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Scalar triple product of vectors

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Question

If the volume of parallelopiped formed by the vectors $\hat{i} + λ \hat{j} + \hat{k}, \hat{j} + λ \hat{k} and λ \hat{i} + \hat{k}$ is minimum, then 2021 $\sqrt{3}$ $λ$ is

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Solution

Given vectors are $\hat{i} + λ \hat{j} + \hat{k}, \hat{j} + λ \hat{k} and λ \hat{i} + \hat{k}$ which forms a parallelopiped.
Volume of the parallelopiped is
$\begin{array}{l} V = ||\begin{array}{lll} 1 & λ & 1 \\ 0 & 1 & λ \\ λ & 0 & 1 \end{array}|| = 1 + λ^{3} - λ \\ \Rightarrow V = λ^{3} - λ + 1 \end{array}$
On differentiating w.r.t. $λ$ , we get
$\frac{dV}{dλ} = 3 λ^{2} - 1$
For maxima or minima, $\frac{dV}{dλ} = 0 \Rightarrow λ = \pm \frac{1}{\sqrt{3}}$
and $\frac{d^{2} V}{{dλ}^{2}} = 6 λ = \{\begin{cases} 2 \sqrt{3} > 0, for λ = \frac{1}{\sqrt{3}} \\ - 2 \sqrt{3} < 0, for λ = - \frac{1}{\sqrt{3}} \end{cases}$
$∵ \frac{d^{2} V}{{dλ}^{2}}$ positive for $λ = \frac{1}{\sqrt{3}}$ , hence volume V is minimum for $λ = \frac{1}{\sqrt{3}}$

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