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+2cosθ=2cos(θ2)|\widehat{A} + \widehat{B}| = \sqrt{2 + 2\cos\theta} = 2\cos\left(\frac{\theta}{2}\right)
  • Magnitude of the difference: A^B^=22cosθ=2sin(θ2)|\widehat{A} - \widehat{B}| = \sqrt{2 - 2\cos\theta} = 2\sin\left(\frac{\theta}{2}\right)

These follow from the identity v2=vv|\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.