The unit vectors perpendicular to the plane d | vedclass

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(C) Let the points be $A(1, -1, 2)$,$B(2, 0, -1)$,and $C(0, 2, 1)$.First,find two vectors in the plane: $\vec{AB} = (2-1)\hat{i} + (0-(-1))\hat{j} + (

Vedclass · Accepted Answer

$\pm\left(\frac{2 \hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}\right)$

Vedclass · Answer

(C) Let the points be $A(1, -1, 2)$,$B(2, 0, -1)$,and $C(0, 2, 1)$.First,find two vectors in the plane: $\vec{AB} = (2-1)\hat{i} + (0-(-1))\hat{j} + (-1-2)\hat{k} = \hat{i} + \hat{j} - 3\hat{k}$ and $\vec{AC} = (0-1)\hat{i} + (2-(-1))\hat{j} + (1-2)\hat{k} = -\hat{i} + 3\hat{j} - \hat{k}$.$A$ vector perpendicular to the plane is given by the cross product $\vec{n} = \vec{AB} 	imes \vec{AC}$.$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & -3 \ -1 & 3 & -1 \end{vmatrix} = \hat{i}(-1 - (-9)) - \hat{j}(-1 - 3) + \hat{k}(3 - (-1)) = 8\hat{i} + 4\hat{j} + 4\hat{k}$.Simplifying,we can take $\vec{n}' = 2\hat{i} + \hat{j} + \hat{k}$.The magnitude is $|\vec{n}'| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$.The unit vectors are $\pm \frac{\vec{n}'}{|\vec{n}'|} = \pm \frac{2\hat{i} + \hat{j} + \hat{k}}{\sqrt{6}}$.Thus,the correct option is $C$.

The unit vectors perpendicular to the plane determined by the points $A(1, -1, 2)$,$B(2, 0, -1)$,and $C(0, 2, 1)$ are:

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