Let u and v be vectors in a real inner product space V and s

Let u and v be vectors in a real inner product space V and suppose that ||U|| = 2 and ||V|| = 7, and <u, v> = -3.

Find <-u + 4v, 3u + 2v>
Find ||3u - v||

So I assume that to do the above I need to figure out was U and V is, but I am not sure how to do that. Any help? It seems easy, and I know the rules, but not sure what to do.

Solution

Given ||U|| = 2 and ||V|| = 7

Given u.v = -3

a) Find <-u + 4v, 3u+2v)

Taking (-u+4v).(3u+2v)

=> -3||U||^2 -2*u.v + 12*u.v + 8||V||^2

=> -3(2)^2 + 10 * -3 + 8 * (7)^2

=> 350

b) ||3U-V||

=> (3u-v).(3u-v)

=> 9||U||^2 -3*u.v - 3*u.v + |V|^2

=> 9 * (2)^2 -6*(-3) + (7)^2

=> 36 + 18 + 49

=> 103