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

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°.

45

step-by-step explanation: