A unit vector perpendicular to the plane dete | vedclass

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Question

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

Vedclass · Accepted Answer

$\pm \frac{1}{\sqrt{6}}(2i + j + k)$

Vedclass · Answer

(A) Let the points be $A(1, -1, 2)$,$B(2, 0, -1)$,and $C(0, 2, 1)$.Define two vectors in the plane: $\vec{AB} = (2-1)i + (0-(-1))j + (-1-2)k = i + j - 3k$ and $\vec{AC} = (0-1)i + (2-(-1))j + (1-2)k = -i + 3j - k$.Alternatively,using the provided solution vectors: $\vec{u} = \vec{AB} = i + j - 3k$ and $\vec{v} = \vec{BC} = (0-2)i + (2-0)j + (1-(-1))k = -2i + 2j + 2k$.The normal vector $\vec{n}$ to the plane is given by the cross product $\vec{u} 	imes \vec{v}$:$\vec{n} = \begin{vmatrix} i & j & k \ 1 & 1 & -3 \ -2 & 2 & 2 \end{vmatrix} = i(2 - (-6)) - j(2 - 6) + k(2 - (-2)) = 8i + 4j + 4k$.The unit vector is $\hat{n} = \pm \frac{\vec{n}}{|\vec{n}|} = \pm \frac{8i + 4j + 4k}{\sqrt{8^2 + 4^2 + 4^2}} = \pm \frac{8i + 4j + 4k}{\sqrt{64 + 16 + 16}} = \pm \frac{8i + 4j + 4k}{\sqrt{96}} = \pm \frac{8i + 4j + 4k}{4\sqrt{6}} = \pm \frac{2i + j + k}{\sqrt{6}}$.

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

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