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If $\vec{\alpha}$ is a unit vector, $\vec{\beta}=\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}, \vec{\gamma}=\hat{\mathrm{i}}+\hat{\mathrm{k}}$, then the maximum value of $[\vec{\alpha} \vec{\beta} \vec{\gamma}]$ is
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The correct answer is:
$\sqrt{6}$
$[\vec{\alpha} \vec{\beta} \vec{\gamma}]=\vec{\alpha} \cdot(\vec{\beta} \times \vec{\gamma})=\vec{\alpha} \cdot\left|\begin{array}{ccc}\hat{i} & \hat{j} & k \\ 1 & 1 & -1 \\ 1 & 0 & 1\end{array}\right|$
$=\vec{\alpha} \cdot(\hat{i}+2 \hat{j}-\hat{k})$ is maximum $\Rightarrow$ angle between $\vec{\alpha} \& \hat{i}+2 \hat{j}-\hat{k}$ will be 0
$\therefore[\vec{\alpha} \vec{\beta} \vec{\gamma}]=|\vec{\alpha}||\hat{i}+2 \hat{j}-\hat{k}|=\sqrt{6}$
$=\vec{\alpha} \cdot(\hat{i}+2 \hat{j}-\hat{k})$ is maximum $\Rightarrow$ angle between $\vec{\alpha} \& \hat{i}+2 \hat{j}-\hat{k}$ will be 0
$\therefore[\vec{\alpha} \vec{\beta} \vec{\gamma}]=|\vec{\alpha}||\hat{i}+2 \hat{j}-\hat{k}|=\sqrt{6}$
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