---
title: "Coplanar lines with point of intersection"
subject: "Mathematics"
chapter: 3
key: "semester_2/mathematics:coplanar-lines-point-intersection-2"
source: https://www.bctengineer.com/q/semester_2/mathematics/coplanar-lines-point-intersection-2
site: BCT Engineering
---

# Coplanar lines with point of intersection

Mathematics · Chapter 3: Three Dimensional Solid Geometry

- Asked in: 2072 Ashwin, 5 marks
- Video: [Coplanar lines/Point of intersection and the plane containing the lines](https://www.youtube.com/watch?v=67E9CE3byQM) — PHYSICSLEARNING
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/coplanar-lines-point-intersection-2
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-3

## Question

Show that the lines (x−1)/2 = (y−2)/3 = (z−3)/4 and 4x−3y+1 = 0 = 5x−3z+2 are coplanar and find their point of intersection.

## Short answer

Point on L1 (1+2t,2+3t,3+4t) satisfies both planes of L2 for t=-1 each, so lines intersect at (-1,-1,-1) and are coplanar.

## Answer

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

#### Method 1: test for a common point
A general point of L₁ is (1 + 2t, 2 + 3t, 3 + 4t). Put it into both planes of L₂:

```text
4(1 + 2t) − 3(2 + 3t) + 1 = 4 + 8t − 6 − 9t + 1 = −1 − t = 0   →  t = −1
5(1 + 2t) − 3(3 + 4t) + 2 = 5 + 10t − 9 − 12t + 2 = −2 − 2t = 0 →  t = −1
```

Both give t = −1, so this point of L₁ lies on L₂. The lines intersect, so they are coplanar.

#### Method 2: determinant (as a check)
Write L₂ in symmetric form. From 4x − 3y + 1 = 0: y = (4x + 1)/3. From 5x − 3z + 2 = 0: z = (5x + 2)/3. So with x = 3m: (x, y, z) = (3m, 4m + 1/3, 5m + 2/3). Direction (3, 4, 5), point C(0, 1/3, 2/3).

With A(1, 2, 3): AC = (−1, −5/3, −7/3) ∝ (−3, −5, −7).

```text
| −3  −5  −7 |
|  2   3   4 | = −3(15 − 16) + 5(10 − 12) − 7(8 − 9)
|  3   4   5 |
             = 3 − 10 + 7 = 0
```

#### Point of intersection
With t = −1: (1 − 2, 2 − 3, 3 − 4) = (−1, −1, −1).

**Answer: the lines are coplanar and intersect at (−1, −1, −1).**

## Check yourself

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

1. How is intersection tested?
   A. Substitute a general point of L1 into both planes of L2 and compare t
   B. Compute the angle between lines
   C. Compute the cross product of the two plane normals only
   D. Set x = y = z
   Answer: A. If the same t works for both planes, the point lies on L2. (Key point: Method)

2. What is t?
   A. 1
   B. -2
   C. 0
   D. -1
   Answer: D. -1 - t = 0 and -2 - 2t = 0. (Key point: Parameter)

3. What is the point of intersection?
   A. (1, 1, 1)
   B. (-1, -1, -1)
   C. (-1, -1, 1)
   D. (3, 5, 7)
   Answer: B. With t = -1: (1 - 2, 2 - 3, 3 - 4). (Key point: Intersection)

4. What does the determinant check give?
   A. 0, confirming the lines are coplanar
   B. 1
   C. -1
   D. a non-zero value
   Answer: A. -3(-1) + 5(-2) - 7(-1) = 3 - 10 + 7 = 0. (Key point: Check)

## Close variants set in other exams

- [Coplanar lines and point of intersection](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Acoplanar-lines-point-intersection) — Mathematics, chapter 3, asked 2×
- [Coplanar lines and common point](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Acoplanar-lines-common-point) — Mathematics, chapter 3, asked 2×

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