If bar{a}=2hat{i}+3hat{j}-hat{k},bar{b}=-hat{ | vedclass

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(A) First,calculate the cross product $\bar{a} 	imes \bar{b}$:$\bar{a} 	imes \bar{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & -1 \

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$-74$

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(A) First,calculate the cross product $\bar{a} 	imes \bar{b}$:$\bar{a} 	imes \bar{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & -1 \ -1 & 2 & -4 \end{vmatrix} = \hat{i}(-12 - (-2)) - \hat{j}(-8 - 1) + \hat{k}(4 - (-3)) = -10\hat{i} + 9\hat{j} + 7\hat{k}$Next,calculate the cross product $\bar{a} 	imes \bar{c}$:$\bar{a} 	imes \bar{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & -1 \ 1 & 1 & 1 \end{vmatrix} = \hat{i}(3 - (-1)) - \hat{j}(2 - (-1)) + \hat{k}(2 - 3) = 4\hat{i} - 3\hat{j} - \hat{k}$Now,compute the dot product of the two resulting vectors:$(\bar{a} 	imes \bar{b}) \cdot (\bar{a} 	imes \bar{c}) = (-10\hat{i} + 9\hat{j} + 7\hat{k}) \cdot (4\hat{i} - 3\hat{j} - \hat{k})$$= (-10)(4) + (9)(-3) + (7)(-1)$$= -40 - 27 - 7 = -74$

If $\bar{a}=2\hat{i}+3\hat{j}-\hat{k}$,$\bar{b}=-\hat{i}+2\hat{j}-4\hat{k}$ and $\bar{c}=\hat{i}+\hat{j}+\hat{k}$,then $(\bar{a} \times \bar{b}) \cdot(\bar{a} \times \bar{c})=$

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