Let vec{alpha}, vec{beta}, vec{gamma} be thre | vedclass

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(C) Since $\vec{\alpha} \cdot \vec{\beta} = 0$ and $\vec{\alpha} \cdot \vec{\gamma} = 0$, the vector $\vec{\alpha}$ is perpendicular to both $\vec{\be

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$\pm 2(\vec{\beta} 	imes \vec{\gamma})$

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(C) Since $\vec{\alpha} \cdot \vec{\beta} = 0$ and $\vec{\alpha} \cdot \vec{\gamma} = 0$, the vector $\vec{\alpha}$ is perpendicular to both $\vec{\beta}$ and $\vec{\gamma}$.Thus, $\vec{\alpha}$ must be parallel to the cross product $\vec{\beta} 	imes \vec{\gamma}$.Let $\vec{\alpha} = \lambda(\vec{\beta} 	imes \vec{\gamma})$ for some scalar $\lambda$.Since $\vec{\alpha}, \vec{\beta}, \vec{\gamma}$ are unit vectors, we have $|\vec{\alpha}| = 1, |\vec{\beta}| = 1, |\vec{\gamma}| = 1$.Taking the magnitude on both sides: $|\vec{\alpha}| = |\lambda| |\vec{\beta} 	imes \vec{\gamma}|$.We know that $|\vec{\beta} 	imes \vec{\gamma}| = |\vec{\beta}| |\vec{\gamma}| \sin(30^{\circ}) = 1 \cdot 1 \cdot \frac{1}{2} = \frac{1}{2}$.Substituting the values: $1 = |\lambda| \cdot \frac{1}{2} \Rightarrow |\lambda| = 2 \Rightarrow \lambda = \pm 2$.Therefore, $\vec{\alpha} = \pm 2(\vec{\beta} 	imes \vec{\gamma})$.

Let $\vec{\alpha}, \vec{\beta}, \vec{\gamma}$ be three unit vectors such that $\vec{\alpha} \cdot \vec{\beta} = \vec{\alpha} \cdot \vec{\gamma} = 0$ and the angle between $\vec{\beta}$ and $\vec{\gamma}$ is $30^{\circ}$. Then $\vec{\alpha}$ is

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