Let $\vec{v}=2 \hat{i}+\hat{j}-\hat{k}, \vec{w}=\hat{i}+3 \hat{k}$ and $\vec{u}$ is a unit vector th

Question

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

Professor PC Thomas Classes · Answer

(c) $: \vec{v} 	imes \vec{w}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -1 \ 1 & 0 & 3\end{array}ight|=3 \hat{i}-7 \hat{j}-\hat{k}$ Now, $[\vec{u} \vec{v} \vec{w}]=\vec{u} \cdot(\vec{v} 	imes \vec{w})$\ $=|\vec{u}||\vec{v} 	imes \vec{w}| \cos (\vec{u}, \vec{v} 	imes \vec{w})$\ is maximum if $(\vec{u}, \vec{v} 	imes \vec{w})=0$\ $	herefore \quad \operatorname{Max} .[\vec{u} \vec{v} \vec{w}]$\ $=1 	imes|3 \hat{i}-7 \hat{j}-\hat{k}| \cos 0=\sqrt{59}$.\