Find the angle between the given vectors to the nearest tenth of a degree.

Question

Find the angle between the given vectors to the nearest tenth of a degree.
a = 7i + 2j,   b = -6i + 6j

Answer 1

Let θ be the angle between vectors a and b.

 

Then, cosθ = (a·b) / [(llall)(llbll)]

 

   a·b = (7)(-6) + (2)(6) = -30

 

   llall = √[(7)²+(2)²] = √53

 

   Similarly, llbll = √72 = 6√2

 

Substituting into the formula, cosθ = -5/√106

 

                                                   ≈119.7°