TITLE: Find the Vector Equation of a line perpendicular to the plane.
QUESTION [6 upvotes]: Question: Find the vector equation $r(t)$ for the line through the point $P = (-1, -5, 2)$ that is perpendicular to the plane $1 x  - 5 y + 1 z = 1$.
Use $t$ as your variable, $t = 0$ should correspond to $P$, and the velocity vector of the line should be the same as the standard normal vector of the plane.
This one is really giving me a hard time.  I know that to find the plane perpendicular to the line I can use the vector n between two points on the line and and the plane.  I cannot wrap my mind around how to reverse this process, particularly because the plane is equal to 1 and not zero.
Any help would be greatly appreciated.

REPLY [4 votes]: In general, the normal vector of the plane $a\cdot x+b\cdot y+c\cdot z+d=0$ is $(a,b,c).$
Hence, the normal vector of the plane $1\cdot x-5\cdot y+1\cdot z-1=0$ is $(1,-5,1).$
Hence, the line can be represented by $t$ as
$$\frac{x-(-1)}{1}=\frac{y-(-5)}{-5}=\frac{z-2}{1}=t.$$
Thus, 
$$r(t)=\left(t-1, -5t-5, t+2\right).$$