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 :

• A. ${\sin }^{-1}\left(\frac{1}{\sqrt{3}}\right)$
• B. ${\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)$
• C. ${\cos }^{-1}\left(\frac{1}{\sqrt{2}}\right)$
• D. ${\sin }^{-1}\left(\frac{1}{\sqrt{6}}\right)$

Answer 1

Answer: (B)

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