---
title: "Prove (a×∇)×r = −2a"
subject: "Mathematics"
chapter: 5
key: "semester_2/mathematics:prove-r-2a"
source: https://www.bctengineer.com/q/semester_2/mathematics/prove-r-2a
site: BCT Engineering
---

# Prove (a×∇)×r = −2a

Mathematics · Chapter 5: Vector Algebra and Calculus

- Asked in: 2074 Bhadra, 5 marks; 2073 Bhadra, 5 marks (asked 2 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/prove-r-2a
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-5

## Question

If a is a constant vector and r be the position vector then prove that (a×∇)×r = −2a.

## Short answer

Expand (a×∇)×r = ∇(a·r) - a(∇·r). With a constant, ∇(a·r) = a and ∇·r = 3, so (a×∇)×r = a - 3a = -2a.

## Answer

Here a is constant, r = xi + yj + zk, and ∇ acts only on r. Write the result as an expansion of the vector triple product.

#### Method 1: triple product expansion
For vectors, (P × Q) × R = Q(P · R) − P(Q · R). Take P = a, Q = ∇, R = r. Because ∇ is differentiating r (a is constant):

```text
(a × ∇) × r = ∇(a · r) − a(∇ · r)
```
- ∇(a · r) with a constant: a · r = a₁x + a₂y + a₃z, so ∇(a · r) = a₁i + a₂j + a₃k = a.
- ∇ · r = ∂x/∂x + ∂y/∂y + ∂z/∂z = 3.

Therefore (a × ∇) × r = a − 3a = **−2a**.

#### Method 2: components (confirmation)
Let a = (a₁, a₂, a₃). Then

a × ∇ = (a₂∂z − a₃∂y) i + (a₃∂x − a₁∂z) j + (a₁∂y − a₂∂x) k = (D₁, D₂, D₃).

The x-component of (a × ∇) × r is D₂ z − D₃ y:

```text
D₂ z = (a₃∂x − a₁∂z) z = −a₁
D₃ y = (a₁∂y − a₂∂x) y = a₁
x-component = −a₁ − a₁ = −2a₁
```
The y and z components follow cyclically: −2a₂ and −2a₃.

**Hence (a × ∇) × r = −2a.**

## Check yourself

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

1. (a × ∇) × r expands to:
   A. a(∇·r) − ∇(a·r)
   B. ∇(a·r) + a(∇·r)
   C. ∇(a·r) − a(∇·r)
   D. a × (∇ × r)
   Answer: C. Using (P×Q)×R = Q(P·R) − P(Q·R). (Key point: Expansion)

2. ∇(a · r) for constant a equals:
   A. 0
   B. a
   C. r
   D. 3a
   Answer: B. a·r = a1x + a2y + a3z. (Key point: Gradient)

3. ∇ · r equals:
   A. 3
   B. 0
   C. 1
   D. r
   Answer: A. ∂x/∂x + ∂y/∂y + ∂z/∂z = 3. (Key point: Divergence)

4. Combining, the value is:
   A. 2a
   B. 4a
   C. −3a
   D. −2a
   Answer: D. a − 3a = −2a. (Key point: Result)

## Close variants set in other exams

- [Prove ∇×(a×r) = 2a](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Aprove-r-2a-2) — Mathematics, chapter 5, asked 6×
- [Gradient of (a·r)/rⁿ](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Agradient-r-r) — Mathematics, chapter 5, asked 6×
- [Divergence of (a×r)/r is zero](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Adivergence-r-r-zero) — Mathematics, chapter 5, asked 6×

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