---
title: "Right circular cone with vertex (1,1,1), semi-vertical 30°"
subject: "Mathematics"
chapter: 3
key: "semester_2/mathematics:right-circular-cone-vertex-1"
source: https://www.bctengineer.com/q/semester_2/mathematics/right-circular-cone-vertex-1
site: BCT Engineering
---

# Right circular cone with vertex (1,1,1), semi-vertical 30°

Mathematics · Chapter 3: Three Dimensional Solid Geometry

- Asked in: 2070 Magh, 5 marks
- Video: [find right circular cone with given conditions | 6 th sem maths important question #degree #mpc](https://www.youtube.com/watch?v=gbqsYq6NuZw) — Degree Empire
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/right-circular-cone-vertex-1
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-3

## Question

Find the equation of a right circular cone with vertex (1,1,1) and axis is the line (x−1)/1 = (y−1)/2 = (z−1)/3 and semi vertical angle 30°.

## Answer

**Given.** Vertex V(1, 1, 1), axis (x − 1)/1 = (y − 1)/2 = (z − 1)/3, semi-vertical angle α = 30°.

#### Condition
For a point P(x, y, z) on the cone, the line VP makes angle α with the axis. Write X = x − 1, Y = y − 1, Z = z − 1, so VP = (X, Y, Z), and the axis direction is (1, 2, 3) with length √14.

```text
cos α = (X + 2Y + 3Z) / ( √14 · √(X² + Y² + Z²) )
(X + 2Y + 3Z)² = 14 cos²α (X² + Y² + Z²)
```

With α = 30°, cos²α = 3/4, so 14 cos²α = 21/2:

```text
2(X + 2Y + 3Z)² = 21 (X² + Y² + Z²)
```

#### Expand
(X + 2Y + 3Z)² = X² + 4Y² + 9Z² + 4XY + 12YZ + 6ZX.

```text
2X² + 8Y² + 18Z² + 8XY + 24YZ + 12ZX = 21X² + 21Y² + 21Z²
19X² + 13Y² + 3Z² − 8XY − 24YZ − 12ZX = 0
```

#### In terms of x, y, z
Replace X, Y, Z by x − 1, y − 1, z − 1 and expand:

**19x² + 13y² + 3z² − 8xy − 24yz − 12zx − 18x + 6y + 30z − 9 = 0**

Check: the vertex (1, 1, 1) gives 19 + 13 + 3 − 8 − 24 − 12 − 18 + 6 + 30 − 9 = 0 ✓.

## Check yourself

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

1. What condition puts a point P on the right circular cone?
   A. VP makes the semi-vertical angle with the axis
   B. VP is perpendicular to the axis
   C. P is at constant distance from the vertex
   D. VP is parallel to the axis
   Answer: A. Every generator makes the angle alpha with the axis. (Key point: Defining condition)

2. Why put X = x-1, Y = y-1, Z = z-1?
   A. To make the axis the z-axis
   B. To measure VP from the vertex (1, 1, 1)
   C. To make the cone symmetric
   D. To remove the semi-vertical angle
   Answer: B. VP = (X, Y, Z) is the vector from the vertex to P. (Key point: Shifted coordinates)

3. For alpha = 30° and axis (1, 2, 3), what equation results?
   A. (X+2Y+3Z)^2 = 14(X^2+Y^2+Z^2)
   B. 4(X+2Y+3Z)^2 = 21(X^2+Y^2+Z^2)
   C. 2(X+2Y+3Z)^2 = 21(X^2+Y^2+Z^2)
   D. 2(X+2Y+3Z)^2 = 7(X^2+Y^2+Z^2)
   Answer: C. (X+2Y+3Z)^2 = 14 cos^2(30°)(...) = (21/2)(...). (Key point: Squared cosine step)

4. A quick check of the final cone equation is that:
   A. The vertex (1, 1, 1) satisfies it
   B. The origin satisfies it
   C. The constant term is 1
   D. It contains no xy term
   Answer: A. The vertex lies on the cone, and substitution gives 0. (Key point: Check)

## Close variants set in other exams

- [Right circular cone, vertical angle 30°](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Aright-circular-cone-vertical-angle) — Mathematics, chapter 3, asked 3×

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