Question

If `veca` and `vecb` are unit vectors and θ is the angle between them, then prove that `sin  θ/2 = 1/2 |veca  - vecb|`.

Answer 1

Given, `|veca|` = 1 and `|vecb|` = 1

Now, we take `|veca - vecb|^2 = (veca - vecb).(veca - vecb)`

= `|veca|^2 + |vecb| - 2|veca|.|vecb|`

= `1 + 1 - 2|veca|.|vecb| cosθ`

= 2 – 2 × 1 × 1 cosθ

= 2(1 – cosθ)

 = `[1 - (1 - 2sin^2  θ/2)]`

= `2(2sin^2  θ/2)`

= `4 sin^2  θ/2`

or, `sin^2  θ/2 = (|veca - vecb|^2)/4`

⇒ `sin  θ/2 = (|veca - vecb|)/4`

Hence proved