What is cartesian equation?

Writing n = (a, b, c) and r = (x, y, z), the vector equation becomes

What are plane through three points?

Given three non-collinear points A, B, C, form two vectors in the plane, AB and AC. Their cross product is normal to both, hence normal to the plane:

QLDSpecialist MathematicsSyllabus dot point

Topic 2: Vectors and matrices

Determine the vector equation and the Cartesian equation of a plane from a point and a normal or from three points, find the distance from a point to a plane, and find the line of intersection or angle between planes

A focused answer to the QCE Specialist Mathematics Unit 3 dot point on planes in three dimensions. Covers the normal vector, the vector equation r dot n equals a dot n, the Cartesian form, building a plane from three points, the distance from a point to a plane and the angle between planes, with a verified worked example and the normal-vector trap.

Generated by Claude Opus 4.76 min answerUpdated 2026-05-29

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

QCAA wants you to describe a plane in three dimensions using a normal vector, write it in both vector and Cartesian form, build it from three points, and answer geometric questions: the distance from a point to the plane and the angle between two planes. The normal vector is the central object, since a plane is fixed by one point on it and a direction perpendicular to it.

The answer

Normal vector and the vector equation

A plane is determined by a point $A$ with position vector $\mathbf{a}$ and a normal vector $\mathbf{n}$ perpendicular to the plane. A point $P$ with position vector $\mathbf{r}$ lies on the plane exactly when $\vec{AP}$ is perpendicular to $\mathbf{n}$ , that is $(\mathbf{r} - \mathbf{a})\cdot\mathbf{n} = 0$ . This rearranges to

\mathbf{r}\cdot\mathbf{n} = \mathbf{a}\cdot\mathbf{n}.

Cartesian equation

Writing $\mathbf{n} = (a, b, c)$ and $\mathbf{r} = (x, y, z)$ , the vector equation becomes

ax + by + cz = d, \qquad d = \mathbf{a}\cdot\mathbf{n},

equivalently $ax + by + cz + d' = 0$ . The coefficients of $x$ , $y$ and $z$ are precisely the components of the normal vector, which is how you read a normal straight off a Cartesian equation.

Plane through three points

Given three non-collinear points $A$ , $B$ , $C$ , form two vectors in the plane, $\vec{AB}$ and $\vec{AC}$ . Their cross product is normal to both, hence normal to the plane:

\mathbf{n} = \vec{AB} \times \vec{AC}.

Then use $\mathbf{r}\cdot\mathbf{n} = \mathbf{a}\cdot\mathbf{n}$ with any one of the three points.

Distance from a point to a plane

The perpendicular distance from a point $P_0 = (x_0, y_0, z_0)$ to the plane $ax + by + cz + d' = 0$ is

\text{distance} = \frac{|a x_0 + b y_0 + c z_0 + d'|}{\sqrt{a^2 + b^2 + c^2}}.

The absolute value gives an unsigned distance, and the denominator is the magnitude of the normal.

Angle between two planes

The angle between two planes equals the angle between their normals. For normals $\mathbf{n}_1$ and $\mathbf{n}_2$ ,

\cos\theta = \frac{|\mathbf{n}_1\cdot\mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|}.

The absolute value selects the acute angle between the planes. Two planes are parallel when their normals are scalar multiples.

Line of intersection

Two non-parallel planes meet in a line whose direction is $\mathbf{n}_1 \times \mathbf{n}_2$ , since the line lies in both planes and so is perpendicular to both normals. Finding one common point fixes the line.

Worked example

Find the Cartesian equation of the plane through $A(1, 0, 2)$ , $B(3, 1, 1)$ and $C(2, -1, 4)$ .

Form two in-plane vectors.

\vec{AB} = (3-1, 1-0, 1-2) = (2, 1, -1), \qquad \vec{AC} = (2-1, -1-0, 4-2) = (1, -1, 2).

Normal via cross product.

\mathbf{n} = \vec{AB}\times\vec{AC} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 2 & 1 & -1\\ 1 & -1 & 2\end{vmatrix}.

$\mathbf{i}$ : $(1)(2) - (-1)(-1) = 2 - 1 = 1.$ $\quad \mathbf{j}$ : $-[(2)(2) - (-1)(1)] = -[4 + 1] = -5.$ $\quad \mathbf{k}$ : $(2)(-1) - (1)(1) = -3.$

So $\mathbf{n} = (1, -5, -3)$ .

Apply $\mathbf{r}\cdot\mathbf{n} = \mathbf{a}\cdot\mathbf{n}$ using $A(1,0,2)$ : $\mathbf{a}\cdot\mathbf{n} = (1)(1) + (0)(-5) + (2)(-3) = 1 - 6 = -5.$

Hence $x - 5y - 3z = -5$ , or $x - 5y - 3z + 5 = 0$ .

Check with $B(3,1,1)$ : $3 - 5(1) - 3(1) + 5 = 3 - 5 - 3 + 5 = 0$ . The point satisfies the equation.

Exam-style practice questions

Practice questions written in the style of QCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2022 QCAA5 marksConsider points A(3, -1, 3) and B(1, 1, 6). a) Determine the vector AB. b) Determine the Cartesian equation of the line that passes through points A and B. Point A lies on the plane phi, and the line through A and B is perpendicular to this plane. c) Determine the Cartesian equation of the plane.

Show worked answer →

a) [1 mark] AB = B - A = (1 - 3, 1 - (-1), 6 - 3) = (-2, 2, 3).

b) [2 marks] A line through A(3, -1, 3) with direction (-2, 2, 3) has symmetric Cartesian form:
(x - 3)/(-2) = (y + 1)/2 = (z - 3)/3.

c) [2 marks] Since the line AB is perpendicular to the plane, the direction (-2, 2, 3) is the plane's normal vector n. Using point A(3, -1, 3) in n dot (r - A) = 0:
-2(x - 3) + 2(y + 1) + 3(z - 3) = 0
-2x + 6 + 2y + 2 + 3z - 9 = 0
So the Cartesian equation of the plane is -2x + 2y + 3z = 1.