Let bar{a},bar{b},and bar{c} be unit vectors. | vedclass

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(B) Since $\bar{a} \cdot \bar{b} = 0$ and $\bar{a} \cdot \bar{c} = 0$,the vector $\bar{a}$ is perpendicular to both $\bar{b}$ and $\bar{c}$.Therefore,

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$\pm 2(\bar{b} 	imes \bar{c})$

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(B) Since $\bar{a} \cdot \bar{b} = 0$ and $\bar{a} \cdot \bar{c} = 0$,the vector $\bar{a}$ is perpendicular to both $\bar{b}$ and $\bar{c}$.Therefore,$\bar{a}$ must be parallel to the cross product $\bar{b} 	imes \bar{c}$.Let $\bar{a} = k(\bar{b} 	imes \bar{c})$ for some scalar $k$.Since $\bar{a}$ is a unit vector,$|\bar{a}| = 1$,so $|k| |\bar{b} 	imes \bar{c}| = 1$.The magnitude $|\bar{b} 	imes \bar{c}| = |\bar{b}| |\bar{c}| \sin(\frac{\pi}{6}) = (1)(1)(\frac{1}{2}) = \frac{1}{2}$.Substituting this,$|k| \cdot \frac{1}{2} = 1$,which implies $|k| = 2$,so $k = \pm 2$.Thus,$\bar{a} = \pm 2(\bar{b} 	imes \bar{c})$.

Let $\bar{a}$,$\bar{b}$,and $\bar{c}$ be unit vectors. Suppose that $\bar{a} \cdot \bar{b} = \bar{a} \cdot \bar{c} = 0$ and the angle between $\bar{b}$ and $\bar{c}$ is $\frac{\pi}{6}$. Then $\bar{a}$ is equal to:

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