If the points P(1, -1, 2),Q(2, 0, -1),and R(0 | vedclass

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(B) To find the unit vector perpendicular to the plane containing points $P(1, -1, 2)$,$Q(2, 0, -1)$,and $R(0, 2, 1)$,we first find two vectors lying

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$\frac{2\hat{i} + \hat{j} + \hat{k}}{\sqrt{6}}$

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(B) To find the unit vector perpendicular to the plane containing points $P(1, -1, 2)$,$Q(2, 0, -1)$,and $R(0, 2, 1)$,we first find two vectors lying in the plane: $\vec{PQ} = (2-1)\hat{i} + (0-(-1))\hat{j} + (-1-2)\hat{k} = \hat{i} + \hat{j} - 3\hat{k}$ $\vec{PR} = (0-1)\hat{i} + (2-(-1))\hat{j} + (1-2)\hat{k} = -\hat{i} + 3\hat{j} - \hat{k}$ Now,find the cross product $\vec{n} = \vec{PQ} 	imes \vec{PR}$: $\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & -3 \ -1 & 3 & -1 \end{vmatrix} = \hat{i}(-1 - (-9)) - \hat{j}(-1 - 3) + \hat{k}(3 - (-1)) = 8\hat{i} + 4\hat{j} + 4\hat{k}$ Simplifying the vector by dividing by $4$,we get a vector parallel to the normal: $2\hat{i} + \hat{j} + \hat{k}$. The magnitude is $\sqrt{2^2 + 1^2 + 1^2} = \sqrt{4+1+1} = \sqrt{6}$. Thus,the unit vector is $\frac{2\hat{i} + \hat{j} + \hat{k}}{\sqrt{6}}$.

If the points $P(1, -1, 2)$,$Q(2, 0, -1)$,and $R(0, 2, 1)$ are coplanar,find the unit vector perpendicular to the plane containing these points.

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