# Prove the 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

$(\vec{a}+\vec{b}+(\vec{a} \times \vec{b}))$ and $\vec{a}$ 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:

$|\overrightarrow{\mathrm{a}}|=|\overrightarrow{\mathrm{b}}|,|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{a}}|, \overrightarrow{\mathrm{a}} \perp \overrightarrow{\mathrm{b}}$

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

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

vectors.

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

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