---
title: "Reciprocal system of 2i−3j+k, i+2j−k, 3i−j+2k"
subject: "Mathematics"
chapter: 5
key: "semester_2/mathematics:reciprocal-system-2i-3j-k"
source: https://www.bctengineer.com/q/semester_2/mathematics/reciprocal-system-2i-3j-k
site: BCT Engineering
---

# Reciprocal system of 2i−3j+k, i+2j−k, 3i−j+2k

Mathematics · Chapter 5: Vector Algebra and Calculus

- Asked in: 2072 Ashwin, 5 marks
- Video: [Finding Reciprocal Vectors of (2i + 3j - k), (i - j - 2k), and (-i + 2j + 2k) - Bhagvati classes](https://www.youtube.com/watch?v=QvNqesoUjoQ) — Bhagvati Classes
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/reciprocal-system-2i-3j-k
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-5

## Question

Find the set of reciprocal system to the set of vectors: 2i−3j+k, i+2j−k and 3i−j+2k.

## Short answer

Scalar triple product is 14. Compute: a' = (3i-5j-7k)/14, b' = (5i+j-7k)/14, c' = (i+3j+7k)/14.

## Answer

For a, b, c with [a b c] ≠ 0, the reciprocal system is a′ = (b × c)/[a b c], b′ = (c × a)/[a b c], c′ = (a × b)/[a b c].

Here a = (2, −3, 1), b = (1, 2, −1), c = (3, −1, 2).

#### Cross products
```text
b × c = (2·2 − (−1)(−1), (−1)(3) − (1)(2), (1)(−1) − (2)(3))
      = (4 − 1, −3 − 2, −1 − 6) = (3, −5, −7)
c × a = ((−1)(1) − (2)(−3), (2)(2) − (3)(1), (3)(−3) − (−1)(2))
      = (−1 + 6, 4 − 3, −9 + 2) = (5, 1, −7)
a × b = ((−3)(−1) − (1)(2), (1)(1) − (2)(−1), (2)(2) − (−3)(1))
      = (3 − 2, 1 + 2, 4 + 3) = (1, 3, 7)
```

#### Triple product
[a b c] = a · (b × c) = 2(3) + (−3)(−5) + (1)(−7) = 6 + 15 − 7 = 14.

#### Reciprocal vectors
```text
a′ = (3i − 5j − 7k)/14
b′ = (5i + j − 7k)/14
c′ = (i + 3j + 7k)/14
```

Check: a · a′ = (6 + 15 − 7)/14 = 1 ✓; a · b′ = (10 − 3 − 7)/14 = 0 ✓; c · c′ = (3 − 3 + 14)/14 = 1 ✓.

**Answer: a′ = (3i − 5j − 7k)/14, b′ = (5i + j − 7k)/14, c′ = (i + 3j + 7k)/14.**

## Check yourself

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

1. For a = (2,−3,1), b = (1,2,−1), c = (3,−1,2), [a b c] is:
   A. −14
   B. 7
   C. 20
   D. 14
   Answer: D. 6 + 15 − 7 = 14. (Key point: Triple product)

2. b × c equals:
   A. (3, 5, −7)
   B. (−3, −5, 7)
   C. (3, −5, −7)
   D. (5, 1, −7)
   Answer: C. Determinant expansion. (Key point: b × c)

3. a' equals:
   A. (3i − 5j − 7k)/3
   B. (3i − 5j − 7k)/14
   C. (5i + j − 7k)/14
   D. (3i + 5j + 7k)/14
   Answer: B. (b×c)/[abc]. (Key point: a')

4. A valid check of the result is:
   A. a · a' = 1 and a · b' = 0
   B. a · a' = 0
   C. a' × a = 1
   D. |a'| = 1
   Answer: A. Reciprocal vectors satisfy these. (Key point: Check)

## Close variants set in other exams

- [Reciprocal system of vectors](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Areciprocal-system-vectors) — Mathematics, chapter 5, asked 4×

## Related in Mathematics

- [Reciprocal system of 2i+3j−k, i−j−2k, −i+2j+2k](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Areciprocal-system-2i-3j-k-2) — Mathematics, chapter 5, asked 2×