Answer to Question #321487 in Analytic Geometry for Simom

Question

Find a unit vector perpendicular to the plane through P(2, 1, -1), Q(-1, 1, 2) and R (1, -1, 2)

Answer 1

$\overrightarrow{PQ}=(-3,0,3), \overrightarrow{PR}=(-1,-2,3)$

$\overrightarrow{PQ}\times \overrightarrow{PR}=\begin{vmatrix} \vec{i} & \vec{j} &\vec{k} \\ -3 & 0 & 3 \\ -1 & -2 & 3 \\ \end{vmatrix}$

$=\vec{i}\begin{vmatrix} 0 & 3 \\ -2& 3 \end{vmatrix}-\vec{j}\begin{vmatrix} -3 & 3 \\ -1 & 3 \end{vmatrix}+\vec{k}\begin{vmatrix} -3 & 0 \\ -1 & -2 \end{vmatrix}$

$=6\vec{i}+6\vec{j}+6\vec{k}$

$|\overrightarrow{PQ}\times \overrightarrow{PR}|=6\sqrt{3}$

$\vec{u}=(\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}})$