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Question
If `vec"a", vec"b"` are unit vectors and q is the angle between them, show that
`sin theta/2 = 1/2|vec"a" - vec"b"|`
Solution
`(vec"a" - vec"b")^2 = vec"a"^2 + vec"b"^2 - 2vec"a" * vec"b"`
`|vec"a" - vec"b"|^2 = |vec"a"|^2 + |vec"b"|^2 - 2|vec"a"|*|vec"b"| cos theta`
`|vec"a" - vec"b"|^2 = 1^2 + 1^2 - 2 xx 1*1 cos theta`
= `2 - 2 cos theta`
= `2(1 - cos theta)`
`|vec"a" - vec"b"|^2 = 2 xx 2 sin^2 theta/2`
`|vec"a" - vec"b"|^2 = 2 sin theta/2`
`sin theta/2 = 1/2|vec"a" - vec"b"|`
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