Question

If vector P,Q,R have magnitude 5,12,13 units and P+Q=R the angle between Q and R is

Answer 1

R = P + Q
13 = sqrt(5^2 + 12^2 + (2 × 5 × 12 × Cosx))
169 = 169 + 120Cosx
0 = 120Cosx
Cosx = 0
Cosx = Cos90
x = 90°

Angle between P and Q is 90°

Tan(y) = P Sinx / (Q + PCosx)
= 5 / (12 + 0)
= 5/12
y = Tan^-1 (5/12)

Angle between Q and R is Tan^-1 (5/12)

Answer 2

A̲ = A cos(θ) i̲ + A sin(θ) j̲  

Now just calculate the dot-product (two perpendicular vectors have as dot product zero):  

(A cos(θ) i̲ + A sin(θ) j̲) • (B sin(θ) i̲ − B cos(θ) j̲)  

= AB cos(θ) sin(θ) − AB sin(θ) cos(θ)  

= 0

Hence option is C