---
title: "Magnitude of shortest distance between lines"
subject: "Mathematics"
chapter: 3
key: "semester_2/mathematics:magnitude-shortest-distance-between-lines"
source: https://www.bctengineer.com/q/semester_2/mathematics/magnitude-shortest-distance-between-lines
site: BCT Engineering
---

# Magnitude of shortest distance between lines

Mathematics · Chapter 3: Three Dimensional Solid Geometry

- Asked in: 2072 Magh, 5 marks
- 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-shortest-distance-between-lines
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-3

## Question

Find the magnitude of the line of the shortest distance between the lines x/4 = (y+1)/3 = (z−2)/2, 5x−2y−3z+6 = 0, x−3y+2z−3 = 0. [Second line as printed: the pair of planes 5x−2y−3z+6 = 0 and x−3y+2z−3 = 0; scan unclear]

## Short answer

Line 1: A(0,-1,2), d1=(4,3,2). Line 2: direction d2=(1,1,1) from cross product, point B(-24/13,-21/13,0). n=d1×d2=(1,-2,1). Distance = |AB·n|/√6 = (34/13)/√6 = 17√6/39 ≈ 1.07 units.

## Answer

**Reading.** The second line is the intersection of the planes 5x − 2y − 3z + 6 = 0 and x − 3y + 2z − 3 = 0, and I take that as printed. The first line is x/4 = (y + 1)/3 = (z − 2)/2.

#### Line 1
Through A(0, −1, 2), direction d₁ = (4, 3, 2).

#### Line 2: direction and a point
Direction = cross product of the plane normals:

```text
d₂ = (5, −2, −3) × (1, −3, 2)
   = ((−2)(2) − (−3)(−3), (−3)(1) − (5)(2), (5)(−3) − (−2)(1))
   = (−13, −13, −13)  ∝ (1, 1, 1)
```

Point: put z = 0: 5x − 2y = −6 and x − 3y = 3. Solving: x = −24/13, y = −21/13. So B(−24/13, −21/13, 0).

#### Common perpendicular direction
```text
n = d₁ × d₂ = (4, 3, 2) × (1, 1, 1) = (3 − 2, 2 − 4, 4 − 3) = (1, −2, 1),   |n| = √6
```

#### Shortest distance
AB = B − A = (−24/13, −21/13 + 1, −2) = (−24/13, −8/13, −2).

```text
AB · n = −24/13 + 16/13 − 2 = −8/13 − 26/13 = −34/13

S.D. = |AB · n| / |n| = (34/13)/√6 = 34/(13√6) = 17√6/39 ≈ 1.07
```

**Answer: the magnitude of the shortest distance is 17√6/39 ≈ 1.07 units.**

## Check yourself

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

1. What is the direction of the line of intersection of the two planes?
   A. (5, -2, -3), the first normal
   B. (1, -3, 2), the second normal
   C. (1, 1, 1), found as the cross product of the plane normals
   D. (4, 3, 2)
   Answer: C. (5, -2, -3) x (1, -3, 2) = (-13, -13, -13). (Key point: Direction of line 2)

2. What is n = d1 x d2 with d1 = (4, 3, 2), d2 = (1, 1, 1)?
   A. (1, -2, 1)
   B. (1, 2, 1)
   C. (5, 4, 3)
   D. (-1, 2, 1)
   Answer: A. (3 - 2, 2 - 4, 4 - 3) = (1, -2, 1). (Key point: Common perpendicular)

3. How is the S.D. computed?
   A. |AB x n|
   B. |AB . n| / |n|
   C. |AB| / |n|
   D. |n| / |AB|
   Answer: B. It is the component of AB along the common perpendicular. (Key point: Formula)

4. What is the shortest distance?
   A. 34/13
   B. 34 sqrt(6)/13
   C. 17/39
   D. 17 sqrt(6)/39
   Answer: D. (34/13)/sqrt(6) = 17 sqrt(6)/39 which is about 1.07. (Key point: Value)

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