Vector and cartesian equations of lines and planes (SACE Stage 2 Specialist Mathematics)

Syllabus dot point

“Write vector, parametric and cartesian equations of lines and planes in three dimensions and find intersections and angles between them.”

9 min readUpdated 2 Jun 2026

Quick answer

A line is $\mathbf{r} = \mathbf{a} + t\mathbf{d}$ , a point $\mathbf{a}$ plus a scalar $t$ times a direction $\mathbf{d}$ . A plane is $\mathbf{r}\cdot\mathbf{n} = \mathbf{a}\cdot\mathbf{n}$ , fixed by a point $\mathbf{a}$ and a normal $\mathbf{n}$ , giving cartesian form $n_1 x + n_2 y + n_3 z = d$ . Find a line-plane intersection by substituting the line's parametric coordinates into the plane equation and solving for $t$ .

Jump to a section

What this dot point is asking
Equations of a line
Equations of a plane
Intersections
Angles
Exam-style questions

What this dot point is asking

You need to write vector, parametric and cartesian equations of lines and planes, and find intersections and angles between them.

Equations of a line

A line through point $\mathbf{a}$ with direction vector $\mathbf{d}$ has the vector equation

\mathbf{r} = \mathbf{a} + t\mathbf{d}, \quad t \in \mathbb{R}.

Writing components gives the parametric equations

x = a_1 + t d_1

y = a_2 + t d_2

z = a_3 + t d_3

. Eliminating

t

gives the cartesian (symmetric) form

\frac{x - a_1}{d_1} = \frac{y - a_2}{d_2} = \frac{z - a_3}{d_3},

valid where the

d_i

are nonzero.

Direction vector from two points

The direction of the line through points $P$ and $Q$ is $\vec{PQ} = Q - P$ . Any nonzero scalar multiple of a direction vector describes the same line, so directions are only defined up to scale.

Equations of a plane

A plane is fixed by a point $\mathbf{a}$ on it and a normal vector $\mathbf{n}$ perpendicular to it. Any point $\mathbf{r}$ lies in the plane exactly when $\mathbf{r} - \mathbf{a}$ is perpendicular to $\mathbf{n}$ :

(\mathbf{r} - \mathbf{a})\cdot\mathbf{n} = 0 \iff \mathbf{r}\cdot\mathbf{n} = \mathbf{a}\cdot\mathbf{n}.

In components, with

\mathbf{n} = (n_1, n_2, n_3)

, this is the cartesian equation

n_1 x + n_2 y + n_3 z = d, \quad \text{where } d = \mathbf{a}\cdot\mathbf{n}.

The coefficients of

x, y, z

are exactly the components of a normal vector - a fact used constantly when finding angles.

Cartesian equation of a plane through three points

Find the plane through $A(1,0,2)$ , $B(2,1,0)$ , $C(0,1,1)$

Step 1: Form two in-plane vectors

A plane through three points is determined once we know a point on it and a normal to it. To find the normal, we first need two vectors that lie in the plane. Subtracting the coordinates of $A$ from those of $B$ and $C$ gives two such vectors.

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

Step 2: Cross the vectors to obtain the normal

The cross product of two vectors lying in the plane is perpendicular to both and hence perpendicular to the plane itself, giving the normal vector $\mathbf{n}$ .

\mathbf{n} = \vec{AB} \times \vec{AC} = \big((1)(-1)-(-2)(1)\big)\mathbf{i} - \big((1)(-1)-(-2)(-1)\big)\mathbf{j} + \big((1)(1)-(1)(-1)\big)\mathbf{k}.

Component by component: $\mathbf{i}: -1 + 2 = 1$ ; $\mathbf{j}: -(-1 - 2) = 3$ ; $\mathbf{k}: 1 + 1 = 2$ . So $\mathbf{n} = (1, 3, 2)$ .

Step 3: Use a known point and the normal to write the cartesian equation

The cartesian equation of the plane is $\mathbf{n}\cdot\mathbf{r}=d$ where $d=\mathbf{n}\cdot\mathbf{a}$ for any point $\mathbf{a}$ on the plane. Substituting point $A$ gives $d$ .

Use point $A$ : $d = \mathbf{a}\cdot\mathbf{n} = (1)(1) + (0)(3) + (2)(2) = 5$ . The plane is

x + 3y + 2z = 5.

Final answer: $x + 3y + 2z = 5$ .

Intersections

To find where a line meets a plane, substitute the line's parametric coordinates into the plane's cartesian equation and solve for the parameter $t$ , then back-substitute to get the point.

Line meeting a plane

Where does $\mathbf{r} = (1, 0, -1) + t(2, 1, 3)$ meet the plane $x + 3y + 2z = 5$ ?

Step 1: Write out the parametric coordinates

Read the $x$ , $y$ , $z$ expressions directly from the vector equation of the line. These parametric coordinates describe every point on the line as $t$ varies.

