Question

Prove that Vector A + B cap = Vector A + Vector B

Answer 1

Since a and b are unit vectors |a| = |b| =1

Let angle be x between vectors a and b (just fr convenience while typing ;-) )

|a-b| = {|a|^2 +|b|^2 + 2 |a||b|cos(180- x)}^0.5

Putting values in rhs

|a-b| = {1+1+2*1*1*(-cos x)}^0.5

= {2–2cosx}^0.5

={2(1-cosx)}^0.5

={2*2(sinx/2)^2}^0.5

= 2 sinx/2

|a-b|= 2 sinx/2

1/2|a-b|= sinx/2

Hence, proved.

Answer 2

Answer:

Since a and b are unit vectors, |a| = 1, |b| = 1

Lets assume angle between the unit vectors, a and b, is x.

Now, Using the law of cosines on the triangle formed by vector a, b and its resultant:

|a - b| = sqrt( |a|^2 + |b|^2 - 2 cosx)

=> |a - b| = sqrt( 1 + 1 - 2 cosx)

Since, cosx = 1 - 2 sin^2 (x/2)

=> |a - b| = sqrt( 2 - 2 + 4 sin^2 (x/2))

=> |a - b| = 2 sin(x/2)

=> sin(x/2) = 1/2 |a - b| Hence proved