Let vec{v} = 2hat{i} + 2hat{j} - hat{k} and v | vedclass

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(D) The scalar triple product is defined as $[\vec{u} \vec{v} \vec{w}] = \vec{u} \cdot (\vec{v} 	imes \vec{w})$.First,calculate the cross product $\v

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$\sqrt{89}$

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(D) The scalar triple product is defined as $[\vec{u} \vec{v} \vec{w}] = \vec{u} \cdot (\vec{v} 	imes \vec{w})$.First,calculate the cross product $\vec{v} 	imes \vec{w}$:$\vec{v} 	imes \vec{w} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 2 & -1 \ 1 & 0 & 3 \end{vmatrix} = \hat{i}(6 - 0) - \hat{j}(6 - (-1)) + \hat{k}(0 - 2) = 6\hat{i} - 7\hat{j} - 2\hat{k}$.The magnitude of this vector is $|\vec{v} 	imes \vec{w}| = \sqrt{6^2 + (-7)^2 + (-2)^2} = \sqrt{36 + 49 + 4} = \sqrt{89}$.Since $\vec{u}$ is a unit vector $(|\vec{u}| = 1)$,the scalar triple product is $\vec{u} \cdot (\vec{v} 	imes \vec{w}) = |\vec{u}| |\vec{v} 	imes \vec{w}| \cos 	heta$,where $	heta$ is the angle between $\vec{u}$ and $(\vec{v} 	imes \vec{w})$.The maximum value occurs when $\cos 	heta = 1$,which gives the maximum value as $|\vec{v} 	imes \vec{w}| = \sqrt{89}$.

Let $\vec{v} = 2\hat{i} + 2\hat{j} - \hat{k}$ and $\vec{w} = \hat{i} + 3\hat{k}$. If $\vec{u}$ is a unit vector,then the maximum value of the scalar triple product $[\vec{u} \vec{v} \vec{w}]$ is

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