Let a=2 hat{i}+hat{j}-3 hat{k} and b=hat{i}+3 | vedclass

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(A) We have, $a=2 \hat{i}+\hat{j}-3 \hat{k}$ and $b=\hat{i}+3 \hat{j}+2 \hat{k}$.Since $c$ is perpendicular to the plane of $a$ and $b$, $c$ is parall

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$2 \sqrt{195}$

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(A) We have, $a=2 \hat{i}+\hat{j}-3 \hat{k}$ and $b=\hat{i}+3 \hat{j}+2 \hat{k}$.Since $c$ is perpendicular to the plane of $a$ and $b$, $c$ is parallel to $a 	imes b$.First, calculate $a 	imes b$:$a 	imes b = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -3 \ 1 & 3 & 2 \end{vmatrix} = \hat{i}(2 - (-9)) - \hat{j}(4 - (-3)) + \hat{k}(6 - 1) = 11 \hat{i} - 7 \hat{j} + 5 \hat{k}$.The magnitude $|a 	imes b| = \sqrt{11^2 + (-7)^2 + 5^2} = \sqrt{121 + 49 + 25} = \sqrt{195}$.Since $c$ is parallel to $a 	imes b$ and $|c|=2$, we have $c = \pm 2 \frac{a 	imes b}{|a 	imes b|} = \pm \frac{2}{\sqrt{195}} (11 \hat{i} - 7 \hat{j} + 5 \hat{k})$.The volume of the parallelopiped is given by the scalar triple product $|[a, b, c]| = |(a 	imes b) \cdot c|$.$|[a, b, c]| = |(a 	imes b) \cdot (\pm 2 \frac{a 	imes b}{|a 	imes b|})| = |\pm 2 \frac{|a 	imes b|^2}{|a 	imes b|}| = 2 |a 	imes b|$.Substituting $|a 	imes b| = \sqrt{195}$, the volume is $2 \sqrt{195}$.

Let $a=2 \hat{i}+\hat{j}-3 \hat{k}$ and $b=\hat{i}+3 \hat{j}+2 \hat{k}$. Then the volume of the parallelopiped having coterminous edges as $a, b$ and $c$, where $c$ is the vector perpendicular to the plane of $a, b$ and $|c|=2$ is

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