\( \DeclareMathOperator{cosec}{cosec} \)

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International Baccalaureate Mathematics

Geometry and Trigonometry

Syllabus Content

Intersections of: a line with a plane; two planes; three planes.
Angle between: a line and a plane; two planes.

Here are some exam-style questions on this statement:

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Furthermore

Official Guidance, clarification and syllabus links:

Finding intersections by solving equations; geometrical interpretation of solutions.

Link to: solutions of systems of linear equations (AHL 1.16).


The angle between a line and a plane is:

$$ \sin^{-1} \frac{|\mathbf{n \cdot d|}}{|\mathbf{n}| |\mathbf{d}|} $$

\(\mathbf{d}\) is the direction vector of the line;

\(\mathbf{n}\) is the normal vector to the plane.


The angle between two planes can be found by finding the angles between their normals:

$$ \cos^{-1} \frac{|\mathbf{n_1 \cdot n_2|}}{|\mathbf{n_1}| |\mathbf{n_2}|} $$

\(\mathbf{n_1}\) is the normal vector to the first plane;

\(\mathbf{n_2}\) is the normal vector to the second plane.


To find where two planes intersect.

For example:

$$x-3y+2z=8$$ $$3x-9y+2z=4$$

Begin by writing the augmented matrix for the system of equations:

$$\left[\begin{array}{ccc|c}1 & -3 & 2 & 8\\3 & -9 & 2 & 4 \end{array} \right]$$

We now perform row operations to transform this matrix into echelon form:

\(R2 = R2 - 3 \times R1\)

$$\left[\begin{array}{ccc|c}1 & -3 & 2 & 8\\0 & 0 & -4 & -20 \end{array}\right]$$

Two planes in space could be intersecting, parallel or coincident.

from \(R2\) it can be seen that \( z=5 \)

from \(R1\) it can be seen that \(x\) and \(y\) can have infinitely many values so let \( y=t \)

from \(R1\) it can be seen that \( x=3t-2 \)

The planes intersect along the line defined by a vector equation, \( \mathbf{r} \). Let \(t = 0\) to find a point on the line: \( (-2,0,0) \)

$$ \mathbf{r} = \begin{pmatrix} -2 \\ 0 \\ 0 \end{pmatrix} \quad + \quad t \begin{pmatrix} 3 \\ 1 \\ 0 \end{pmatrix} $$

To find where three planes intersect solve their equations simultaneously.

For example:

$$x+y+z=3$$ $$2x-y+z=0$$ $$x-2y-z=-3$$

We begin by writing the augmented matrix for the system of equations:

$$\left[\begin{array}{ccc|c}1 & 1 & 1 & 3\\2 & -1 & 1 & 0\\1 & -2 & -1 & -3\end{array}\right]$$

We now perform row operations to transform this matrix into echelon form:

\(R2 = R2 - 2 \times R1\)

$$\left[\begin{array}{ccc|c}1 & 1 & 1 & 3\\0 & -3 & -1 & -6\\1 & -2 & -1 & -3\end{array}\right]$$

\(R3 = R3 - R1\)

$$\left[\begin{array}{ccc|c}1 & 1 & 1 & 3\\0 & -3 & -1 & -6\\0 & -3 & -2 & -6\end{array}\right]$$

\(R3 = R3 - R2\)

$$\left[\begin{array}{ccc|c}1 & 1 & 1 & 3\\0 & -3 & -1 & -6\\0 & 0 & -1 & 0\end{array}\right]$$

The augmented matrix is now in the form:

$$\left[\begin{array}{ccc|c}a & b & c & d\\0 & e & f & g\\0 & 0 & h & i\end{array}\right]$$

If \( h \neq 0\) there is a unique solution.

If \( h = 0 \text{ and } i \neq 0 \) there is no solution

If \( h = 0 \text{ and } i = 0 \) there are infinitely many solutions (let \(z=t\)).

In the example above ...

from \(R3\) it can be seen that \( z=0 \)

from \(R2\) it can be seen that \( y=2 \)

from \(R1\) it can be seen that \( x=1 \)

The planes intersect at \( (1,2,0) \).

Echelon Matrix

Aide Memoire

Intersections of a line with a plane: Substitute parametric expressions into plane equation then solve to find the paramater.

Intersections of two planes: Solve simultaneous equations by introducing a parameter or find vector product of normals if a common point is known.

Intersections of three planes: Solve simultaneous equations using augmented matrix row reduction.

Angle between a line and a plane: 90° - angle between direction of line and normal to the plane.

Angle between two planes: Find angle between the normals of the plane.


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