---
title: "Identity with four vectors (a×b)×(c×d)+..."
subject: "Mathematics"
chapter: 5
key: "semester_2/mathematics:identity-four-vectors-b-c"
source: https://www.bctengineer.com/q/semester_2/mathematics/identity-four-vectors-b-c
site: BCT Engineering
---

# Identity with four vectors (a×b)×(c×d)+...

Mathematics · Chapter 5: Vector Algebra and Calculus

- Asked in: 2067 Mangsir, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/identity-four-vectors-b-c
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-5

## Question

Prove that (a×b)×(c×d) + (a×c)×(d×b) + (a×d)×(b×c) = −2[b c d]a.

## Short answer

Apply (P×Q)×(R×S) = [P R S]Q − [Q R S]P to each term. Summing yields −3[b c d]a + [a c d]b − [a b d]c + [a b c]d. The linear expansion identity [b c d]a = [a c d]b − [a b d]c + [a b c]d reduces the sum to −2[b c d]a.

## Answer

We use (P × Q) × (R × S) = [P R S] Q − [Q R S] P, which follows from (P × Q) × Y = Q(P · Y) − P(Q · Y) with Y = R × S.

#### Expand each term
```text
(a×b)×(c×d) = [a c d] b − [b c d] a
(a×c)×(d×b) = [a d b] c − [c d b] a
(a×d)×(b×c) = [a b c] d − [d b c] a
```
In the second and third terms, the triple products multiplying a are [c d b] = [b c d] and [d b c] = [b c d] (cyclic permutation).

#### Add
The a-terms give −3[b c d] a. With the remaining terms:

```text
Sum = −3[b c d] a + [a c d] b + [a d b] c + [a b c] d
```
Now [a d b] = −[a b d], so the last three terms are [a c d] b − [a b d] c + [a b c] d.

#### Key relation for any four vectors
Take b, c, d non-coplanar. Then a = x b + y c + z d. Taking the dot product with c × d, d × b and b × c in turn gives

x = [a c d]/[b c d], y = [b a d]/[b c d], z = [b c a]/[b c d],

so [b c d] a = [a c d] b + [b a d] c + [b c a] d = [a c d] b − [a b d] c + [a b c] d.

Hence [a c d] b − [a b d] c + [a b c] d = [b c d] a. (Both sides are polynomials in the components, so the relation also holds when b, c, d are coplanar, by continuity.)

#### Finish
Sum = −3[b c d] a + [b c d] a = **−2[b c d] a**, as required.

## Check yourself

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

1. Which result is used to expand (a×b)×(c×d)?
   A. [a c d] b − [b c d] a
   B. [a b c] d − [a b d] c
   C. (a·c)(b·d) − (a·d)(b·c)
   D. [a b c] a − [b c d] d
   Answer: A. Using (P×Q)×Y = Q(P·Y) − P(Q·Y) with Y = R×S gives [P R S]Q − [Q R S]P, so the term is [a c d] b − [b c d] a. (Key point: Formula for (P×Q)×(R×S))

2. Why can the coefficients of a in the 2nd and 3rd terms, [c d b] and [d b c], both be replaced by [b c d]?
   A. Swapping any two vectors leaves a triple product unchanged
   B. A cyclic permutation of the three vectors does not change the triple product
   C. Each of them is zero
   D. They are equal in magnitude but opposite in sign
   Answer: B. Cyclic permutation keeps the triple product the same, so [c d b] = [d b c] = [b c d]. This makes the a-terms add up to −3[b c d]a. (Key point: Cyclic permutation of triple products)

3. After adding the three expansions, what is the coefficient of a in the sum?
   A. −[b c d]
   B. −2[b c d]
   C. −3[b c d]
   D. +3[b c d]
   Answer: C. Each expansion contributes −[b c d]a, so the total is −3[b c d]a plus the terms in b, c and d. (Key point: Sum after collecting a-terms)

4. Which relation (for non-coplanar b, c, d) turns the remaining terms into [b c d] a so the sum becomes −2[b c d] a?
   A. [a c d] b − [a b d] c + [a b c] d = [b c d] a
   B. [a c d] b + [a b d] c + [a b c] d = [b c d] a
   C. [a c d] b − [a b d] c + [a b c] d = −[b c d] a
   D. [a c d] b − [a b d] c + [a b c] d = 3[b c d] a
   Answer: A. Writing a = xb + yc + zd and dotting with c×d, d×b, b×c gives this relation. Then −3[b c d]a + [b c d]a = −2[b c d]a. (Key point: Key relation for four vectors)

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