Find the angle between vectors u = [3,0] and v= [-1,1].

Question

I found the following solution:

cos⁡θ=(u⋅v) /‖u‖‖v‖=

(3⋅(-1) +0.1) / (√ (3^2+0^2) ⋅√((-1) ^2+1^2)) =-3/ (√9 √2). 

However, the correct solution is -1/√2. What did I wrong? How do they find this soultion?

Answer 1

B= ‖v‖ = √2, ‖u‖ = √9, ‖u‖‖v‖ = 3√2; you made a mistake here.

A= (3⋅(-1) +0⋅1) = -3; it is multiplication not a decimal pt--just cosmetic.

A/B =  -1/√2