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

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(C) 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 produ

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$\sqrt{59}$

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(C) 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 $\vec{V} 	imes \vec{W}$:$\vec{V} 	imes \vec{W} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -1 \ 1 & 0 & 3 \end{vmatrix} = \hat{i}(3 - 0) - \hat{j}(6 - (-1)) + \hat{k}(0 - 1) = 3\hat{i} - 7\hat{j} - \hat{k}$.Let $\vec{A} = \vec{V} 	imes \vec{W} = 3\hat{i} - 7\hat{j} - \hat{k}$.Then $[\vec{U} \, \vec{V} \, \vec{W}] = \vec{U} \cdot \vec{A} = |\vec{U}| |\vec{A}| \cos 	heta$,where $	heta$ is the angle between $\vec{U}$ and $\vec{A}$.Since $\vec{U}$ is a unit vector,$|\vec{U}| = 1$.Thus,the scalar triple product is $|\vec{A}| \cos 	heta$.The maximum value occurs when $\cos 	heta = 1$,which is $|\vec{A}|$.$|\vec{A}| = \sqrt{3^2 + (-7)^2 + (-1)^2} = \sqrt{9 + 49 + 1} = \sqrt{59}$.Therefore,the maximum value is $\sqrt{59}$.

Let $\vec{V} = 2\hat{i} + \hat{j} - \hat{k}$ 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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