---
title: "Scalar triple product squared identity"
subject: "Mathematics"
chapter: 5
key: "semester_2/mathematics:scalar-triple-product-squared-identity"
source: https://www.bctengineer.com/q/semester_2/mathematics/scalar-triple-product-squared-identity
site: BCT Engineering
---

# Scalar triple product squared identity

Mathematics · Chapter 5: Vector Algebra and Calculus

- Asked in: 2073 Bhadra, 5 marks; 2073 Magh, 5 marks; 2070 Magh, 5 marks (asked 3 times)
- Video: [L-30| vector| square of scalar triple product of vectors](https://www.youtube.com/watch?v=s2UfXJdrS7g) — JEE Made Easy - by Rajesh sir
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/scalar-triple-product-squared-identity
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-5

## Question

Prove that [b×c c×a a×b] = [a b c]².

## Short answer

Use (c×a)×(a×b) = [a b c] a. Then [b×c c×a a×b] = (b×c)·([a b c] a) = [a b c](b×c)·a = [a b c]².

## Answer

The scalar triple product is [a b c] = a · (b × c), and it is unchanged by cyclic permutation: [a b c] = [b c a] = [c a b].

We need the identity P × (Q × R) = Q(P · R) − R(P · Q).

#### Step 1: Simplify (c × a) × (a × b)
Take P = c × a, Q = a, R = b:

```text
(c × a) × (a × b) = a[(c × a) · b] − b[(c × a) · a]
```
Now (c × a) · b = [c a b] = [a b c], and (c × a) · a = 0 (a cross product is perpendicular to its factors). So

(c × a) × (a × b) = [a b c] a.

#### Step 2: Evaluate the triple product
[b×c  c×a  a×b] = (b × c) · [(c × a) × (a × b)]

```text
= (b × c) · [a b c] a
= [a b c] (b × c) · a
= [a b c] [a b c]
```
because a · (b × c) = [a b c].

**Hence [b×c c×a a×b] = [a b c]².**

(Numerical check with a = i, b = j, c = k: b×c = i, c×a = j, a×b = k, so the left side is [i j k] = 1 and [a b c]² = 1.)

## Check yourself

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

1. The identity used is:
   A. P × (Q × R) = P(Q·R) − R(P·Q)
   B. P × (Q × R) = Q(P·R) − R(P·Q)
   C. P × (Q × R) = (P·Q)(P·R)
   D. P × (Q × R) = Q(P·Q) − R(P·R)
   Answer: B. BAC-CAB form. (Key point: Vector triple identity)

2. (c × a) × (a × b) equals:
   A. [a b c] a
   B. [a b c] b
   C. 0
   D. [a b c]^2 a
   Answer: A. The b-term vanishes since (c×a)·a = 0. (Key point: Key simplification)

3. Why is (c × a) · a = 0?
   A. a × a = 0
   B. c is parallel to a
   C. Triple products are always zero
   D. A cross product is perpendicular to its factors
   Answer: D. c × a ⟂ a. (Key point: Why zero)

4. After Step 1, [b×c c×a a×b] = (b×c)·[a b c]a = ?
   A. [a b c]
   B. 2[a b c]
   C. [a b c]^2
   D. 0
   Answer: C. (b×c)·a = [abc], multiplied by [abc]. (Key point: Final step)

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