if Vectors A+B=100 (A-B), and both vectors have equal magnitudes, what is the angle between them?

Question

Let me rephrase the question:
Two vectors A and B have precisely equal magnitudes. In order for the magnitude of A+B to be 100 times larger than the magnitude of A-B, what must be the angle between them. Also what if it is 110 times larger.

Answer 1

Get B on one side and A on the other:
101B=99A
Take the dot product of both sides with A:
101A·B=99||A||²
||A||=||B||, so:
101A·B=99||A||*||B||
Divide by 101 ||A||*||B||:
A·B/(||A||*||B||) = 99/101
By the definition of the dot product, A·B/(||A||*||B||) = cos θ, where θ is the angle between A and B. So:
cos θ=99/101
θ = arccos (99/101) ≈ 11.42° or 0.1993 rad

Answer 2

The magnitude of A+B = 100 The magnitude of A-B, i.e.

|A+B| = 100 |A-B|

if x is the angle between A and B , then because |A|=|B| , we have

| A+B | = 2 | A | cos(x/2), and
| A - B | = 2 | A | sin(x/2)

Thus 2 | A | cos(x/2) = (100) ( 2 | A | sin(x/2) )

dividing both sides by 2 |A| we get

cos(x/2) = 100 sin(x/2)

dividing both sides by cos(x/2),

1 = 100 tan(x/2)

thus tan(x/2) = 1/100 = 0.01

from which x/2 = arctan(0.01) = 0.5729 degrees ,

therefore x = 2 (0.5729) = 1.1458 degrees

Answer 3

90

Answer 4

I did something like this before.