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JEE Main 202431 JanuaryEvening Shift
Exam
JEE Main
Year
2024
Date
31 January
Shift
Evening Shift
Subject
Physics
Chapter
Mathematics in Physics

JEE Main 2024 · 31 January · Evening Shift · Physics · Mathematics in Physics

If two vectors → A and → B having equal magnitude R are inclined at...

If two vectors A→\overset{\to}{A} and B→\overset{\to}{B} having equal magnitude RR are inclined at an angle θ\theta, then

Options

  1. A∣A→−B→∣=2Rsin(θ2)\left|\overset{\to}{A} - \overset{\to}{B}\right| = \sqrt{2}Rsin\left(\dfrac{\theta}{2}\right)
  2. B∣A→+B→∣=2Rsin(θ2)\left|\overset{\to}{A} + \overset{\to}{B}\right| = 2Rsin\left(\dfrac{\theta}{2}\right)
  3. C∣A→+B→∣=2Rcos(θ2)\left|\overset{\to}{A} + \overset{\to}{B}\right| = 2Rcos\left(\dfrac{\theta}{2}\right)
  4. D∣A→−B→∣=2Rcos(θ2)\left|\overset{\to}{A} - \overset{\to}{B}\right| = 2Rcos\left(\dfrac{\theta}{2}\right)

Correct answer

C. ∣A→+B→∣=2Rcos(θ2)\left|\overset{\to}{A} + \overset{\to}{B}\right| = 2Rcos\left(\dfrac{\theta}{2}\right)

Step-by-step solution

The magnitude of resultant vector A→+B→\overset{\to}{A} + \overset{\to}{B} is given by, 

R′=A2+B2+2ABcosθR ' = \sqrt{{A}^{2} + {B}^{2} + 2ABcos\theta}

Here A=B=RA = B = R

Then R′=R2+R2+2R2cosθR ' = \sqrt{{R}^{2} + {R}^{2} + 2{R}^{2}cos\theta}

=R21+cosθ= R\sqrt{2}\sqrt{1 + cos\theta}

=2R2cos⁡2θ2= \sqrt{2}R\sqrt{2\cos^{2}\dfrac{\theta}{2}}

=2Rcosθ2= 2Rcos\dfrac{\theta}{2}

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