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

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(A) Given vectors are $\vec{a}=2 \hat{i}+\hat{j}-3 \hat{k}$,$\vec{b}=\hat{i}-2 \hat{j}+\hat{k}$,$\vec{c}=-\hat{i}+\hat{j}-4 \hat{k}$,and $\vec{d}=\hat

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$5 \sqrt{114}$

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(A) Given vectors are $\vec{a}=2 \hat{i}+\hat{j}-3 \hat{k}$,$\vec{b}=\hat{i}-2 \hat{j}+\hat{k}$,$\vec{c}=-\hat{i}+\hat{j}-4 \hat{k}$,and $\vec{d}=\hat{i}+\hat{j}+\hat{k}$.First,calculate the cross product $\vec{a} 	imes \vec{b}$:$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -3 \ 1 & -2 & 1 \end{vmatrix} = \hat{i}(1 - 6) - \hat{j}(2 + 3) + \hat{k}(-4 - 1) = -5 \hat{i} - 5 \hat{j} - 5 \hat{k}$.Next,calculate the cross product $\vec{c} 	imes \vec{d}$:$\vec{c} 	imes \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & 1 & -4 \ 1 & 1 & 1 \end{vmatrix} = \hat{i}(1 + 4) - \hat{j}(-1 + 4) + \hat{k}(-1 - 1) = 5 \hat{i} - 3 \hat{j} - 2 \hat{k}$.Now,calculate the cross product of the two resulting vectors:$(\vec{a} 	imes \vec{b}) 	imes(\vec{c} 	imes \vec{d}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -5 & -5 & -5 \ 5 & -3 & -2 \end{vmatrix} = \hat{i}(10 - 15) - \hat{j}(10 + 25) + \hat{k}(15 + 25) = -5 \hat{i} - 35 \hat{j} + 40 \hat{k}$.Finally,calculate the magnitude:$|(\vec{a} 	imes \vec{b}) 	imes(\vec{c} 	imes \vec{d})| = \sqrt{(-5)^2 + (-35)^2 + (40)^2} = \sqrt{25 + 1225 + 1600} = \sqrt{2850} = 5 \sqrt{114}$.

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

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