---
title: "Triple product times (a×b) as determinant"
subject: "Mathematics"
chapter: 5
key: "semester_2/mathematics:triple-product-times-b-as"
source: https://www.bctengineer.com/q/semester_2/mathematics/triple-product-times-b-as
site: BCT Engineering
---

# Triple product times (a×b) as determinant

Mathematics · Chapter 5: Vector Algebra and Calculus

- Asked in: 2069 Bhadra, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/triple-product-times-b-as
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-5

## Question

Prove if l, m, n be three non-coplanar vectors then [l m n](a×b) = | l·a  l·b  l ; m·a  m·b  m ; n·a  n·b  n |. [3×3 determinant: rows (l·a, l·b, l), (m·a, m·b, m), (n·a, n·b, n)]

## Short answer

Write a×b = xl+ym+zn; dotting with m×n etc. gives x = [(m·a)(n·b)−(n·a)(m·b)]/[l m n], and similarly for y, z. Then [l m n](a×b) equals the expansion of the determinant with rows (l·a, l·b, l), (m·a, m·b, m), (n·a, n·b, n) along the third column.

## Answer

[Reading: [l m n](a × b) equals the 3 × 3 determinant with rows (l·a, l·b, l), (m·a, m·b, m), (n·a, n·b, n), whose third column holds the vectors l, m, n, expanded along that column.]

l, m, n are non-coplanar, so V = [l m n] ≠ 0 and they form a basis. Write

a × b = x l + y m + z n.

#### Find x, y, z
Dot with m × n. Since m · (m × n) = 0 = n · (m × n),

(a × b) · (m × n) = x l · (m × n) = x V.

By Lagrange's identity, (a × b) · (m × n) = (a·m)(b·n) − (a·n)(b·m). So

x V = (m·a)(n·b) − (n·a)(m·b).

Dotting with n × l and with l × m in the same way:

```text
y V = (n·a)(l·b) − (l·a)(n·b)
z V = (l·a)(m·b) − (m·a)(l·b)
```

#### Combine
```text
V (a × b) = [(m·a)(n·b) − (n·a)(m·b)] l
          + [(n·a)(l·b) − (l·a)(n·b)] m
          + [(l·a)(m·b) − (m·a)(l·b)] n
```

#### The determinant
Expand along the third column (l, m, n):

```text
| l·a  l·b  l |
| m·a  m·b  m | = l[(m·a)(n·b) − (m·b)(n·a)]
| n·a  n·b  n |   − m[(l·a)(n·b) − (l·b)(n·a)]
                  + n[(l·a)(m·b) − (l·b)(m·a)]
```
Compare term by term with V(a × b): the coefficients of l, m, n are identical (for m: −[(l·a)(n·b) − (l·b)(n·a)] = (n·a)(l·b) − (l·a)(n·b)).

**Hence [l m n](a × b) = | l·a l·b l ; m·a m·b m ; n·a n·b n |.**

## Check yourself

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

1. Why can a × b = x l + y m + z n be written?
   A. Because a × b is parallel to l
   B. l, m, n are non-coplanar, so they form a basis
   C. Because l, m, n are orthogonal
   D. Because a, b are unit vectors
   Answer: B. Non-coplanar means [l m n] ≠ 0. (Key point: Basis)

2. x is found by dotting with:
   A. m × n
   B. l × m
   C. l
   D. a × b
   Answer: A. m·(m×n) = n·(m×n) = 0. (Key point: Finding x)

3. (a × b) · (m × n) equals:
   A. (a·m)(b·n) + (a·n)(b·m)
   B. (a·b)(m·n)
   C. 0
   D. (a·m)(b·n) − (a·n)(b·m)
   Answer: D. Lagrange's identity. (Key point: Identity)

4. The determinant is expanded along:
   A. its first row
   B. the diagonal
   C. its third column (l, m, n)
   D. its first column
   Answer: C. That column holds the vectors. (Key point: Determinant)

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