Question

Find all the unit vectors in the plane determined by the vectors u = (-1,-2,1) and v = (2,3,0), that are perpendicular to the vector w = (-1,0,3).

Answer 1

I don't know if what I did is correct, but this is what I have done:

The plane determined by the vectors u and v is all the vectors in the form of au + bv
So: au = a(-1,-2,1) = (-a,-2a,a)
bv = b(2,3,0) = (2b,3b,0)

So au + bv = (-a + 2b, -2a + 3b, a)

Now, (au+bv).w = 0
(-a + 2b, -2a + 3b, a) . (-1,0,3) = 0
(-1)(-a+2b) + (0)(-2a+3b) + 3(a) = 0
a - 2b +3a = 0
4a = 2b
2a = b

Sub b = 2a into (-a + 2b, -2a + 3b, a):
(-a + 2b, -2a + 3b, a)
=(-a + 2(2a), -2a + 3(2a), a)
= (3a, 4a, a)

So: a(3,4,1) is the sum of all the vectors that determine the plane

Therefore, find the norm

+/- sqrt(3^2 + 4^2 + 1^2)
+/- sqrt(9+16+1)
+/- sqrt(26)

So, the unit vectors are:

(3,4,1)/sqrt(26)
= (3/sqrt(26), 4/sqrt(26), 1/sqrt(26))

And

- (3,4,1)/sqrt(26)
= (-3/sqrt(26), -4/sqrt(26), -1/sqrt(26))

Answer 2

It's +/- sqrt(26) because |a|*norm = 1 to be a unit vector
So +/- 1/norm is a