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Question

Physics Question on Motion in a plane

If unit vectors A^\hat{A} and B^\hat{B} are inclined at an angle θ\theta, then ∣A^−B^∣\left|\hat{A}-\hat{B}\right| is

A

2 sinθ22\,sin \frac{\theta}{2}

B

2 cosθ22\,cos \frac{\theta}{2}

C

2 tanθ22\,tan \frac{\theta}{2}

D

tanθtan\theta

Answer

2 sinθ22\,sin \frac{\theta}{2}

Explanation

Solution

∣A^−B^∣2=∣A^∣2+∣B^∣2−2 A^⋅B^\left|\hat{A}-\hat{B}\right|^{2}=\left|\hat{A}\right|^{2}+\left|\hat{B}\right|^{2}-2\,\hat{A}\cdot\hat{B} =2−2cosθ=2(1−cosθ) (∵∣A^∣=∣B^∣=1)=2-2cos\theta=2\left(1-cos\theta\right)\,\left(\because \left|\hat{A}\right|=\left|\hat{B}\right|=1\right) =2(sin2θ2)=2\left(sin^{2} \frac{\theta}{2}\right) =4 sin2θ2=4\,sin^{2} \frac{\theta}{2} or ∣A^−B^∣=2 sinθ2\left|\hat{A}-\hat{B}\right|=2\,sin \frac{\theta}{2}