Equation of a Cone?

We are going to use Geogebra 3D calculator to draw. https://www.geogebra.org/calculator

How to draw a cone in a 3 dimensional space of x, y and z axis?

Starting with an equation of a straight line, r(λ), passing through the corner of the cone to center of the base.

r(λ) = (c₁, c₂, c₃) + λ (d₁, d₂, d₃), where (c₁, c₂, c₃) is the coordinate of the corner of the cone, (d₁, d₂, d₃) is a directional vector, λ is a parameter.

Given (x, y, z) = any point on the curved surface of the cone. Their distance, D, would be the perpendicular distance measured from point (x, y, z) to the line r(λ).

Let vector A = (d₁, d₂, d₃), vector B = (x - c₁, y - c₂, z - c₃)

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Using formula |A x B| = |A||B| sin Θ, distance D = |B| sin Θ, rearrange:
|A x B| = |A||B| sin Θ
|A x B| = |A|D
D = |A x B| / |A|

A x B = (d₂(z - r) - d₃(y - q), d₃(x - p) - d₁(z - r), d₁(y - q) - d₂(x - p))
|A x B| = √((d₂ (z - r) - d₃(y - q))² + (d₃(x - p) - d₁(z - r))² + (d₁(y - q) - d₂(x - p))²)
|A| = √(d₁² + d₂² + d₃²)

D = √((d₂ (z - r) - d₃(y - q))² + (d₃(x - p) - d₁(z - r))² + (d₁(y - q) - d₂(x - p))²) / √(d₁² + d₂² + d₃²)