If vec{a}, vec{b}, vec{c} are unit vectors su | vedclass

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(C) Given that $\vec{a}, \vec{b}, \vec{c}$ are unit vectors,so $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$. Since $\vec{a} \cdot \vec{b} = 0$ and $\vec{a}

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$\pm \frac{2}{\sqrt{3}}(\vec{b} 	imes \vec{c})$

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(C) Given that $\vec{a}, \vec{b}, \vec{c}$ are unit vectors,so $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$. Since $\vec{a} \cdot \vec{b} = 0$ and $\vec{a} \cdot \vec{c} = 0$,$\vec{a}$ is perpendicular to both $\vec{b}$ and $\vec{c}$. Therefore,$\vec{a}$ must be parallel to the cross product $\vec{b} 	imes \vec{c}$. Let $\vec{a} = k(\vec{b} 	imes \vec{c})$ for some scalar $k$. Taking the magnitude on both sides,$|\vec{a}| = |k| |\vec{b} 	imes \vec{c}|$. Since $|\vec{a}| = 1$,we have $1 = |k| |\vec{b}| |\vec{c}| \sin(\pi / 3)$. Substituting the values,$1 = |k| (1)(1)(\sqrt{3} / 2)$,which gives $|k| = 2 / \sqrt{3}$. Thus,$\vec{a} = \pm \frac{2}{\sqrt{3}}(\vec{b} 	imes \vec{c})$.

If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} = 0$ and the angle between $\vec{b}$ and $\vec{c}$ is $\pi / 3$,then $\vec{a}$ is equal to

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