If bar{V} = 2bar{i} + bar{j} - bar{k} and bar | vedclass

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(C) Given $\bar{V} = (2, 1, -1)$ and $\bar{W} = (1, 0, 3)$.The scalar triple product is defined as $[\bar{U} \bar{V} \bar{W}] = \bar{U} \cdot (\bar{V}

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$\sqrt{59}$

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(C) Given $\bar{V} = (2, 1, -1)$ and $\bar{W} = (1, 0, 3)$.The scalar triple product is defined as $[\bar{U} \bar{V} \bar{W}] = \bar{U} \cdot (\bar{V} 	imes \bar{W})$.First,calculate the cross product $\bar{V} 	imes \bar{W}$:$\bar{V} 	imes \bar{W} = \begin{vmatrix} \bar{i} & \bar{j} & \bar{k} \ 2 & 1 & -1 \ 1 & 0 & 3 \end{vmatrix} = \bar{i}(3 - 0) - \bar{j}(6 - (-1)) + \bar{k}(0 - 1) = 3\bar{i} - 7\bar{j} - \bar{k}$.The magnitude of the cross product is $|\bar{V} 	imes \bar{W}| = \sqrt{3^2 + (-7)^2 + (-1)^2} = \sqrt{9 + 49 + 1} = \sqrt{59}$.Since $\bar{U}$ is a unit vector,$|\bar{U}| = 1$. The scalar triple product is:$[\bar{U} \bar{V} \bar{W}] = \bar{U} \cdot (\bar{V} 	imes \bar{W}) = |\bar{U}| |\bar{V} 	imes \bar{W}| \cos 	heta = (1)(\sqrt{59}) \cos 	heta$,where $	heta$ is the angle between $\bar{U}$ and $(\bar{V} 	imes \bar{W})$.The maximum value occurs when $\cos 	heta = 1$,which gives the maximum value as $\sqrt{59}$.

If $\bar{V} = 2\bar{i} + \bar{j} - \bar{k}$ and $\bar{W} = \bar{i} + 3\bar{k}$,and if $\bar{U}$ is a unit vector,then the maximum value of $[\bar{U} \bar{V} \bar{W}]$ is ...

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