Let vec{a} and vec{b} be two non-zero vectors | vedclass

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(B) Given that $|\vec{a}|=|\vec{b}|$ and $\vec{a} \perp \vec{b}$.Also,$|\vec{a} 	imes \vec{b}| = |\vec{a}|$.Since $\vec{a} \perp \vec{b}$,we have $|\

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

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(B) Given that $|\vec{a}|=|\vec{b}|$ and $\vec{a} \perp \vec{b}$.Also,$|\vec{a} 	imes \vec{b}| = |\vec{a}|$.Since $\vec{a} \perp \vec{b}$,we have $|\vec{a} 	imes \vec{b}| = |\vec{a}||\vec{b}| \sin 90^{\circ} = |\vec{a}||\vec{b}|$.Substituting this into the given equation: $|\vec{a}||\vec{b}| = |\vec{a}|$.Since $\vec{a}$ is a non-zero vector,$|\vec{a}| 
eq 0$,so $|\vec{b}| = 1$. Consequently,$|\vec{a}| = 1$.Thus,$\vec{a}$ and $\vec{b}$ are mutually perpendicular unit vectors.Let $\vec{a} = \hat{i}$ and $\vec{b} = \hat{j}$. Then $\vec{a} 	imes \vec{b} = \hat{i} 	imes \hat{j} = \hat{k}$.Let $\vec{v} = \vec{a} + \vec{b} + (\vec{a} 	imes \vec{b}) = \hat{i} + \hat{j} + \hat{k}$.The angle $	heta$ between $\vec{v}$ and $\vec{a}$ is given by $\cos 	heta = \frac{\vec{v} \cdot \vec{a}}{|\vec{v}| |\vec{a}|}$.$\vec{v} \cdot \vec{a} = (\hat{i} + \hat{j} + \hat{k}) \cdot \hat{i} = 1$.$|\vec{v}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$ and $|\vec{a}| = 1$.So,$\cos 	heta = \frac{1}{\sqrt{3} 	imes 1} = \frac{1}{\sqrt{3}}$.Therefore,$	heta = \cos^{-1}\left(\frac{1}{\sqrt{3}}ight)$.

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

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