---
title: "Determinant form of plane through point and line"
subject: "Mathematics"
chapter: 3
key: "semester_2/mathematics:determinant-form-plane-through-point"
source: https://www.bctengineer.com/q/semester_2/mathematics/determinant-form-plane-through-point
site: BCT Engineering
---

# Determinant form of plane through point and line

Mathematics · Chapter 3: Three Dimensional Solid Geometry

- Asked in: 2068 Bhadra, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/determinant-form-plane-through-point
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-3

## Question

Prove that the plane through the point (α,β,γ) and the line x = py+q = rz+s is given by the determinant | x  py+q  rz+s ; α  pβ+q  rγ+s ; 1  1  1 | = 0.

## Short answer

The required plane is given by the determinant | x py+q rz+s ; α pβ+q rγ+s ; 1 1 1 | = 0, which is linear and vanishes on the line (rows 1 and 3 proportional) and at the point (rows 1 and 2 equal).

## Answer

A plane through a given line is a member of the pencil of planes through that line; making it pass through the given point fixes the member.

#### The line as two planes
x = py + q = rz + s means x − (py + q) = 0 and x − (rz + s) = 0. Write

u = x, v = py + q, w = rz + s.

The line is u = v = w.

#### Plane through the line
Any plane through the line is (u − v) + λ(u − w) = 0, i.e.

(1 + λ)u − v − λw = 0.

Its coefficients of u, v, w are (1 + λ, −1, −λ), and they add up to 0. It is a plane because u, v, w are linear in x, y, z.

#### The determinant
Let u₀ = α, v₀ = pβ + q, w₀ = rγ + s for the point (α, β, γ). Expand

```text
| u   v   w  |
| u₀  v₀  w₀ |  = u(v₀ − w₀) − v(u₀ − w₀) + w(u₀ − v₀) = 0
| 1   1   1  |
```
Properties of this equation:
1. It is linear in u, v, w, so it represents a plane.
2. The coefficients (v₀ − w₀), −(u₀ − w₀), (u₀ − v₀) add up to 0, like the pencil above, so it belongs to the pencil of planes through the line (put u = v = w = t: the first and third rows become proportional, so the determinant is zero for every point of the line).
3. Put (u, v, w) = (u₀, v₀, w₀): rows 1 and 2 are equal, so the determinant is 0. The point (α, β, γ) lies on the plane.

A plane through the line and the point is unique, hence the plane required is

**| x   py+q   rz+s ; α   pβ+q   rγ+s ; 1   1   1 | = 0.**

## Check yourself

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

1. Planes through the line u = v = w have the form:
   A. u + v + w = λ
   B. (u - v) + λ(u - w) = 0
   C. uvw = λ
   D. (u - v)(u - w) = 0
   Answer: B. A combination of the two planes defining the line. (Key point: Pencil of planes)

2. Why does the determinant equation represent a plane?
   A. It is quadratic in x
   B. It contains xyz
   C. It is linear in u, v, w, which are linear in x, y, z
   D. It equals 1
   Answer: C. Expanding along row 1 gives a linear combination of u, v, w. (Key point: Linear plane)

3. Why does (α, β, γ) satisfy the equation?
   A. Substituting gives two equal rows (row 1 = row 2), so the determinant is 0
   B. The third row becomes zero
   C. The columns are all zero
   D. The point is the origin
   Answer: A. A determinant with two identical rows vanishes. (Key point: Point on plane)

4. Why does every point of the line lie on the plane?
   A. The first row becomes zero
   B. When u = v = w the first row is proportional to the third, so the determinant is 0
   C. The second row is zero
   D. Because α = β = γ
   Answer: B. Rows (t, t, t) and (1, 1, 1) are proportional. (Key point: Line on plane)

## Related in Mathematics

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- [Distance of point from plane parallel to line](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Adistance-point-from-plane-parallel) — Mathematics, chapter 3
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- [Sphere through circle touching a plane](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Asphere-through-circle-touching-plane) — Mathematics, chapter 3

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