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.