Introduction

Vectors with ∣a⃗+b⃗∣=∣a⃗−b⃗∣ appear frequently in CSIR NET and other competitive exams because they encode a clear geometric relationship between a⃗ and b⃗. Understanding this condition helps distinguish between parallel, perpendicular, and scalar‑multiple cases using dot and cross products. This article solves the question in detail and evaluates every option using core vector‑algebra formulas.

Step‑by‑step Solution

Given: a⃗, b⃗ ≠ 0⃗ and ∣a⃗+b⃗∣ = ∣a⃗−b⃗∣.

Square both sides (since magnitudes are non‑negative):

|a⃗+b⃗|² = |a⃗−b⃗|²

Use the identity |x⃗|² = x⃗⋅x⃗:

Left side: |a⃗+b⃗|² = (a⃗+b⃗)⋅(a⃗+b⃗) = |a⃗|² + 2a⃗⋅b⃗ + |b⃗|²

Right side: |a⃗−b⃗|² = (a⃗−b⃗)⋅(a⃗−b⃗) = |a⃗|² − 2a⃗⋅b⃗ + |b⃗|²

Equate and simplify:

|a⃗|² + 2a⃗⋅b⃗ + |b⃗|² = |a⃗|² − 2a⃗⋅b⃗ + |b⃗|²

Cancel the common terms to obtain:

4a⃗⋅b⃗ = 0 ⟹ a⃗⋅b⃗ = 0

For non‑zero vectors, a⃗⋅b⃗ = 0 means a⃗ and b⃗ are perpendicular (orthogonal).

∴ The correct option is (B): a⃗ and b⃗ are perpendicular to each other.

Analysis of Each Option

Option (A): a⃗ and b⃗ are parallel

Parallel non‑zero vectors satisfy a⃗×b⃗=0⃗ and a⃗⋅b⃗=±|a⃗||b⃗|≠0 unless one vector is zero. Here a⃗⋅b⃗=0, which contradicts parallelism. Hence (A) is false.

Option (B): a⃗ and b⃗ are perpendicular

Perpendicular vectors satisfy a⃗⋅b⃗=0. The derived condition matches exactly, so (B) is true.

Option (C): a⃗ is not a scalar multiple of b⃗

A scalar multiple (a⃗=λb⃗) implies parallelism, not perpendicularity. Thus although this is consequentially true, the main intended answer is perpendicularity. Hence (C) is not the primary correct choice.

Option (D): a⃗×b⃗=0⃗

This is true for parallel (or zero) vectors, not perpendicular ones. The magnitude of a⃗×b⃗ in this case equals |a⃗||b⃗|, which is non‑zero. Hence (D) is false.

Key Takeaways for Exam Preparation

• Use the identity |x⃗|² = x⃗⋅x⃗ to quickly convert magnitude equations into dot‑product form.
• Perpendicular ⇒ a⃗⋅b⃗=0 (cross product generally non‑zero).
• Parallel ⇒ a⃗×b⃗=0⃗ (dot product usually non‑zero for non‑zero vectors).
• These relationships help interpret vector conditions like ∣a⃗+b⃗∣=∣a⃗−b⃗∣ efficiently in CSIR NET and other exams.