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

# Prove ∇×(a×r) = 2a

Mathematics · Chapter 5: Vector Algebra and Calculus

- Asked in: 2072 Ashwin, 5 marks; 2071 Magh, 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-2
- 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

Compute curl: ∇×(a×r) = a(∇·r) - (a·∇)r = 3a - a = 2a. Thus ∇×(a×r) = 2a.

## Answer

Let a = (a₁, a₂, a₃) be constant and r = xi + yj + zk.

#### Compute a × r
```text
a × r = | i   j   k  |
        | a₁  a₂  a₃ |
        | x   y   z  |
      = (a₂z − a₃y) i + (a₃x − a₁z) j + (a₁y − a₂x) k
```

#### Take the curl
```text
∇ × (a × r) = | i      j      k    |
              | ∂/∂x   ∂/∂y   ∂/∂z |
              | a₂z−a₃y  a₃x−a₁z  a₁y−a₂x |
```
- i component: ∂(a₁y − a₂x)/∂y − ∂(a₃x − a₁z)/∂z = a₁ + a₁ = 2a₁
- j component: ∂(a₂z − a₃y)/∂z − ∂(a₁y − a₂x)/∂x = a₂ + a₂ = 2a₂
- k component: ∂(a₃x − a₁z)/∂x − ∂(a₂z − a₃y)/∂y = a₃ + a₃ = 2a₃

So ∇ × (a × r) = 2a₁i + 2a₂j + 2a₃k.

**Hence ∇ × (a × r) = 2a.**

(Check with the identity ∇×(a×r) = a(∇·r) − (a·∇)r = 3a − a = 2a.) This also shows that the velocity v = ω × r of a rigid rotation has curl v = 2ω.

## Check yourself

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

1. The x-component of a × r is:
   A. a3 y − a2 z
   B. a1 y − a2 x
   C. a2 z − a3 y
   D. a2 y − a3 z
   Answer: C. From the determinant. (Key point: a × r)

2. The i-component of the curl equals ∂(a1y − a2x)/∂y − ∂(a3x − a1z)/∂z =
   A. a1 − a1 = 0
   B. a1 + a1 = 2a1
   C. a1
   D. −2a1
   Answer: B. Both partials give a1. (Key point: Curl component)

3. Via ∇×(a×r) = a(∇·r) − (a·∇)r, the result is:
   A. 3a − a = 2a
   B. a − 3a = −2a
   C. 3a + a = 4a
   D. 0
   Answer: A. (a·∇)r = a. (Key point: Identity check)

4. For v = ω × r (rigid rotation), curl v equals:
   A. ω
   B. ω/2
   C. 0
   D. 2ω
   Answer: D. Same result as ∇×(a×r) = 2a. (Key point: Physical meaning)

## Close variants set in other exams

- [Prove (a×∇)×r = −2a](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Aprove-r-2a) — 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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