Introduction

Finding the magnitude of |a−b| for unit vectors with angle π/3 (60°) is a core Vector Algebra concept for CSIR NET Life Sciences and engineering exams. This problem tests dot product and magnitude formulas—key for molecular modeling and biomechanics applications in biology. The correct magnitude is 1.

Solution Steps

The magnitude of |a−b| uses the formula:

|a−b|² = |a|² + |b|² − 2|a||b|cosθ

Since a and b are unit vectors, |a| = |b| = 1. With θ = π/3, cos(π/3) = 1/2:

|a−b|² = 1 + 1 − 2(1)(1)(1/2) = 2 − 1 = 1

|a−b| = √1 = 1

Common Options Explained

• 1: Correct, as derived above.
• √3: Appears when |a+b| = 1 implies θ = 120°, then difference is √3.
• √2: For θ = 90°, |a−b| = √2.
• 0: Only if a = b (θ = 0).
• 2: Maximum for θ = 180°.

Detailed Derivation

Formula for unit vectors: |a−b| = √(2−2cosθ).

θ = π/3, cosθ = 1/2

|a−b| = √(2−2(1/2)) = √(2−1) = √1 = 1[web:7]

Geometric Insight: The difference vector forms a parallelogram; at 60°, sides subtract partially along the angle bisector.

Exam Tips

• Always square first to simplify.
• Verify with |a+b| = √(2+2cosθ) = √3 for consistency.
• Practice variations: θ=90° → √2, 120° → √3.