---
title: "Line of intersection lies on cone"
subject: "Mathematics"
chapter: 3
key: "semester_2/mathematics:line-intersection-lies-cone"
source: https://www.bctengineer.com/q/semester_2/mathematics/line-intersection-lies-cone
site: BCT Engineering
---

# Line of intersection lies on cone

Mathematics · Chapter 3: Three Dimensional Solid Geometry

- Asked in: 2073 Bhadra, 5 marks; 2072 Magh, 5 marks; 2067 Mangsir, 5 marks (asked 3 times)
- Video: [The line of intersection lies on the cone equation, planes through ox, oy includes on angle alpha](https://www.youtube.com/watch?v=5pQbk7iELA8) — Knowledge Never Betrays
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/line-intersection-lies-cone
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-3

## Question

The plane through OX and OY includes an angle α, prove that their line of intersection lies on the cone z²(x²+y²+z²) = x²y² tan²α.

## Short answer

Planes through x-axis (y=pz) and y-axis (x=qz) intersect along x=qz, y=pz. The angle α satisfies cos α = pq/√((1+p²)(1+q²)). Substituting p=y/z, q=x/z and eliminating roots yields z²(x²+y²+z²) = x²y² tan²α, the line lies on this cone.

## Answer

**How I read the question.** "The plane through OX and OY" is a misprint for two planes, one through OX and one through OY, that include an angle α. Only one plane (z = 0) contains both axes, so there would be no line of intersection. With this reading the result is exactly the stated cone.

#### Setting up the planes
- A plane through the x-axis OX has equation y = p z (it contains every point (x, 0, 0)).
- A plane through the y-axis OY has equation x = q z.

Normals: n₁ = (0, 1, −p) and n₂ = (1, 0, −q).

#### Angle between the planes
```text
cos α = n₁·n₂ / (|n₁||n₂|) = pq / √((1 + p²)(1 + q²))
```

#### Line of intersection
Both equations hold on the line of intersection, x = q z, y = p z. For any point (x, y, z) on it, p = y/z and q = x/z. Put these in cos α:

```text
cos α = (xy/z²) / √((1 + y²/z²)(1 + x²/z²))
      = xy / √((y² + z²)(x² + z²))
```

#### Eliminate the roots
```text
cos²α = x²y² / [(y² + z²)(x² + z²)]

tan²α = (1 − cos²α)/cos²α
      = [(y² + z²)(x² + z²) − x²y²] / (x²y²)
      = z²(x² + y² + z²) / (x²y²)
```

since (y² + z²)(x² + z²) − x²y² = z²(x² + y² + z²).

So the line of intersection lies on

**z²(x² + y² + z²) = x²y² tan²α.** ∎

## Check yourself

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

1. Which equations represent a plane through OX and a plane through OY?
   A. y = p x and z = q y
   B. y = p z and x = q z
   C. x = p z and y = q z both through OX
   D. z = p y and z = q x
   Answer: B. y = pz contains every point (x, 0, 0), and x = qz contains every point (0, y, 0). (Key point: Plane equations)

2. What is cos alpha between normals (0, 1, -p) and (1, 0, -q)?
   A. (p + q) / sqrt((1 + p^2)(1 + q^2))
   B. 1 / sqrt((1 + p^2)(1 + q^2))
   C. pq / sqrt((1 + p^2)(1 + q^2))
   D. pq / ((1 + p^2)(1 + q^2))
   Answer: C. The dot product is pq and the magnitudes are sqrt(1 + p^2) and sqrt(1 + q^2). (Key point: Angle between planes)

3. Which substitution turns the condition into the cone?
   A. p = z/y and q = z/x
   B. p = y/z and q = x/z on the line of intersection, then compute tan^2 alpha
   C. p = y and q = x
   D. p = x/y and q = y/x
   Answer: B. On x = qz, y = pz, this gives tan^2 alpha = z^2(x^2 + y^2 + z^2)/(x^2 y^2). (Key point: Eliminating p and q)

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