If e is a unit vector perpendicular to the pl | vedclass

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Question

(A) Let the points be $A(2, 1, 1)$, $B(1, -1, 1)$, and $C(-1, 1, -1)$.The normal vector $\vec{n}$ to the plane is given by $\vec{AB} 	imes \vec{AC}$.

Vedclass · Accepted Answer

$\frac{11}{14}(-2 \hat{i}+\hat{j}+3 \hat{k})$

Vedclass · Answer

(A) Let the points be $A(2, 1, 1)$, $B(1, -1, 1)$, and $C(-1, 1, -1)$.The normal vector $\vec{n}$ to the plane is given by $\vec{AB} 	imes \vec{AC}$.$\vec{AB} = (1-2)\hat{i} + (-1-1)\hat{j} + (1-1)\hat{k} = -\hat{i} - 2\hat{j}$.$\vec{AC} = (-1-2)\hat{i} + (1-1)\hat{j} + (-1-1)\hat{k} = -3\hat{i} - 2\hat{k}$.$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & -2 & 0 \ -3 & 0 & -2 \end{vmatrix} = \hat{i}(4-0) - \hat{j}(2-0) + \hat{k}(0-6) = 4\hat{i} - 2\hat{j} - 6\hat{k}$.Dividing by $2$, we get the normal vector $\vec{n}' = 2\hat{i} - \hat{j} - 3\hat{k}$.The unit vector $e$ perpendicular to the plane is $e = \pm \frac{2\hat{i} - \hat{j} - 3\hat{k}}{\sqrt{2^2 + (-1)^2 + (-3)^2}} = \pm \frac{1}{\sqrt{14}}(2\hat{i} - \hat{j} - 3\hat{k})$.Given $a = 2\hat{i} - 3\hat{j} + 6\hat{k}$, the projection vector of $a$ on $e$ is $(a \cdot e)e$.$a \cdot e = \pm \frac{1}{\sqrt{14}}(2(2) + (-3)(-1) + 6(-3)) = \pm \frac{1}{\sqrt{14}}(4 + 3 - 18) = \mp \frac{11}{\sqrt{14}}$.Projection vector $= (a \cdot e)e = \left(\mp \frac{11}{\sqrt{14}}ight) \left(\pm \frac{1}{\sqrt{14}}(2\hat{i} - \hat{j} - 3\hat{k})ight) = -\frac{11}{14}(2\hat{i} - \hat{j} - 3\hat{k}) = \frac{11}{14}(-2\hat{i} + \hat{j} + 3\hat{k})$.

If $e$ is a unit vector perpendicular to the plane determined by the points $2 \hat{i}+\hat{j}+\hat{k}$, $\hat{i}-\hat{j}+\hat{k}$ and $-\hat{i}+\hat{j}-\hat{k}$. If $a=2 \hat{i}-3 \hat{j}+6 \hat{k}$, then the projection vector of $a$ on $e$ is

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