If vec{a}=hat{i}-hat{j}+hat{k},vec{b}=hat{i}+ | vedclass

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(D) First,calculate the cross product $\vec{a} 	imes \vec{c}$:$\vec{a} 	imes \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 1 \

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$-8 \hat{i}+\hat{j}-29 \hat{k}$

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(D) First,calculate the cross product $\vec{a} 	imes \vec{c}$:$\vec{a} 	imes \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 1 \ 2 & -3 & -1 \end{vmatrix} = \hat{i}(1 - (-3)) - \hat{j}(-1 - 2) + \hat{k}(-3 - (-2)) = 4 \hat{i} + 3 \hat{j} - \hat{k}$.Next,calculate the cross product $\vec{b} 	imes \vec{d}$:$\vec{b} 	imes \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & -2 \ 2 & 1 & 1 \end{vmatrix} = \hat{i}(1 - (-2)) - \hat{j}(1 - (-4)) + \hat{k}(1 - 2) = 3 \hat{i} - 5 \hat{j} - \hat{k}$.Finally,calculate the cross product of the two resulting vectors:$(\vec{a} 	imes \vec{c}) 	imes(\vec{b} 	imes \vec{d}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 4 & 3 & -1 \ 3 & -5 & -1 \end{vmatrix} = \hat{i}(-3 - 5) - \hat{j}(-4 - (-3)) + \hat{k}(-20 - 9) = -8 \hat{i} + \hat{j} - 29 \hat{k}$.

If $\vec{a}=\hat{i}-\hat{j}+\hat{k}$,$\vec{b}=\hat{i}+\hat{j}-2 \hat{k}$,$\vec{c}=2 \hat{i}-3 \hat{j}-\hat{k}$,and $\vec{d}=2 \hat{i}+\hat{j}+\hat{k}$ are four vectors,then find the value of $(\vec{a} \times \vec{c}) \times(\vec{b} \times \vec{d})$.

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