JEE PYQ: Vector Algebra - Question ID a34421321612 (JEE Main 2022)

ID: a34421321612JEE Main 2022Single Correct MCQ

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

Select Option

Step-by-step Explanation

Core Formula & Concept:

When two unit vectors A^\widehat{A} and B^\widehat{B} enclose an angle θ\theta, their sum and difference can be expressed using the following fundamental results from vector algebra:

  • Magnitude of the sum: ∣A^+B^∣=2+2cos⁡θ=2cos⁡(θ2)|\widehat{A} + \widehat{B}| = \sqrt{2 + 2\cos\theta} = 2\cos\left(\frac{\theta}{2}\right)
  • Magnitude of the difference: ∣A^−B^∣=2−2cos⁡θ=2sin⁡(θ2)|\widehat{A} - \widehat{B}| = \sqrt{2 - 2\cos\theta} = 2\sin\left(\frac{\theta}{2}\right)

These follow from the identity ∣v⃗∣2=v⃗⋅v⃗|\vec{v}|^2 = \vec{v} \cdot \vec{v} and the double-angle formulas for cosine and sine.

Step-by-Step Derivation:

Step 1: Express the magnitudes explicitly

∣A^+B^∣=2cos⁡(θ2),∣A^−B^∣=2sin⁡(θ2).|\widehat{A} + \widehat{B}| = 2\cos\left(\frac{\theta}{2}\right), \quad |\widehat{A} - \widehat{B}| = 2\sin\left(\frac{\theta}{2}\right).

Step 2: Form the ratio of the magnitudes

∣A^−B^∣∣A^+B^∣=2sin⁡(θ2)2cos⁡(θ2)=tan⁡(θ2).\frac{|\widehat{A} - \widehat{B}|}{|\widehat{A} + \widehat{B}|} = \frac{2\sin\left(\frac{\theta}{2}\right)}{2\cos\left(\frac{\theta}{2}\right)} = \tan\left(\frac{\theta}{2}\right).

Step 3: Rearrange to match the given options

Multiplying both sides by ∣A^+B^∣|\widehat{A} + \widehat{B}| gives ∣A^−B^∣=∣A^+B^∣tan⁡(θ2).|\widehat{A} - \widehat{B}| = |\widehat{A} + \widehat{B}| \tan\left(\frac{\theta}{2}\right).

This exactly matches option B.

Common Traps & Exam Tip:

Many students confuse the roles of sine and cosine in the magnitudes of the sum and difference. A frequent error is to swap the expressions, leading to incorrect options like A or D. Always derive the magnitudes from first principles using the dot product to avoid this pitfall.