Question

How do you find the parametric equations for the line through the point P = (2, -2, -1) that is perpendicular to the plane 1x + 3y - 2z = 1?

Answer 1

`vec r = ((2), (-2), (-1)) + lambda ((1),(3),(-2))`

Explanation:

for the plane

`pi: qquad x + 3y - 2z = 1`

the normal vector is `((1),(3),(-2))`

so the line that runs through the point `((2), (-2), (-1))` and normal to plane `pi` is

`vec r = ((2), (-2), (-1)) + lambda ((1),(3),(-2))`