Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 15/Solution 1

Recall that a line is perpendicular to a plane means that a direction vector of the line is proportional to a normal vector of the plane.

First, we find the normal vector of the plane.

Since ${\vec {AB}}=[0,1,1]-[1,0,1]=[-1,1,0],{\vec {AC}}=[-1,0,1]-[1,0,1]=[-2,0,0]$, the normal vector of the plane of A, B, C points is

${\vec {n}}={\vec {AB}}\times {\vec {AC}}=[0,0,2]$

Denoting a point on the line $L$ by $(x,y,z)$, the direction vector of the line is $[x-3,y-2,z-2]$.

Thus the parametric equation for $L$ is $[x-3,y-2,z-2]=k[0,0,2]$, or

$\color {blue}{\begin{bmatrix}x\\y\\z\end{bmatrix}}=k{\begin{bmatrix}0\\0\\2\end{bmatrix}}+{\begin{bmatrix}3\\2\\2\end{bmatrix}}$.