---
title: "(b×c)×(c×a)=[a b c]c and deduction"
subject: "Mathematics"
chapter: 5
key: "semester_2/mathematics:b-c-c-b-c"
source: https://www.bctengineer.com/q/semester_2/mathematics/b-c-c-b-c
site: BCT Engineering
---

# (b×c)×(c×a)=[a b c]c and deduction

Mathematics · Chapter 5: Vector Algebra and Calculus

- Asked in: 2069 Poush, 5 marks; 2068 Magh, 5 marks; 2067 Chaitra, 5 marks (asked 3 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/b-c-c-b-c
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-5

## Question

Show that (b×c)×(c×a) = [a b c] c and deduce [b×c c×a a×b] = [a b c]².

## Short answer

Using vector triple product, (b×c)×(c×a) = c[(b×c)·a] - a[(b×c)·c] = [a b c] c. Then [b×c c×a a×b] = ([a b c] c)·(a×b) = [a b c][c a b] = [a b c]².

## Answer

#### Part 1: (b × c) × (c × a) = [a b c] c
Use P × (Q × R) = Q(P · R) − R(P · Q) with P = b × c, Q = c, R = a:

```text
(b × c) × (c × a) = c[(b × c) · a] − a[(b × c) · c]
```
- (b × c) · a = [a b c] (the cyclic form of a · (b × c)).
- (b × c) · c = 0, because b × c is perpendicular to c.

**So (b × c) × (c × a) = [a b c] c.**

#### Part 2: Deduce [b×c c×a a×b] = [a b c]²
By definition [u v w] = (u × v) · w. With u = b × c, v = c × a, w = a × b:

```text
[b×c  c×a  a×b] = [(b × c) × (c × a)] · (a × b)
                = [a b c] c · (a × b)       (using Part 1)
                = [a b c] [c a b]
                = [a b c] [a b c]            (cyclic permutation)
```

**Hence [b×c c×a a×b] = [a b c]².**

Check with a = i, b = j, c = k: (j×k)×(k×i) = i × j = k = [i j k] k ✓.

## Check yourself

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

1. Taking P = b × c, Q = c, R = a gives:
   A. a[(b×c)·c] − c[(b×c)·a]
   B. c[(b×c)·a] − a[(b×c)·c]
   C. c[(b×c)·c] − a[(b×c)·a]
   D. b[(c·a)] − c[(b·a)]
   Answer: B. P×(Q×R) = Q(P·R) − R(P·Q). (Key point: Apply identity)

2. (b × c) · c = 0 because:
   A. b × c is perpendicular to c
   B. b is parallel to c
   C. c = 0
   D. the dot product is cyclic
   Answer: A. A cross product is perpendicular to both factors. (Key point: Zero term)

3. [b×c c×a a×b] = [(b×c)×(c×a)]·(a×b) = ?
   A. [a b c]·0
   B. [a b c]
   C. [a b c]^3
   D. [a b c][c a b] = [a b c]^2
   Answer: D. c · (a × b) = [c a b] = [a b c] by cyclic permutation. (Key point: Deduction)