Matrix diagonalisation

A didactic proposal for applying matrix diagonalisation:

Parametrising the intersection of a sphere and plane with the aid of GeoGebra

Work done in collaboration with Bradley Welch

Part 1:

Context and problem formulation

Main Problem: Find the parametrisation of the intersection curve of a sphere and a plane.

What are the conditions for a sphere and a plane to intersect?

How can we determine the parametric equations of the intersection curve of these geometric objects?

Part 1: Context and problem formulation

What are the conditions for a sphere and a plane to intersect?

S: (x-x_0)^2+(y-y_0)^2+(z-z_0)^2=R^2

S: (x-x_0)^2+(y-y_0)^2+(z-z_0)^2=R^2

\Pi: Ax+By+Cz=D

\Pi: Ax+By+Cz=D

Sphere of radius $R>0,$ centred at $(x_0,y_0,z_0)$ :

Plane defined by:

$A,B$ and $C$ real constants, not all simultaneaously zero.

Part 1: Context and problem formulation

What are the conditions for a sphere and a plane to intersect?

S: (x-x_0)^2+(y-y_0)^2+(z-z_0)^2=R^2

S: (x-x_0)^2+(y-y_0)^2+(z-z_0)^2=R^2

\Pi: Ax+By+Cz=D

\Pi: Ax+By+Cz=D

Let $\rho$ be the shortest signed distance between the centre of the sphere and the plane.

\rho = \dfrac{Ax_0+By_0+ Cy_0-D}{\sqrt{A^2+B^2+C^2}}

\rho = \dfrac{Ax_0+By_0+ Cy_0-D}{\sqrt{A^2+B^2+C^2}}

That is,

Part 1: Context and problem formulation

What are the conditions for a sphere and a plane to intersect?

S: (x-x_0)^2+(y-y_0)^2+(z-z_0)^2=R^2

S: (x-x_0)^2+(y-y_0)^2+(z-z_0)^2=R^2

\Pi: Ax+By+Cz=D

\Pi: Ax+By+Cz=D

\rho = \dfrac{Ax_0+By_0+ Cy_0-D}{\sqrt{A^2+B^2+C^2}}

\rho = \dfrac{Ax_0+By_0+ Cy_0-D}{\sqrt{A^2+B^2+C^2}}

We can set the following cases for the intersection $\Pi\cap S$ :

1. If $|\rho|<R,$ then $\Pi\cap S$ is a circle.

2. If $|\rho|=R,$ then $\Pi\cap S$ is a point.

3. If $|\rho|>R,$ then $\Pi\cap S$ is a empty.

Part 1: Context and problem formulation

What are the conditions for a sphere and a plane to intersect?

Part 1: Context and problem formulation

Find the parametrisation of the intersection curve of a sphere and a plane.

X=\left(x_c,y_c,z_c\right)+\left(C_1 \cos(t)+ C_2\sin(t), C_3 \cos(t)+ C_4\sin(t), C_5\sin(t)\right)

X=\left(x_c,y_c,z_c\right)+\left(C_1 \cos(t)+ C_2\sin(t), C_3 \cos(t)+ C_4\sin(t), C_5\sin(t)\right)

\left\{ \begin{array}{l} x=C_1 \cos(t)+ C_2\sin(t)+x_c\\ y= C_3 \cos(t)+ C_4\sin(t)+y_c\\ z=C_5\sin(t)+z_c \end{array} \right.

\left\{ \begin{array}{l} x=C_1 \cos(t)+ C_2\sin(t)+x_c\\ y= C_3 \cos(t)+ C_4\sin(t)+y_c\\ z=C_5\sin(t)+z_c \end{array} \right.

Conjecture: The solution provided by Geogebra takes the general vector form given as

The dynamic exploration in GeoGebra with different semiotic representations provides some clues to solve our main problem:

Part 1: Context and problem formulation

\left\{ \begin{array}{l} x=C_1 \cos(t)+ C_2\sin(t)+x_c\\ y= C_3 \cos(t)+ C_4\sin(t)+y_c\\ z=C_5\sin(t)+z_c \end{array} \right.

\left\{ \begin{array}{l} x=C_1 \cos(t)+ C_2\sin(t)+x_c\\ y= C_3 \cos(t)+ C_4\sin(t)+y_c\\ z=C_5\sin(t)+z_c \end{array} \right.

But how does GeoGebra compute this parametrisation?

What are the mathematical concepts or relationships necessary to determine the values for the constants $C_k, \;x_c, y_c, z_c$ ?

