---
title: "Magnitude and equation of S.D. between two lines"
subject: "Mathematics"
chapter: 3
key: "semester_2/mathematics:magnitude-equation-s-d-between"
source: https://www.bctengineer.com/q/semester_2/mathematics/magnitude-equation-s-d-between
site: BCT Engineering
---

# Magnitude and equation of S.D. between two lines

Mathematics · Chapter 3: Three Dimensional Solid Geometry

- Asked in: 2071 Magh, 5 marks; 2068 Bhadra, 5 marks (asked 2 times)
- Video: [Find the shortest distance between the lines: (x + 1)/2 = (y - 1)/1 = (z - 9)/- 3 and (x - 3)/2 = (y](https://www.youtube.com/watch?v=84HrN5WvKPc) — Mathematics Untold
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/magnitude-equation-s-d-between
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-3

## Question

Find the magnitude and the equation of S.D. between the lines (x−1)/2 = (y−2)/3 = (z−3)/4 and (x−2)/3 = (y−4)/4 = (z−5)/5.

## Short answer

The shortest distance is 1/√6 units, and the equation is given by the planes 11x+2y−7z+6=0 and 7x+y−5z+7=0.

## Answer

**Lines.**
- L₁: (x − 1)/2 = (y − 2)/3 = (z − 3)/4, through A(1, 2, 3), direction d₁ = (2, 3, 4).
- L₂: (x − 2)/3 = (y − 4)/4 = (z − 5)/5, through B(2, 4, 5), direction d₂ = (3, 4, 5).

#### Direction of the common perpendicular
```text
n = d₁ × d₂ = (3·5 − 4·4, 4·3 − 2·5, 2·4 − 3·3)
            = (−1, 2, −1)  ∝ (1, −2, 1),   |n| = √6
```

#### Magnitude
AB = (2 − 1, 4 − 2, 5 − 3) = (1, 2, 2).

```text
S.D. = |AB · n| / |n| = |1(1) + 2(−2) + 2(1)| / √6 = |1 − 4 + 2|/√6 = 1/√6
```

#### Equation of the shortest distance
It is the intersection of the plane through L₁ parallel to n and the plane through L₂ parallel to n.

- Plane through L₁: normal = d₁ × n = (2, 3, 4) × (1, −2, 1) = (11, 2, −7). Through A(1, 2, 3):
  11(x − 1) + 2(y − 2) − 7(z − 3) = 0, so **11x + 2y − 7z + 6 = 0**.
- Plane through L₂: normal = d₂ × n = (3, 4, 5) × (1, −2, 1) = (14, 2, −10) ∝ (7, 1, −5). Through B(2, 4, 5):
  7(x − 2) + (y − 4) − 5(z − 5) = 0, so **7x + y − 5z + 7 = 0**.

Hence the line of shortest distance is

```text
11x + 2y − 7z + 6 = 0 = 7x + y − 5z + 7
```

(Its feet are (5/3, 3, 13/3) on L₁ and (3/2, 10/3, 25/6) on L₂; the distance between them is 1/√6.)

**Answer: S.D. = 1/√6 ≈ 0.408 units.**

## Check yourself

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

1. What is n = d1 x d2 for d1 = (2,3,4), d2 = (3,4,5)?
   A. (-1, 2, -1), proportional to (1, -2, 1)
   B. (1, 2, 1)
   C. (5, 7, 9)
   D. (2, 12, 20)
   Answer: A. (3*5 - 4*4, 4*3 - 2*5, 2*4 - 3*3) = (-1, 2, -1). (Key point: Common perpendicular direction)

2. With AB = (1, 2, 2), what is the S.D.?
   A. 1/6
   B. 1
   C. 1/sqrt(6)
   D. sqrt(6)
   Answer: C. |1 - 4 + 2| / sqrt(6) = 1/sqrt(6). (Key point: S.D. value)

3. How is the plane containing L1 and parallel to n found?
   A. Its normal is d1 itself
   B. Its normal is d1 x n, giving 11x + 2y - 7z + 6 = 0
   C. Its normal is d1 . n
   D. Its normal is n, giving x - 2y + z = 0
   Answer: B. The plane contains d1 and n, so d1 x n is its normal. (Key point: Plane through L1)

4. What is the line of shortest distance?
   A. 11x + 2y - 7z + 6 = 0 = 7x + y - 5z + 7
   B. 11x + 2y - 7z + 6 = 0 = 7x + y - 5z - 7
   C. 7x + 2y - 11z + 6 = 0 = 11x + y - 5z + 7
   D. x - 2y + z = 0 = 7x + y - 5z + 7
   Answer: A. It is the intersection of the plane through L1 and the plane through L2, both parallel to n. (Key point: Equation of S.D.)

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