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Let $\mathbf{v}=2 \mathbf{i}+\mathbf{j}-\mathbf{k}$ and $\mathbf{w}=\mathbf{i}+3 \mathbf{k}$. If $\mathbf{u}$ is any unit vector, then the m$\sqrt{59}$
aximum value of the scalar triple product $[\mathbf{u} \mathbf{v} \mathbf{w}]$ is
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aximum value of the scalar triple product $[\mathbf{u} \mathbf{v} \mathbf{w}]$ is
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Verified Answer
The correct answer is:
$\sqrt{59}$
$\begin{aligned} & \text { Since, }[\mathbf{u} \mathbf{v} \mathbf{w}]=\mathbf{u} \cdot(\mathbf{v} \times \mathbf{w}) \\ & \Rightarrow[\mathbf{u} \mathbf{v} \mathbf{w}] \leq|\mathbf{u} \| \mathbf{v} \times \mathbf{w}| \\ & \Rightarrow[\mathbf{u} \mathbf{v} \mathbf{w}] \leq|\mathbf{v} \times \mathbf{w}| \\ & \text { Now, } \mathbf{v} \times \mathbf{w}=\left|\begin{array}{rrr}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & -1 \\ 1 & 0 & 3\end{array}\right|=3 \mathbf{i}-7 \mathbf{j}-\mathbf{k} \\ & \therefore \quad \mathbf{v} \times \mathbf{w}=\sqrt{9+49+1}=\sqrt{59}\end{aligned}$
$\therefore \quad[\mathbf{u} \mathbf{v} \mathbf{w}] \leq \sqrt{59}$
$\therefore \quad[\mathbf{u} \mathbf{v} \mathbf{w}] \leq \sqrt{59}$
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