, $\phi=45^{\circ}$. The above is shown in the below figure, From parallelogram law, $\phi \& \theta$ is related as, $\tan (\phi)=\frac{|\overrightarrow{\mathrm{b}}| \sin (\theta)}{|\overrightarrow{\mathrm{a}}|+|\overrightarrow{\mathrm{b}}| \cos (\theta)} \text {. }$ Substituting all the corresponding given values in above relation, we have, $\begin{aligned} & \Rightarrow \tan \left(45^{\circ}\right)=\frac{2 \sin \left(60^{\circ}\right)}{|\overrightarrow{\mathrm{a}}|+2 \cos \left(60^{\circ}\right)}=\frac{\sqrt{3}}{a+1},(\because|\overrightarrow{\mathrm{a}}|=a) \\ & \Rightarrow 1=\frac{\sqrt{3}}{a+1} \\ & \Rightarrow a+1=\sqrt{3} \\ & \Rightarrow a=\sqrt{3}-1 . \end{aligned}$

Hence, the correct answer is option: 'B'.