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QJEEM24Ph542a: JEE Main 2024 Physics - Class 11 JEEM24Ph542a Solution: If two vectors and having equal magnitude are inclined at angle …

If two vectors A⃗\vec{A} and B⃗\vec{B} having equal magnitude RR are inclined at angle θ\theta , then

Multiple Choice Options

  • A ∣A⃗+B⃗∣=2Rcos⁡(θ2)|\vec{A}+\vec{B}|=2 R \cos \left(\frac{\theta}{2}\right)
  • B ∣A⃗−B⃗∣=2Rcos⁡(θ2)|\vec{A}-\vec{B}|=2 R \cos \left(\frac{\theta}{2}\right)
  • C ∣A⃗−B⃗∣=2Rsin⁡(θ2)|\vec{A}-\vec{B}|=\sqrt{2} R \sin \left(\frac{\theta}{2}\right)
  • D ∣A⃗+B⃗∣=2Rsin⁡(θ2)|\vec{A}+\vec{B}|=2 R \sin \left(\frac{\theta}{2}\right)

Step-by-Step Verified Solution

The magnitude of resultant vector

R′=a2+b2+2abcos⁡θR^{\prime}=\sqrt{a^2+b^2+2 a b \cos \theta}

Here a=b=Ra=b=R

Then R′=R2+R2+2R2cos⁡θR^{\prime}=\sqrt{R^2+R^2+2 R^2 \cos \theta}

amp;=R21+cos⁡θamp;=2R2cos⁡2θ2amp;=2Rcos⁡θ2\begin{aligned}& =R \sqrt{2} \sqrt{1+\cos \theta} \\& =\sqrt{2} R \sqrt{2 \cos ^2 \frac{\theta}{2}} \\& =2 R \cos \frac{\theta}{2}\end{aligned}

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