---
title: "Prove [a+b b+c c+a] = 2[a b c]"
subject: "Mathematics"
chapter: 5
key: "semester_2/mathematics:prove-b-b-c-c"
source: https://www.bctengineer.com/q/semester_2/mathematics/prove-b-b-c-c
site: BCT Engineering
---

# Prove [a+b b+c c+a] = 2[a b c]

Mathematics · Chapter 5: Vector Algebra and Calculus

- Asked in: 2071 Bhadra, 5 marks
- Video: [scalar triple product /prove that  [ a+b, b+c, c+a ] = 2 [abc]](https://www.youtube.com/watch?v=kNn2mNItHgE) — Arrow To Success
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/prove-b-b-c-c
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-5

## Question

Show that [a+b b+c c+a] = 2[a b c].

## Short answer

Expand (b+c)×(c+a) = b×c + b×a + c×a. Taking dot with (a+b) yields [a b c] + [a b c] = 2[a b c].

## Answer

[a + b  b + c  c + a] = (a + b) · [(b + c) × (c + a)].

#### Expand the cross product
```text
(b + c) × (c + a) = b × c + b × a + c × c + c × a
                  = b × c + b × a + 0 + c × a
```

#### Dot with (a + b)
```text
(a + b) · (b × c) = a · (b × c) + b · (b × c) = [a b c] + 0
(a + b) · (b × a) = a · (b × a) + b · (b × a) = 0 + 0
(a + b) · (c × a) = a · (c × a) + b · (c × a) = 0 + [b c a] = [a b c]
```
(a triple product with a repeated vector is zero, and [b c a] = [a b c] by cyclic permutation.)

#### Add
[a + b  b + c  c + a] = [a b c] + 0 + [a b c].

**Hence [a + b  b + c  c + a] = 2[a b c].**

## Check yourself

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

1. (b + c) × (c + a) equals:
   A. b×c + a×b + c×a
   B. b×c + c×a + c×c
   C. b×c − c×a
   D. b×c + b×a + c×a
   Answer: D. c × c = 0. (Key point: Expansion)

2. Which dot products vanish?
   A. a·(b×c)
   B. None
   C. Those with a repeated vector, e.g. b·(b×c)
   D. [b c a]
   Answer: C. A repeated vector in a triple product gives zero. (Key point: Zero terms)

3. The surviving terms are:
   A. [a b c] − [b c a] = 0
   B. [a b c] + [b c a] = 2[a b c]
   C. [a b c]²
   D. 3[a b c]
   Answer: B. [b c a] = [a b c] cyclically. (Key point: Total)