Conjecture: The solution provided by Geogebra takes the general vector form given as

Part 2: Learning trajectory

Problem 1: Find the parametric equation of the curve obtained by intersecting the sphere $S:x^2 +y^2 +z^2 =1$ and the plane $\Pi:z= \dfrac{1}{2}.$

Find the parametrisation of the intersection curve of a sphere and a plane.

Before solving the general problem:

Let's try to relax the conditions and solve the simplest case:

Part 2: Learning trajectory

Problem 1: Find the parametric equation of the curve obtained by intersecting the sphere $S:x^2 +y^2 +z^2 =1$ and the plane $\Pi:z= \dfrac{1}{2}.$

Substitute $z=\dfrac{1}{2}$ in the equation of the sphere:

$x^2+y^2+\left(\dfrac{1}{2}\right)^2=1$

Using the identity $\;\cos^2(t)+\sin^2(t)=1$

$\Rightarrow\;\dfrac{4}{3}x^2+\dfrac{4}{3}y^2=1$

\Rightarrow \left\{ \begin{array}{rcl} x&=&\dfrac{\sqrt{3}}{2}\cos(t) \\ \\ y&=&\dfrac{\sqrt{3}}{2}\sin(t) \\ \\ z&=&\dfrac{1}{2} \end{array} \right. \qquad t\in[0, 2\pi].

\Rightarrow \left\{ \begin{array}{rcl} x&=&\dfrac{\sqrt{3}}{2}\cos(t) \\ \\ y&=&\dfrac{\sqrt{3}}{2}\sin(t) \\ \\ z&=&\dfrac{1}{2} \end{array} \right. \qquad t\in[0, 2\pi].

Part 2: Learning trajectory

\Rightarrow \left\{ \begin{array}{rcl} x&=&\dfrac{\sqrt{3}}{2} \cos(t) \\ y&=&\dfrac{\sqrt{3}}{2}\sin(t) \\ z&=&\dfrac{1}{2} \end{array} \right. \qquad t\in[0, 2\pi].

\Rightarrow \left\{ \begin{array}{rcl} x&=&\dfrac{\sqrt{3}}{2} \cos(t) \\ y&=&\dfrac{\sqrt{3}}{2}\sin(t) \\ z&=&\dfrac{1}{2} \end{array} \right. \qquad t\in[0, 2\pi].

Part 2: Learning trajectory

\Rightarrow \left\{ \begin{array}{rcl} x&=&\dfrac{\sqrt{3}}{2} \cos(t) \\ y&=&\dfrac{\sqrt{3}}{2}\sin(t) \\ z&=&\dfrac{1}{2} \end{array} \right. \qquad t\in[0, 2\pi].

\Rightarrow \left\{ \begin{array}{rcl} x&=&\dfrac{\sqrt{3}}{2} \cos(t) \\ y&=&\dfrac{\sqrt{3}}{2}\sin(t) \\ z&=&\dfrac{1}{2} \end{array} \right. \qquad t\in[0, 2\pi].

Part 2: Learning trajectory

Problem 2: Compute the parametric equations of the curve obtained by intersecting the sphere and plane given by

$S:x^2 +y^2 +z^2 =4\;$ and $\;\Pi:2x+2y+z=-3,$

respectively.

Rewrite the plane as $\;z=-3-2x-2y$

Then we obtain $\;x^2+y^2+\left(-3-2-x-2y\right)^2=4$

That is $\;5x^2+8xy+5y^2+12x+12y+5=0$

This is a conic section, also known as a quadratic form in $\mathbb R^2,$ and can be written in matrix form as

$\mathbf x^T M \mathbf x+ K \mathbf x+f=0$

Now let's try a slightly more difficult problem!

Part 2: Learning trajectory

Problem 2: Compute the parametric equations of the curve obtained by intersecting the sphere and plane given by

$S:x^2 +y^2 +z^2 =4\;$ and $\;\Pi:2x+2y+z=-3,$

respectively.

