Find a unit vector perpendicular to a = (-3, 6, 2)

thanks in advance!

A vector, v, perpendicular to a will have the following property:

v · a = 0

If v has coordinates [x, y, z] then:

[x, y, z] · [-3, 6, 2] = 0
-3x + 6y + 2z = 0

Since we only need an arbitrary vector, let x = 0. Then:

6y + 2z = 0
6y = -2z
-3y = z

Then let z = 1, so y = -1/3. Thus our vector is:

v = [0, -1/3, 1]

But we require that this be a unit vector, so divide it by its magnitude:

|v| = sqrt(0^2 + (-1/3)^2 + 1^2) = sqrt(10/9) = (1/3)sqrt(10)

v unit = v / |v| = [0, -1/3, 1] = 3/sqrt(10) [0, -1/3, 1]

Done!