The vector a=3u-v is perpendicular to the vector b=u+3v. If |u|=1 and |v|=2 what is the cosine of the angle made by vectors u and v?

We'll recall that the dot product of two orthogonal
vectors is cancelling out. Therefore, the dot product of a and b is
0.

a*b = 0

We'll multiply the
two vectors:

(3u - v)(u + 3v) = 3u^2 + 9u*v - u*v -
3v^2

We'll substitute u^2 = |u|^2 = 1 and v^2 = |v|^2 = 2^2
= 4

(3u - v)(u + 3v) = 3*1 + 8u*v -
3*4

(3u - v)(u + 3v) = 8u*v -
9

Since the dor product is cancelling, then (3u - v)(u +
3v) = 0 => 8u*v - 9 = 0.

8u*v =
9

u*v = 9/8

We'll apply the
formula of dot product:

u*v = |u|*|v|*cos
(u,v)

9/8 = 1*2*cos (u,v)

cos
(u,v) = 9/16

The cosine of the angle between
the vectors u and v is: cos (u,v) = 9/16.