Question

How do you find the vector `C` that is perpendicular to `A-> -3x+9y-z=0` and which vector `C` has a magnitude of `1`?

Answer 1

`{{-3,9,-1}}/sqrt(91)`

Explanation:

One of the usual plane representations is

`Pi-> << p - p_0, vec n >> = 0`

where

`p = {x,y,z}` a generic point pertaining to `Pi`
`p_0 = {x_0,y_0,z_0}` a given point in `Pi`
`vec n = {a.b.c}` a normal vector to `Pi`

So

`Pi -> (x-x_0)a+(y-y_0)b+(z-z_0)c = 0`

or

`a x + b y + c z = a x_0+b y_0 + c z_0`

After that we can recognize

`vec n = {-3,9,-1}`
`p_0 = {0,0,0}`

Now, normalizing `vec n` we have

`hat n = (vec n)/norm vec n = {{-3,9,-1}}/sqrt(3^2+9^2+1) = {{-3,9,-1}}/sqrt(91)`