Let vec{b} = -hat{i} + 4hat{j} + 6hat{k} and | vedclass

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(C) The scalar triple product is defined as $[\vec{a} \ \vec{b} \ \vec{c}] = \vec{a} \cdot (\vec{b} 	imes \vec{c})$.For this value to be maximum,$\ve

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$\frac{1}{3} (2\hat{i} + 2\hat{j} - \hat{k})$

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(C) The scalar triple product is defined as $[\vec{a} \ \vec{b} \ \vec{c}] = \vec{a} \cdot (\vec{b} 	imes \vec{c})$.For this value to be maximum,$\vec{a}$ must be in the direction of the vector $\vec{b} 	imes \vec{c}$.First,we calculate the cross product $\vec{v} = \vec{b} 	imes \vec{c}$:$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & 4 & 6 \ 2 & -7 & -10 \end{vmatrix} = \hat{i}(-40 - (-42)) - \hat{j}(10 - 12) + \hat{k}(7 - 8) = 2\hat{i} + 2\hat{j} - \hat{k}$.The magnitude of this vector is $|\vec{v}| = \sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.Since $\vec{a}$ is a unit vector in the direction of $\vec{v}$,we have $\vec{a} = \frac{\vec{v}}{|\vec{v}|} = \frac{2\hat{i} + 2\hat{j} - \hat{k}}{3}$.

Let $\vec{b} = -\hat{i} + 4\hat{j} + 6\hat{k}$ and $\vec{c} = 2\hat{i} - 7\hat{j} - 10\hat{k}$. If $\vec{a}$ is a unit vector and the scalar triple product $[\vec{a} \ \vec{b} \ \vec{c}]$ has the greatest value,then $\vec{a}$ is equal to:

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