If vec{A} = hat{i} - 2hat{j} - 3hat{k},vec{B} | vedclass

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

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$5(-\hat{i} + 3\hat{j} + 4\hat{k})$

Vedclass · Answer

(A) First,calculate the cross product $\vec{A} 	imes \vec{B}$:$\vec{A} 	imes \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -2 & -3 \ 2 & 1 & -1 \end{vmatrix}$$= \hat{i}(2 - (-3)) - \hat{j}(-1 - (-6)) + \hat{k}(1 - (-4))$$= \hat{i}(5) - \hat{j}(5) + \hat{k}(5) = 5\hat{i} - 5\hat{j} + 5\hat{k}$Now,calculate $(\vec{A} 	imes \vec{B}) 	imes \vec{C}$:$(\vec{A} 	imes \vec{B}) 	imes \vec{C} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 5 & -5 & 5 \ 1 & 3 & -2 \end{vmatrix}$$= \hat{i}(10 - 15) - \hat{j}(-10 - 5) + \hat{k}(15 - (-5))$$= \hat{i}(-5) - \hat{j}(-15) + \hat{k}(20)$$= -5\hat{i} + 15\hat{j} + 20\hat{k}$$= 5(-\hat{i} + 3\hat{j} + 4\hat{k})$

If $\vec{A} = \hat{i} - 2\hat{j} - 3\hat{k}$,$\vec{B} = 2\hat{i} + \hat{j} - \hat{k}$,and $\vec{C} = \hat{i} + 3\hat{j} - 2\hat{k}$,then $(\vec{A} \times \vec{B}) \times \vec{C} = \dots$

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