Question

Two vectors A and B have precisely equal magnitudes. In order for the magnitude of A+ B to be 110 times larger than the magnitude of A-B , what must be the angle between them?,

Answer 1

Due to the magnitudes, as mentioned in the problem, being equal, B has components Acos(θ)x along the x-axis and Asin(θ) along the y-axis.

The solution is as follows:

A+B has:
total x-component = A + Acos(θ) = A(1+cos(θ))
total y-component = 0 + Asin(θ) = Asin(θ)
Resultant = √[( A(1+cos(θ)))² + (Asin(θ))²]
= A√[1 + 2cos(θ) + cos²(θ) + sin²(θ)]
=A√[2 + 2cos(θ)]

A-B has:
total x-component = A - Acos(θ) = A(1-cos(θ))
total y-component = 0 - Asin(θ) = -Asin(θ)
Resultant = √[( A(1-cos(θ)))² + (-Asin(θ))²]
= A√[1 - 2cos(θ) + cos²(θ) + sin²(θ)]
=A√[2 - 2cos(θ)]

A+B/A-B = [A√[2 + 2cos(θ)]] / A√[2 - 2cos(θ)]
= [√[2 + 2cos(θ)]] / [√[2 - 2cos(θ)]]

110 = [√[1 + cos(θ)]] / [√[1 - cos(θ)]]
12100 = [1 + cos(θ)] / [1 - cos(θ)]

12100 - 12100cos(θ) = 1 + cos(θ)
12101cos(θ)] = 12099
cos(θ)= 0.999834724
θ = 1.04º