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

# Coplanar lines and point of intersection

Mathematics · Chapter 3: Three Dimensional Solid Geometry

- Asked in: 2075 Baisakh, 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
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-3

## Question

Show that (x+4)/3 = (y+6)/5 = (z−1)/(−2) and 3x−2y+z+5 = 0 = 2x+3y+4z−4 are coplanar lines and find the point of intersection.

## Short answer

Putting the parametric point of the first line into the planes defining the second yields t=2; the lines intersect at (2, 4, −3), thus are coplanar.

## Answer

**Lines.**
- L₁: (x + 4)/3 = (y + 6)/5 = (z − 1)/(−2).
- L₂: 3x − 2y + z + 5 = 0 and 2x + 3y + 4z − 4 = 0 (a line given as two planes).

#### Approach
Two lines are coplanar if they intersect (or are parallel). We test whether L₁ meets L₂ by putting a general point of L₁ into both planes of L₂.

A general point of L₁: (3t − 4, 5t − 6, 1 − 2t).

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

Plane 2: 2(3t − 4) + 3(5t − 6) + 4(1 − 2t) − 4 = 0
         6t − 8 + 15t − 18 + 4 − 8t − 4 = 0
         13t − 26 = 0  →  t = 2
```

Both planes give the same value t = 2, so the point of L₁ with t = 2 lies on both planes, i.e. on L₂. The lines meet, so they are coplanar. (They are not parallel: L₂ has direction (3, −2, 1) × (2, 3, 4) = (−11, −10, 13), which is not proportional to (3, 5, −2).)

#### Point of intersection
With t = 2: x = 3(2) − 4 = 2, y = 5(2) − 6 = 4, z = 1 − 4 = −3.

Check in L₂: 3(2) − 2(4) + (−3) + 5 = 0 ✓ and 2(2) + 3(4) + 4(−3) − 4 = 0 ✓.

**Answer: the lines are coplanar and intersect at (2, 4, −3).**

## Check yourself

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

1. How is coplanarity shown here?
   A. By showing that the lines intersect at a common point
   B. By showing d1 . d2 = 0
   C. By showing both lines pass through the origin
   D. By showing both directions are equal
   Answer: A. Two lines that meet lie in a common plane. (Key point: Coplanarity test)

2. How is the test applied?
   A. Solve the two planes for x only
   B. Put a general point of L1 into both planes of L2
   C. Compute the angle between L1 and the planes
   D. Put the origin into both planes
   Answer: B. If both planes give the same t, that point is on L2. (Key point: Method)

3. What do both planes give for t?
   A. t = 1
   B. t = -2
   C. t = 2
   D. t = 0
   Answer: C. -3t + 6 = 0 and 13t - 26 = 0 both give t = 2. (Key point: Parameter)

4. What is the point of intersection?
   A. (2, 4, -3)
   B. (2, 4, 3)
   C. (-2, 4, -3)
   D. (4, 2, -3)
   Answer: A. x = 3(2) - 4, y = 5(2) - 6, z = 1 - 4. (Key point: Intersection)

## Close variants set in other exams

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