# Find a nonzero vector normal to a plane

1. Sep 22, 2013

### PsychonautQQ

1. The problem statement, all variables and given/known data
Find a nonzero vector normal to the plane -5x -y +z +9 = 0

2. Relevant equations

3. The attempt at a solution
so the direction of the vector would be (a,b,c) = (-5,-1,1)
I'm not exactly sure what the final form should look like..
is r = ri + tv on the right track? where i plug (-5,-1,1) in for v and for ri i can plug in any numbers that make -5x - 6 + z +9 = 0?
If so I could theoretically plug in (1,4,0)
so
r= (1i + 4j + 0k) + t(-5i -1j +1k)
r = (1-5t)i + (4-t)j + (t)k

The internet program says i'm wrong.. halp?

2. Sep 22, 2013

### Staff: Mentor

Your normal looks fine, but there are an infinite number of vectors that are perpendicular to (normal to) a given plane. However, all of the vectors are scalar multiples of one another. I suspect that the answer the program was looking for was <5, 1, -1>. Or maybe <1, .2, -.2>.

Did it mention anything about wanting a unit vector?