Solve this following

Question:

Let $\vec{a}$ and $\vec{b}$ be two non-zero vectors perpendicular to each other and $|\vec{a}|=|\vec{b}|$. If $|\vec{a} \times \vec{b}|=|\vec{a}|$, then the angle between the vectors $(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}))$ and $\overrightarrow{\mathrm{a}}$ is equal to :

1. $\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)$

2. $\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)$

3. $\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)$

4. $\sin ^{-1}\left(\frac{1}{\sqrt{6}}\right)$

Correct Option: , 2

Solution:

$|\vec{a}|=|\vec{b}|,|\vec{a} \times \vec{b}|=|\vec{a}|, \vec{a} \perp \vec{b}$

$|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{a}}| \Rightarrow|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \sin 90^{\circ}=|\overrightarrow{\mathrm{a}}| \Rightarrow|\overrightarrow{\mathrm{b}}|=1=|\overrightarrow{\mathrm{a}}|$

$\vec{a}$ and $\vec{b}$ are mutually perpendicular unit vectors.

Let $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}, \overrightarrow{\mathrm{b}}=\hat{\mathrm{j}} \Rightarrow \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\hat{\mathrm{k}}$

$\cos \theta=\frac{(\hat{i}+\hat{j}+\hat{k}) \cdot \hat{i}}{\sqrt{3} \sqrt{1}}=\frac{1}{\sqrt{3}} \Rightarrow \theta=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)$