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Physics Question on Motion in a plane

Which of the following relations is true for two unit vector A^\hat A and B^\hat B making an angle θ to each other?

A

∣A^+B^∣=∣A^−B^∣ tanθ2|\hat A+\hat B| = |\hat A-\hat B|\ tan\frac {θ}{2}

B

∣A^−B^∣=∣A^+B^∣ tanθ2|\hat A-\hat B| = |\hat A+\hat B|\ tan\frac {θ}{2}

C

∣A^+B^∣=∣A^−B^∣ cosθ2|\hat A+\hat B| = |\hat A-\hat B|\ cos\frac {θ}{2}

D

∣A^−B^∣=∣A^+B^∣ cosθ2|\hat A-\hat B| = |\hat A+\hat B|\ cos\frac {θ}{2}

Answer

∣A^−B^∣=∣A^+B^∣ tanθ2|\hat A-\hat B| = |\hat A+\hat B|\ tan\frac {θ}{2}

Explanation

Solution

∣A^−B^∣=2sin (θ2)|\hat A-\hat B| = 2sin\ (\frac θ2)
and
∣A^+B^∣=2cos (θ2)|\hat A+\hat B| = 2cos\ (\frac θ2)

⇒ ∣A^−B^∣∣A^+B^∣=tan(θ2)\frac {|\hat A-\hat B|}{|\hat A+\hat B|} = tan (\frac θ2)
⇒ |A^−B^∣=∣A^+B^∣ tan(θ2)\hat A-\hat B| = |\hat A+\hat B|\ tan (\frac θ2)

So, the correct option is (B): A^−B^∣=∣A^+B^∣ tan(θ2)\hat A-\hat B| = |\hat A+\hat B|\ tan (\frac θ2)