That is $\;5x^2+8xy+5y^2+12x+12y+5=0$

Let $\mathbf{x}=\begin{pmatrix}x\\y\end{pmatrix}.$

$\mathbf x^T M \mathbf x+ K \mathbf x+f=0$

Then we have

\begin{pmatrix} x & y \end{pmatrix} \begin{pmatrix} 5&4\\4&5 \end{pmatrix} \begin{pmatrix}x\\y \end{pmatrix}+ \begin{pmatrix} 12&12\end{pmatrix} \begin{pmatrix}x\\ y \end{pmatrix}+5=0.

\begin{pmatrix} x & y \end{pmatrix} \begin{pmatrix} 5&4\\4&5 \end{pmatrix} \begin{pmatrix}x\\y \end{pmatrix}+ \begin{pmatrix} 12&12\end{pmatrix} \begin{pmatrix}x\\ y \end{pmatrix}+5=0.

By means of a rotation of the plane about the origin, a translation of the plane, or both, it is possible to represent every conic in a simplified standard, or canonical, form.

Part 2: Learning trajectory

That is $\;5x^2+8xy+5y^2+12x+12y+5=0$

$\mathbf x^T M \mathbf x+ K \mathbf x+f=0$

By means of a rotation of the plane about the origin, a translation of the plane, or both, it is possible to represent every conic in a simplified standard, or canonical, form.

This process is known as diagonalisation of quadratic forms.

Let $\mathbf{x}=\begin{pmatrix}x\\y\end{pmatrix}.$

Then we have

\begin{pmatrix} x & y \end{pmatrix} \begin{pmatrix} 5&4\\4&5 \end{pmatrix} \begin{pmatrix}x\\y \end{pmatrix}+ \begin{pmatrix} 12&12\end{pmatrix} \begin{pmatrix}x\\ y \end{pmatrix}+5=0.

\begin{pmatrix} x & y \end{pmatrix} \begin{pmatrix} 5&4\\4&5 \end{pmatrix} \begin{pmatrix}x\\y \end{pmatrix}+ \begin{pmatrix} 12&12\end{pmatrix} \begin{pmatrix}x\\ y \end{pmatrix}+5=0.

Problem 2: Compute the parametric equations of the curve obtained by intersecting the sphere and plane given by

$S:x^2 +y^2 +z^2 =4\;$ and $\;\Pi:2x+2y+z=-3,$

respectively.

Part 2: Learning trajectory

That is $\;5x^2+8xy+5y^2+12x+12y+5=0$

$\mathbf x^T M \mathbf x+ K \mathbf x+f=0$

Let $\mathbf{x}=\begin{pmatrix}x\\y\end{pmatrix}.$

Then we have

The diagonalisation process and a change of variables provides the following expression:

9u^2+v^2+12\sqrt{2}u+5=0

9u^2+v^2+12\sqrt{2}u+5=0

the simplified standard form of

Problem 2: Compute the parametric equations of the curve obtained by intersecting the sphere and plane given by

$S:x^2 +y^2 +z^2 =4\;$ and $\;\Pi:2x+2y+z=-3,$

respectively.

\begin{pmatrix} x & y \end{pmatrix} \begin{pmatrix} 5&4\\4&5 \end{pmatrix} \begin{pmatrix}x\\y \end{pmatrix}+ \begin{pmatrix} 12&12\end{pmatrix} \begin{pmatrix}x\\ y \end{pmatrix}+5=0.

\begin{pmatrix} x & y \end{pmatrix} \begin{pmatrix} 5&4\\4&5 \end{pmatrix} \begin{pmatrix}x\\y \end{pmatrix}+ \begin{pmatrix} 12&12\end{pmatrix} \begin{pmatrix}x\\ y \end{pmatrix}+5=0.

Part 2: Learning trajectory

Using again the identity $\cos^2(t)+\sin^2(t)=1$ and rewriting everything back with the variables $x,y$ and $z;$

\left\{ \begin{array} {rcl} x &=&\dfrac{\sqrt{6}}{6}\cos(t)-\dfrac{\sqrt{6}}{2}\sin(t)-\dfrac{2}{3} \\ \\ y &=&\dfrac{\sqrt{6}}{6}\cos(t)+\dfrac{\sqrt{6}}{2}\sin(t)-\dfrac{2}{3} \\ \\ z &=&-\dfrac{2\sqrt{6}}{3}\cos(t)-\dfrac{1}{3} \end{array} \right. \qquad t\in[0, 2\pi].

\left\{ \begin{array} {rcl} x &=&\dfrac{\sqrt{6}}{6}\cos(t)-\dfrac{\sqrt{6}}{2}\sin(t)-\dfrac{2}{3} \\ \\ y &=&\dfrac{\sqrt{6}}{6}\cos(t)+\dfrac{\sqrt{6}}{2}\sin(t)-\dfrac{2}{3} \\ \\ z &=&-\dfrac{2\sqrt{6}}{3}\cos(t)-\dfrac{1}{3} \end{array} \right. \qquad t\in[0, 2\pi].

