Let vec{a}, vec{b}, vec{c} be unit vectors. S | vedclass

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

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

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(D) Since $\vec{a} \cdot \vec{b} = 0$ and $\vec{a} \cdot \vec{c} = 0$, the vector $\vec{a}$ 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 $\vec{a} = \lambda(\vec{b} 	imes \vec{c})$ for some scalar $\lambda$.Since $\vec{a}, \vec{b}, \vec{c}$ are unit vectors, $|\vec{a}| = 1$.Taking the magnitude on both sides: $|\vec{a}| = |\lambda| |\vec{b} 	imes \vec{c}|$.$|\vec{b} 	imes \vec{c}| = |\vec{b}| |\vec{c}| \sin(\frac{\pi}{6}) = (1)(1)(\frac{1}{2}) = \frac{1}{2}$.Thus, $1 = |\lambda| \cdot \frac{1}{2}$, which implies $|\lambda| = 2$, so $\lambda = \pm 2$.Therefore, $\vec{a} = \pm 2(\vec{b} 	imes \vec{c})$.

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

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