vec{b}=hat{i}-hat{j}+2 hat{k} and vec{c}=hat{ | vedclass

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(A) First, calculate the cross product $\vec{v} = \vec{b} 	imes \vec{c}$:$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 2 \ 1 &

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$3$

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(A) First, calculate the cross product $\vec{v} = \vec{b} 	imes \vec{c}$:$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 2 \ 1 & 2 & -1 \end{vmatrix} = \hat{i}(1-4) - \hat{j}(-1-2) + \hat{k}(2+1) = -3\hat{i} + 3\hat{j} + 3\hat{k}$The magnitude is $|\vec{v}| = \sqrt{(-3)^2 + 3^2 + 3^2} = \sqrt{9+9+9} = \sqrt{27} = 3\sqrt{3}$.Let $	heta$ be the angle between $\vec{a}$ and $\vec{v}$. Given $\cos 	heta = \sqrt{\frac{2}{3}}$, we find $\sin 	heta = \sqrt{1 - \cos^2 	heta} = \sqrt{1 - \frac{2}{3}} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}}$.The magnitude of the cross product is $|\vec{a} 	imes \vec{v}| = |\vec{a}| |\vec{v}| \sin 	heta$.Since $|\vec{a}| = 1$, we have $|\vec{a} 	imes \vec{v}| = 1 \cdot (3\sqrt{3}) \cdot \frac{1}{\sqrt{3}} = 3$.

$\vec{b}=\hat{i}-\hat{j}+2 \hat{k}$ and $\vec{c}=\hat{i}+2 \hat{j}-\hat{k}$ are two vectors and $\vec{a}$ is a unit vector such that $\cos (\vec{a}, \vec{b} \times \vec{c})=\sqrt{\frac{2}{3}}$. Then $|\vec{a} \times(\vec{b} \times \vec{c})|=$

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