Problem 2: Compute the parametric equations of the curve obtained by intersecting the sphere and plane given by

$S:x^2 +y^2 +z^2 =4\;$ and $\;\Pi:2x+2y+z=-3,$

respectively.

we obtain the solution:

Part 2: Learning trajectory

\left\{ \begin{array} {rcl} x &=&\dfrac{\sqrt{6}}{6}\cos(t)-\dfrac{\sqrt{6}}{2}\sin(t)-\dfrac{2}{3} \\ y &=&\dfrac{\sqrt{6}}{6}\cos(t)+\dfrac{\sqrt{6}}{2}\sin(t)-\dfrac{2}{3} \\ z &=&-\dfrac{2\sqrt{6}}{3}\cos(t)-\dfrac{1}{3} \end{array} \right. \qquad t\in[0, 2\pi].

\left\{ \begin{array} {rcl} x &=&\dfrac{\sqrt{6}}{6}\cos(t)-\dfrac{\sqrt{6}}{2}\sin(t)-\dfrac{2}{3} \\ y &=&\dfrac{\sqrt{6}}{6}\cos(t)+\dfrac{\sqrt{6}}{2}\sin(t)-\dfrac{2}{3} \\ z &=&-\dfrac{2\sqrt{6}}{3}\cos(t)-\dfrac{1}{3} \end{array} \right. \qquad t\in[0, 2\pi].

Part 2: Learning trajectory

Problem 3: Find the parametrisation of the intersection curve of the sphere and plane $\left\{ \begin{array} {rl} S: &(x-x_0)^2+(y-y_0)^2+(z-z_0)^2=R^2 \\ \\ \Pi: & Ax+By+Cz=D, \\ \end{array} \right.$ where $A,B,C$ are not all simultaneously zero.

General problem!

To simplify some calculations, first we translate both objects to the origin: $\left\{ \begin{array} {rl} S_0: &x^2+y^2+z^2=R^2 \\ \\ \Pi_0: & A(x+x_0)+B(y+y_0)+C(z+z_0)=D \\ \end{array} \right.$

Part 2: Learning trajectory

Substituing the equation of the plane $\Pi_0$ into the sphere $S_0$ yields

$\left\{ \begin{array} {rl} S_0: &x^2+y^2+z^2=R^2 \\ \\ \Pi_0: & A(x+x_0)+B(y+y_0)+C(z+z_0)=D \\ \end{array} \right.$

x^2+y^2+(\alpha_0-\alpha_1x-\alpha_2y -\alpha_3)^2=R^2

x^2+y^2+(\alpha_0-\alpha_1x-\alpha_2y -\alpha_3)^2=R^2

\text{where}\quad \alpha_0=\frac{D}{C}, \quad \alpha_1=\frac{A}{C}, \quad \alpha_2=\frac{B}{C} \quad \text{and} \quad \alpha_3=z_0+ \frac{A}{C} x_0 + \frac{B}{C}{y_0}

\text{where}\quad \alpha_0=\frac{D}{C}, \quad \alpha_1=\frac{A}{C}, \quad \alpha_2=\frac{B}{C} \quad \text{and} \quad \alpha_3=z_0+ \frac{A}{C} x_0 + \frac{B}{C}{y_0}

Part 2: Learning trajectory

x^2+y^2+(\alpha_0-\alpha_1x-\alpha_2y -\alpha_3)^2=R^2

x^2+y^2+(\alpha_0-\alpha_1x-\alpha_2y -\alpha_3)^2=R^2

This expression is in fact a quadratic form:

$ax^2+bxy+cy^2+dx+ey+f=0$

So we can rewrite it as

$\mathbf{x}^{T} M\mathbf{x}+K\mathbf{x}+f=0$

Thus we can apply the process of diagonalisation

to obtain the solution of the general problem.

Part 2: Learning trajectory

Thus we can apply the process of diagonalisation

to obtain the solution of the general problem.

\left\{ \begin{array} {rcl} x &=& p_{1,1}u+p_{1,2}v \\ \\ y &=&p_{2,1}u+p_{2,2}v \\ \\ z &=&\alpha_0-\alpha_1x-\alpha_2y -\alpha_3 \end{array} \right. \quad\quad t\in[0, 2\pi].

