# How to find an equation for the plane that is perpendicular to a line and passes through a point?

I'm struggling with a problem on my Calculus III homework assignment. The question asks:

Find an equation for the plane that is perpendicular to the line l(t) = (8, 0, 3)t + (8, −1, 1) and passes through (6, −1, 0).

I only know how to find an equation for a plane that is perpendicular to a vector and passes through a point, but I don't understand how to find a vector through a line. Can you please explain to me how to find or choose a vector that passes through a line and a point?

## 2 Answers

HINT.

The perpendicular vector in this case is just $(8,0,3)$, that is the direction vector of the line.

Very simple: if the coordinates of a directing vector of the line are $(\alpha,\beta,\gamma)$ and if the plane passes through the point $(x_0,y_0,z_0)$, an equation is $$\alpha x+\beta y+\gamma z=\alpha x_0+\beta y_0+\gamma z_0.$$

Thus here, you obtain $$\color{red}{x+y}=6-1\color{red}{=5}$$