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

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(C) First,calculate $\vec{P} = \vec{A} + \vec{B} + \vec{C} = (1-1+2)\hat{i} + (1+1-2)\hat{j} + (3+4-8)\hat{k} = 2\hat{i} + 0\hat{j} - 1\hat{k}$.Next,c

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$90$

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(C) First,calculate $\vec{P} = \vec{A} + \vec{B} + \vec{C} = (1-1+2)\hat{i} + (1+1-2)\hat{j} + (3+4-8)\hat{k} = 2\hat{i} + 0\hat{j} - 1\hat{k}$.Next,calculate $\vec{Q} = \vec{A} 	imes \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 3 \ -1 & 1 & 4 \end{vmatrix} = \hat{i}(4-3) - \hat{j}(4+3) + \hat{k}(1+1) = 1\hat{i} - 7\hat{j} + 2\hat{k}$.Now,find the dot product $\vec{P} \cdot \vec{Q} = (2)(1) + (0)(-7) + (-1)(2) = 2 + 0 - 2 = 0$.Since the dot product is $0$,the vectors $\vec{P}$ and $\vec{Q}$ are perpendicular to each other.Therefore,the angle between them is $90^{\circ}$.

If $\vec{A}=\hat{i}+\hat{j}+3 \hat{k}$,$\vec{B}=-\hat{i}+\hat{j}+4 \hat{k}$ and $\vec{C}=2 \hat{i}-2 \hat{j}-8 \hat{k}$,then the angle between the vectors $\vec{P}=\vec{A}+\vec{B}+\vec{C}$ and $\vec{Q}=(\vec{A} \times \vec{B})$ is (in degree) (in $^{\circ}$)

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