\left\{ \begin{array} {rcl} x &=& p_{1,1}u+p_{1,2}v \\ \\ y &=&p_{2,1}u+p_{2,2}v \\ \\ z &=&\alpha_0-\alpha_1x-\alpha_2y -\alpha_3 \end{array} \right. \quad\quad t\in[0, 2\pi].

+\,x_0

+\,x_0

+\,y_0

+\,y_0

+\,z_0

+\,z_0

Part 3: Final comments

The sequence of problems mentioned before can be adapted within a problem-solving learning environment for:
- discussing the geometric properties of the intersection curve of the sphere and plane,

S: (x-x_0)^2+(y-y_0)^2+(z-z_0)^2=R^2

S: (x-x_0)^2+(y-y_0)^2+(z-z_0)^2=R^2

\Pi: Ax+By+Cz=D

\Pi: Ax+By+Cz=D

\rho = \dfrac{Ax_0+By_0+ Cy_0-D}{\sqrt{A^2+B^2+C^2}}

\rho = \dfrac{Ax_0+By_0+ Cy_0-D}{\sqrt{A^2+B^2+C^2}}

Part 3: Final comments

The sequence of problems mentioned before can be adapted within a problem-solving learning environment for:
- discussing the geometric properties of the intersection curve of the sphere and plane,
- establishing mathematical connections, in this case, with the process of diagonalisation of quadratic forms to solve a geometric problem.

S: (x-x_0)^2+(y-y_0)^2+(z-z_0)^2=R^2

S: (x-x_0)^2+(y-y_0)^2+(z-z_0)^2=R^2

\Pi: Ax+By+Cz=D

\Pi: Ax+By+Cz=D

\rho = \dfrac{Ax_0+By_0+ Cy_0-D}{\sqrt{A^2+B^2+C^2}}

\rho = \dfrac{Ax_0+By_0+ Cy_0-D}{\sqrt{A^2+B^2+C^2}}

$\mathbf{x}^{T} M\mathbf{x}+K\mathbf{x}+f=0$

\left\{ \begin{array} {rcl} x &=& p_{1,1}u+p_{1,2}v + x_0 \\ y &=&p_{2,1}u+p_{2,2}v + y_0 \\ z &=&\alpha_0-\alpha_1x-\alpha_2y -\alpha_3 + z_0 \end{array} \right. \\t\in[0, 2\pi].

\left\{ \begin{array} {rcl} x &=& p_{1,1}u+p_{1,2}v + x_0 \\ y &=&p_{2,1}u+p_{2,2}v + y_0 \\ z &=&\alpha_0-\alpha_1x-\alpha_2y -\alpha_3 + z_0 \end{array} \right. \\t\in[0, 2\pi].

Part 3: Final comments

GeoGebra offers different semiotic representations (e.g. geometric, algebraic, CAS) and a set of affordances for students to explore and model these problems dynamically.

Algebraic view

CAS view

3D view

Part 3: Final comments

The exploration of these models opens up novel approaches for learners to engage in mathematical thinking since they can generate different representations of a problem to visualise and contrast how numeric, graphic, and algebraic approaches are connected.

Algebraic view

CAS view

3D view

Part 3: Final comments

Collaboration with Bradley Welch, a UQ student soon to become a senior maths and chemistry teacher.

Future work: Implement the didactic proposal in the classroom to analise how this can help students to develop their problem-solving skills and create more meaningful mathematical connections.

The general solution, together with the didactic proposal, has been recently published in the International Journal of Mathematical Education in Science and Technology:
- Welch, B. G. & Ponce Campuzano, J. C. (2023). Applying matrix diagonalisation in the classroom with GeoGebra: parametrising the intersection of a sphere and plane.
- https://doi.org/10.1080/0020739X.2023.2233513

But how does GeoGebra compute this parametrisation?

What are the mathematical concepts or relationships necessary to determine the values for the constants $C_k, \;x_c, y_c, z_c$ ?

\left\{ \begin{array}{l} x=C_1 \cos(t)+ C_2\sin(t)+x_c\\ y= C_3 \cos(t)+ C_4\sin(t)+y_c\\ z=C_5\sin(t)+z_c \end{array} \right.

\left\{ \begin{array}{l} x=C_1 \cos(t)+ C_2\sin(t)+x_c\\ y= C_3 \cos(t)+ C_4\sin(t)+y_c\\ z=C_5\sin(t)+z_c \end{array} \right.

/* 
GeoGebra - Dynamic Mathematics for Everyone
http://www.geogebra.org

This file is part of GeoGebra.

