---
title: "Prove lines coplanar"
subject: "Mathematics"
chapter: 3
key: "semester_2/mathematics:prove-lines-coplanar"
source: https://www.bctengineer.com/q/semester_2/mathematics/prove-lines-coplanar
site: BCT Engineering
---

# Prove lines coplanar

Mathematics · Chapter 3: Three Dimensional Solid Geometry

- Asked in: 2071 Bhadra, 5 marks
- Video: [Condition for two lines to be coplanar -Three Dimensional Geometry - COPLANAR LINES](https://www.youtube.com/watch?v=yn3EGG_f8Tc) — dy by dx
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/prove-lines-coplanar
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-3

## Question

Prove that the lines (x−1)/2 = (y−2)/3 = (z−3)/4 and 4x−3y+1 = 0 = 5x−3z+2 are coplanar.

## Short answer

L2 symmetric form: x/3 = (y-1/3)/4 = (z-2/3)/5. Points A(1,2,3) and C(0,1/3,2/3). Scalar triple product of AC, d1(2,3,4), d2(3,4,5) equals zero, so the lines are coplanar.

## Answer

**Lines.**
- L₁: (x − 1)/2 = (y − 2)/3 = (z − 3)/4, through A(1, 2, 3) with direction (2, 3, 4).
- L₂: 4x − 3y + 1 = 0 and 5x − 3z + 2 = 0.

#### Symmetric form of L₂
From 4x − 3y + 1 = 0: y = (4x + 1)/3 = 4x/3 + 1/3. From 5x − 3z + 2 = 0: z = (5x + 2)/3.

Put x = 3m: x = 3m, y = 4m + 1/3, z = 5m + 2/3. So L₂ passes through C(0, 1/3, 2/3) with direction (3, 4, 5). In symmetric form:

x/3 = (y − 1/3)/4 = (z − 2/3)/5.

#### Coplanarity condition
Two lines through A and C with directions d₁ and d₂ are coplanar if [AC, d₁, d₂] = 0.

AC = (0 − 1, 1/3 − 2, 2/3 − 3) = (−1, −5/3, −7/3) ∝ (−3, −5, −7) (multiplying by 3).

```text
| −3  −5  −7 |
|  2   3   4 | = −3(3·5 − 4·4) + 5(2·5 − 4·3) − 7(2·4 − 3·3)
|  3   4   5 |
             = −3(−1) + 5(−2) − 7(−1)
             = 3 − 10 + 7
             = 0
```

The scalar triple product is zero, so the lines are coplanar. ∎

(They also intersect, at (−1, −1, −1): this is the point of L₁ with parameter −1, and it satisfies both planes of L₂.)

## Check yourself

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

1. Which condition shows two lines through A and C with directions d1, d2 are coplanar?
   A. d1 . d2 = 0
   B. d1 x d2 = 0
   C. [AC, d1, d2] = 0
   D. AC . d1 = 0
   Answer: C. Coplanar lines need the join vector AC to lie in the plane of d1 and d2, so the scalar triple product vanishes. (Key point: Coplanarity test)

2. How is the second line (given as two planes) converted to symmetric form?
   A. Express y and z in terms of x (or a parameter), read a point and direction (3, 4, 5)
   B. Add the two plane equations to get one direction
   C. Take the normals (4, -3, 0) and (5, 0, -3) as the direction directly
   D. Set x = y = z = 0 and use the origin as the point
   Answer: A. Solving y = (4x+1)/3, z = (5x+2)/3 with x = 3m gives point (0, 1/3, 2/3) and direction (3, 4, 5). (Key point: Symmetric form of L2)

3. With AC ∝ (-3, -5, -7), d1 = (2, 3, 4), d2 = (3, 4, 5), what is the determinant and the conclusion?
   A. 0, so the lines are skew
   B. 0, so the lines are coplanar
   C. Nonzero, so the lines are coplanar
   D. Nonzero, so the lines are parallel
   Answer: B. The expansion gives 3 - 10 + 7 = 0, which is the coplanarity condition. (Key point: Result of the determinant)

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