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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|`.
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Solution
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
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