If vec{u} = hat{i} + 2hat{j} - 2hat{k}, vec{v | vedclass

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(C) The scalar triple product is given by $[\vec{u} \vec{v} \vec{w}] = (\vec{u} 	imes \vec{v}) \cdot \vec{w}$.First, calculate the cross product $\ve

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$3\sqrt{5}$

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(C) The scalar triple product is given by $[\vec{u} \vec{v} \vec{w}] = (\vec{u} 	imes \vec{v}) \cdot \vec{w}$.First, calculate the cross product $\vec{u} 	imes \vec{v}$:$\vec{u} 	imes \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & -2 \ 2 & 0 & 1 \end{vmatrix} = \hat{i}(2 - 0) - \hat{j}(1 - (-4)) + \hat{k}(0 - 4) = 2\hat{i} - 5\hat{j} - 4\hat{k}$.Let $\vec{n} = \vec{u} 	imes \vec{v} = 2\hat{i} - 5\hat{j} - 4\hat{k}$.The magnitude $|\vec{n}| = \sqrt{2^2 + (-5)^2 + (-4)^2} = \sqrt{4 + 25 + 16} = \sqrt{45} = 3\sqrt{5}$.Since $[\vec{u} \vec{v} \vec{w}] = \vec{n} \cdot \vec{w} = |\vec{n}| |\vec{w}| \cos 	heta$, and $|\vec{w}| = 1$, the maximum value is $|\vec{n}| = 3\sqrt{5}$ when $\vec{w}$ is in the direction of $\vec{n}$.

If $\vec{u} = \hat{i} + 2\hat{j} - 2\hat{k}$, $\vec{v} = 2\hat{i} + \hat{k}$ and $\vec{w}$ is a unit vector, then the maximum value of the scalar triple product $[\vec{u} \vec{v} \vec{w}]$ is

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