Vector A = (Vector B)×(1∠α)
|A+B| = 110×|A-B|
angle α
Substituting for vector A in the given relation, we have
... |B×(1∠α) + B| = 110|B×(1∠α) -B|
Now 1∠α in polar coordinates is (cos(α), sin(α)) in rectangular coordinates.
Dividing by |B|, we have the relation between sum and difference vector magnitudes is
... √((1 +cos(α))² +sin(α)²) = 110√((cos(α) -1)² +(-sin(α))²)
Squaring both sides gives
... 1 +2cos(α) +cos(α)² +sin(α)² = 12100(1 -2cos(α) +cos(α)² +sin(α)²)
... 1 +cos(α) = 12100(1 -cos(α)) . . . . using sin²+cos²=1 and dividing by 2
And solving for cos(α) gives
... cos(α)(1+12100) = 12100 -1
... α = arccos(12099/12101) ≈ 1.0417°
The angle between the vectors is about 1.0417°.
