Question

How do you find the parametric equations for the line through the point P = (3, 5, -2) that is perpendicular to the plane −5x + 1y + 3z = 1?

Answer 1

`vec l(t) = ((3),( 5), (-2)) + t ((-5),(1),(3))`

Explanation:

a plane in vector form can be written as

`(vec r - vec r_o)* vec n = 0`; or, as it is here, as `vec r * vec n = vec r_o* vec n = const`

where, in Cartesian, `vec r = ((x),(y),(x))` and normal vector `vec n = ((a),(b),(c))`

so, here, the plane's normal vector is `vec n = ((-5),(1),(3))`

that is to say, the plane is: `((x),(y),(x)) * ((-5),(1),(3)) = 1`

and the line through P , in terms of parameter t, can be written as

`vec l(t) = ((3),( 5), (-2)) + t ((-5),(1),(3))`