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Q.
A unit vector perpendicular to the plane $ABC$, where $A , B$ and $C$ are respectively the points $(3,-1,2),(1,-1,-3)$ and $(4,-3,1)$, is
Vector Algebra
Solution:
Let $O$ be the origin. Then $\overrightarrow{ OA }=3 \hat{ i }-\hat{ j }+2 \hat{ k }$,
$\overrightarrow{ OB }=\hat{ i }-\hat{ j }-3 \hat{ k }$ and $\overrightarrow{ OC }=4 \hat{ i }-3 \hat{ j }+\hat{ k }$
$\therefore \overrightarrow{ AB }=\overrightarrow{ OB }-\overrightarrow{ OA }=(\hat{ i }-\hat{ j }-3 \hat{ k })-(3 \hat{ i }-\hat{ j }+2 \hat{ k })=-2 \hat{ i }+0 \hat{ j }-5 \hat{ k } $
$\overrightarrow{ AC }=\overrightarrow{ OC }-\overrightarrow{ OA }=(4 \hat{ i }-3 \hat{ j }+\hat{ k })-(3 \hat{ i }-\hat{ j }+2 \hat{ k })=\hat{ i }-2 \hat{ j }-\hat{ k } $
Now, $\overrightarrow{ AB } \times \overrightarrow{ AC }=\begin{vmatrix}
\hat{ i } & \hat{ j } & \hat{ k } \\ -2 & 0 & -5 \\1 & -2 & -1\end{vmatrix}$
$=(0-10) \hat{ i }-(2+5) \hat{ j }+(4-0) \hat{ k }=-10 \hat{ i }-7 \hat{ j }+4 \hat{ k } $
$\Rightarrow |\overrightarrow{A B} \times \overrightarrow{A C}|=\sqrt{(-10)^{2}+(-7)^{2}+(4)^{2}}=\sqrt{100+49+16}=\sqrt{165}$
A unit vector perpendicular to the plane of $\Delta ABC$ is perpendicular to both $\overrightarrow{ AB }$ and $\overrightarrow{ AC }$. Hence, a unit vector perpendicular, to the plane of
$\Delta ABC =\frac{\overrightarrow{ AB } \times \overrightarrow{ AC }}{|\overrightarrow{ AB } \times \overrightarrow{ AC }|}=\frac{-10 \hat{ i }-7 \hat{ j }+4 \hat{ k }}{\sqrt{165}}=-\frac{1}{\sqrt{165}}(10 \hat{ i }+7 \hat{ j }-4 \hat{ k }) $