This program is free software; you can redistribute it and/or modify it 
under the terms of the GNU General Public License as published by 
the Free Software Foundation.

 */

/*
 * AlgoJoinPointsSegment
 *
 * Created on 21. August 2003
 */

package org.geogebra.common.geogebra3D.kernel3D.algos;

import org.geogebra.common.geogebra3D.kernel3D.geos.GeoConic3D;
import org.geogebra.common.kernel.Construction;
import org.geogebra.common.kernel.StringTemplate;
import org.geogebra.common.kernel.commands.Commands;
import org.geogebra.common.kernel.geos.GeoElement;
import org.geogebra.common.kernel.kernelND.GeoCoordSys2D;
import org.geogebra.common.kernel.kernelND.GeoQuadricND;
import org.geogebra.common.kernel.matrix.CoordMatrix;

/**
 *
 * @author ggb3D
 * 
 */
public class AlgoIntersectPlaneQuadric extends AlgoElement3D {

	// inputs
	/** plane */
	protected GeoCoordSys2D plane;
	/** second coord sys */
	protected GeoQuadricND quadric;

	// output
	/** intersection */
	protected GeoConic3D conic;

	private CoordMatrix cm = new CoordMatrix(3, 3);
	private CoordMatrix tmpMatrix = new CoordMatrix(3, 4);
	private CoordMatrix parametricMatrix;

	/**
	 * Creates new AlgoIntersectLinePlane
	 * 
	 * @param cons
	 *            the construction
	 * @param plane
	 *            plane
	 * @param quadric
	 *            quadric
	 * @param addToCons
	 *            whether to add to cons
	 */
	AlgoIntersectPlaneQuadric(Construction cons, GeoCoordSys2D plane,
			GeoQuadricND quadric, boolean addToCons) {

		super(cons, addToCons);

		this.plane = plane;
		this.quadric = quadric;

		conic = newConic(cons);

		// end
		if (addToCons) {
			end();
		}
	}

	/**
	 * Creates new AlgoIntersectLinePlane
	 * 
	 * @param cons
	 *            the construction
	 * @param plane
	 *            plane
	 * @param quadric
	 *            quadric
	 */
	AlgoIntersectPlaneQuadric(Construction cons, GeoCoordSys2D plane,
			GeoQuadricND quadric) {

		this(cons, plane, quadric, true);

	}

	/**
	 * end of contructor for this algo
	 */
	protected void end() {
		setInputOutput(new GeoElement[] { plane.toGeoElement(), quadric },
				new GeoElement[] { conic });
	}

	/**
	 * 
	 * @param cons1
	 *            construction
	 * @return new conic for intersection
	 */
	protected GeoConic3D newConic(Construction cons1) {
		return new GeoConic3D(cons1, true);
	}

	/**
	 * return the intersection
	 * 
	 * @return the intersection
	 */
	public GeoConic3D getConic() {
		return conic;
	}

	// /////////////////////////////////////////////
	// COMPUTE

	@Override
	public void compute() {
		conic.setCoordSys(plane.getCoordSys());
		if (!quadric.isDefined() || !plane.isDefined()) {
			conic.setUndefined();
			return;
		}
		intersectPlaneQuadric(plane, quadric, conic);
	}

	private void intersectPlaneQuadric(GeoCoordSys2D inputPlane,
			GeoQuadricND inputQuad, GeoConic3D outputConic) {
		if (parametricMatrix == null) {
			parametricMatrix = new CoordMatrix(4, 3);
		}
		CoordMatrix qm = inputQuad.getSymetricMatrix();
		CoordMatrix pm = inputPlane.getCoordSys()
				.getParametricMatrix(parametricMatrix);

		// sets the conic matrix from plane and quadric matrix
		cm.setMul(tmpMatrix.setMulT1(pm, qm), pm);

		outputConic.setCoordSys(inputPlane.getCoordSys());
		outputConic.setMatrix(cm);

	}

	@Override
	public Commands getClassName() {
		return Commands.IntersectPath;
	}

	@Override
	final public String toString(StringTemplate tpl) {
		StringBuilder sb = new StringBuilder();

		sb.append(getLoc().getPlain("IntersectionCurveOfAB",
				plane.getLabel(tpl), quadric.getLabel(tpl)));

		return sb.toString();
	}

}

We need to study the source code of GeoGebra, which is written in Java.

For example this:

Future work: Investigate more in depth the source code of GeoGebra