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

# Coplanar lines and plane containing them

Mathematics · Chapter 3: Three Dimensional Solid Geometry

- Asked in: 2069 Poush, 5 marks
- Video: [Show that the lines `(x-3)/2=(y+1)/(-3)=(z+2)/1 and (x-7)/(-3)=y/1=(z+7)/2` are coplanar. Also find](https://www.youtube.com/watch?v=hQk75QbNNdc) — Doubtnut
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/coplanar-lines-plane-containing-them
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-3

## Question

Prove that the lines (x+1)/(−3) = (y−3)/2 = (z+2)/1 and x = (y−7)/(−3) = (z+7)/2 are coplanar and find the equation of plane in which they lie.

## Short answer

The lines are coplanar because the scalar triple product AB·(a×b)=0. The plane containing them has normal a×b = (1,1,1) and passes through (-1,3,-2), giving x+y+z=0.

## Answer

Two lines are coplanar if the vector joining a point on each is perpendicular to the cross product of their directions, i.e. the scalar triple product is zero.

Line 1: (x+1)/(−3) = (y−3)/2 = (z+2)/1, point A(−1, 3, −2), direction a = (−3, 2, 1).
Line 2: x/1 = (y−7)/(−3) = (z+7)/2, point B(0, 7, −7), direction b = (1, −3, 2).

#### Proof of coplanarity
AB = B − A = (1, 4, −5).

```text
| 1   4  −5 |
|−3   2   1 |  = 1(4+3) − 4(−6−1) + (−5)(9−2)
| 1  −3   2 |  = 7 + 28 − 35 = 0
```
The determinant is zero, so the lines are coplanar. (They are not parallel since a and b are not proportional, so they intersect.)

#### Equation of the plane
The normal is perpendicular to both directions:

```text
a × b = (2·2 − 1·(−3), 1·1 − (−3)·2, (−3)(−3) − 2·1)
      = (7, 7, 7)  ∥  (1, 1, 1)
```
The plane through A with normal (1, 1, 1):

(x + 1) + (y − 3) + (z + 2) = 0, i.e. **x + y + z = 0**.

Check: B(0, 7, −7) gives 0 + 7 − 7 = 0, so the second line's point is also in the plane.

**Answer: the lines are coplanar and lie in the plane x + y + z = 0.**

## Check yourself

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

1. Two lines are coplanar when:
   A. The scalar triple product of AB, a and b is zero
   B. a . b = 0
   C. AB = a x b
   D. a and b are parallel to AB
   Answer: A. AB must lie in the plane spanned by a and b. (Key point: Coplanarity criterion)

2. With AB = (1, 4, -5), a = (-3, 2, 1), b = (1, -3, 2), the determinant is:
   A. 7
   B. 0
   C. -35
   D. 35
   Answer: B. 7 + 28 - 35 = 0. (Key point: Determinant)

3. The normal to the plane is:
   A. a . b
   B. AB x a
   C. a x b = (7, 7, 7), parallel to (1, 1, 1)
   D. a + b
   Answer: C. A plane containing both lines has normal perpendicular to both directions. (Key point: Normal vector)

4. The plane through A(-1, 3, -2) with normal (1, 1, 1) is:
   A. x + y + z = 0
   B. x + y + z = 1
   C. x - y + z = 0
   D. x + y - z = 0
   Answer: A. (x+1)+(y-3)+(z+2) = x+y+z = 0. (Key point: Plane)

## Close variants set in other exams

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

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