---
title: "Condition for constant direction of vector"
subject: "Mathematics"
chapter: 5
key: "semester_2/mathematics:condition-constant-direction-vector"
source: https://www.bctengineer.com/q/semester_2/mathematics/condition-constant-direction-vector
site: BCT Engineering
---

# Condition for constant direction of vector

Mathematics · Chapter 5: Vector Algebra and Calculus

- Asked in: 2075 Baisakh, 5 marks; 2072 Ashwin, 5 marks; 2071 Magh, 5 marks; 2070 Magh, 5 marks; 2069 Poush, 5 marks; 2068 Magh, 5 marks (asked 6 times)
- Video: [the necessary and sufficient condition for the vector a(t) to have constant direction is](https://www.youtube.com/watch?v=rATC4aNleDk) — Vishal Jihuli
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/condition-constant-direction-vector
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-5

## Question

Prove that the necessary and sufficient conditions for the vector function a of scalar variable t to have constant direction is a × da/dt = 0.

## Short answer

Write a = r â with r = |a|. Then a × da/dt = r² (â × dâ/dt). Constant direction implies dâ/dt = 0 so cross product zero. Conversely, a × da/dt = 0 gives â × dâ/dt = 0; since â·dâ/dt = 0, |dâ/dt| = 0, so direction constant.

## Answer

A vector a(t) has constant direction when its unit vector â = a/|a| does not change with t. Assume a ≠ 0.

Write a = r â, where r = |a| is a scalar function of t and â is a unit vector.

Then da/dt = r′â + r (dâ/dt), where r′ = dr/dt, so

a × da/dt = r r′ (â × â) + r² (â × dâ/dt) = r² (â × dâ/dt).   ... (1)

#### Necessary condition
If the direction is constant, dâ/dt = 0. Then from (1), a × da/dt = 0.

(Directly: a = f(t)â₀ with â₀ constant gives da/dt = f′â₀, parallel to a, so the cross product is zero.)

#### Sufficient condition
Let a × da/dt = 0. By (1), â × dâ/dt = 0 (as r ≠ 0).

Since â · â = 1, differentiating gives 2â · dâ/dt = 0, so dâ/dt is perpendicular to â. Therefore

|â × dâ/dt| = |â||dâ/dt| sin 90° = |dâ/dt|.

But â × dâ/dt = 0, so |dâ/dt| = 0, i.e. dâ/dt = 0. Hence â is constant: a has constant direction.

**Therefore a has constant direction if and only if a × da/dt = 0.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Writing a = r â, a × da/dt simplifies to:
   A. r^2 (â × dâ/dt)
   B. r r' (â × â)
   C. r' â × dâ/dt
   D. r^2 â
   Answer: A. The â × â term is zero, leaving r^2 (â × dâ/dt). (Key point: Setup)

2. If the direction is constant, a × da/dt = 0 because:
   A. r' = 0
   B. â is perpendicular to a
   C. a = 0
   D. dâ/dt = 0
   Answer: D. With dâ/dt = 0, equation (1) gives zero. (Key point: Necessary)

3. Why is dâ/dt perpendicular to â?
   A. â is a constant vector
   B. â × â = 0
   C. â · â = 1, differentiating gives 2â · dâ/dt = 0
   D. |â| = 0
   Answer: C. Differentiating the unit length condition. (Key point: Perpendicular)

4. Given â × dâ/dt = 0 and dâ/dt ⟂ â, we get dâ/dt = 0 because:
   A. |â × dâ/dt| = 0 forces â = 0
   B. |â × dâ/dt| = |dâ/dt|
   C. |â × dâ/dt| = |dâ/dt|^2
   D. sin 90° = 0
   Answer: B. For perpendicular vectors the cross product magnitude equals |â||dâ/dt| = |dâ/dt|. (Key point: Sufficient)

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