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

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(A) Given vectors are $a = \hat{i} - 2\hat{j} + 3\hat{k}$ and $b = 2\hat{i} + \hat{j} + \hat{k}$.The scalar triple product $[a \ b \ c]$ is defined as

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$\frac{-\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$

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(A) Given vectors are $a = \hat{i} - 2\hat{j} + 3\hat{k}$ and $b = 2\hat{i} + \hat{j} + \hat{k}$.The scalar triple product $[a \ b \ c]$ is defined as $(a 	imes b) \cdot c$.First, calculate the cross product $a 	imes b$:$a 	imes b = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -2 & 3 \ 2 & 1 & 1 \end{vmatrix} = \hat{i}(-2 - 3) - \hat{j}(1 - 6) + \hat{k}(1 + 4) = -5\hat{i} + 5\hat{j} + 5\hat{k}$.The magnitude is $|a 	imes b| = \sqrt{(-5)^2 + 5^2 + 5^2} = \sqrt{25 + 25 + 25} = \sqrt{75} = 5\sqrt{3}$.Since $[a \ b \ c] = (a 	imes b) \cdot c = |a 	imes b| |c| \cos 	heta$, where $	heta$ is the angle between $(a 	imes b)$ and $c$.For the scalar triple product to be maximum, $\cos 	heta$ must be $1$ (i.e.,$	heta = 0^\circ$), which means $c$ must be in the direction of $(a 	imes b)$.Since $c$ is a unit vector, $c = \frac{a 	imes b}{|a 	imes b|} = \frac{-5\hat{i} + 5\hat{j} + 5\hat{k}}{5\sqrt{3}} = \frac{-\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$.

Let $a = \hat{i} - 2\hat{j} + 3\hat{k}$ and $b = 2\hat{i} + \hat{j} + \hat{k}$. If $c$ is a unit vector such that $[a \ b \ c]$ is maximum, then $c =$

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