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:
Correct Option: , 2
$|\overrightarrow{\mathrm{a}}|=|\overrightarrow{\mathrm{b}}|,|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{a}}|, \overrightarrow{\mathrm{a}} \perp \overrightarrow{\mathrm{b}}$
$|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{a}}| \Rightarrow|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \sin 90^{\circ}=|\overrightarrow{\mathrm{a}}| \Rightarrow|\overrightarrow{\mathrm{b}}|=1=|\overrightarrow{\mathrm{a}}|$
$\vec{a}$ and $\vec{b}$ are mutually perpendicular unit
vectors.
Let $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}, \overrightarrow{\mathrm{b}}=\hat{\mathrm{j}} \Rightarrow \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\hat{\mathrm{k}}$
$\cos \theta=\frac{(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}) \hat{\mathrm{i}}}{\sqrt{3} \sqrt{1}}=\frac{1}{\sqrt{3}} \Rightarrow \theta=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)$
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