Parametric coordinates: $x = 1 + 2t$ , $y = t$ , $z = -1 + 3t$ .

Step 2: Substitute into the plane equation and solve for $t$

Replacing $x$ , $y$ , $z$ in the plane equation with the parametric expressions gives a single equation in $t$ . Solving it identifies the unique value of $t$ at which the line meets the plane.

Substitute:

(1 + 2t) + 3(t) + 2(-1 + 3t) = 5.

Expand: $1 + 2t + 3t - 2 + 6t = 5$ , so $11t - 1 = 5$ , giving $11t = 6$ , $t = \dfrac{6}{11}$ .

Step 3: Back-substitute to find the intersection point

Plug $t = \dfrac{6}{11}$ into each parametric expression to recover the actual coordinates of the intersection point.

\left(1 + \tfrac{12}{11}, \ \tfrac{6}{11}, \ -1 + \tfrac{18}{11}\right) = \left(\tfrac{23}{11}, \ \tfrac{6}{11}, \ \tfrac{7}{11}\right).

Final answer: The intersection point is $\left(\tfrac{23}{11},\ \tfrac{6}{11},\ \tfrac{7}{11}\right)$ .

Angles

Between two lines: use the dot product of their direction vectors: $\cos\theta = \dfrac{\mathbf{d}_1\cdot\mathbf{d}_2}{|\mathbf{d}_1||\mathbf{d}_2|}$ .
Between two planes: the angle equals the angle between their normals, found the same way.
Between a line and a plane: find the angle $\phi$ between the line's direction and the plane's normal, then the line-plane angle is $90^{\circ} - \phi$ , since the normal is perpendicular to the plane.

Common errors

Mixing up direction and normal: A line uses a direction vector along it; a plane uses a normal perpendicular to it. The coefficients in $n_1 x + n_2 y + n_3 z = d$ are the normal, not a direction in the plane.
Line-plane angle formula slip: The dot product of direction and normal gives the complement; the line-plane angle is $90^{\circ}$ minus that.
Assuming a line must meet a plane: A line parallel to a plane (direction perpendicular to the normal) either misses it entirely or lies inside it. Check whether the equation for $t$ has a solution.
Forgetting directions are only defined up to scale: Two lines with proportional direction vectors are parallel; do not treat a scalar multiple as a different direction.

Exam-style questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

SACE 2023-style3 marks

Calculator-assumed. Consider the plane

P: x + 2y - z = 4

and the line

l

given by

x = 1 + t

y = 2 - 2t

z = 3 - t

. Show that the point of intersection of

P

and

l

A(0, 4, 4)

Show worked answer →

Substitute the parametric coordinates into the plane equation:

x + 2y - z = (1 + t) + 2(2 - 2t) - (3 - t) = 2 - 2t.

[1 mark]

Set equal to $4$ : $2 - 2t = 4$ , so $t = -1$ . [1 mark]

Substitute $t = -1$ : $x = 0$ , $y = 4$ , $z = 4$ . So the intersection is $A(0, 4, 4)$ , as required. [1 mark]

SACE 2023-style2 marks

Calculator-assumed. For the plane

P: x + 2y - z = 4

and line

l

with direction from

x = 1 + t

y = 2 - 2t

z = 3 - t

, find the equation of the plane through

A(0, 4, 4)

that is perpendicular to

l

Show worked answer →

A plane perpendicular to $l$ has the line's direction as its normal: $\mathbf{d} = (1, -2, -1)$ . [1 mark]

The plane through $A(0, 4, 4)$ with normal $(1, -2, -1)$ is $1(x - 0) - 2(y - 4) - 1(z - 4) = 0$ , which simplifies to $x - 2y - z = -12$ . [1 mark]

SACE 2021-style2 marks

Calculator-free. Three planes are

P_1: x + 2y + 2z = 4

P_2: 2x + y - 2z = 5

. Show that

P_1

and

P_2

are perpendicular.

Show worked answer →

Two planes are perpendicular when their normals are perpendicular, i.e. the dot product of the normals is zero.

The normals are the coefficients of $x, y, z$ : $\mathbf{n}_1 = (1, 2, 2)$ and $\mathbf{n}_2 = (2, 1, -2)$ . [1 mark]

$\mathbf{n}_1 \cdot \mathbf{n}_2 = (1)(2) + (2)(1) + (2)(-2) = 2 + 2 - 4 = 0$ . Since the dot product is zero and neither normal is zero, the planes are perpendicular. [1 mark]

Practise this

Sources & how we know this

Stage 2 Specialist Mathematics subject outline (2025 onwards) — SACE Board
Stage 2 Specialist Mathematics: external assessment — SACE Board

What this dot point is asking

Equations of a line

Equations of a plane

Find the plane through A(1,0,2)A(1,0,2)A(1,0,2), B(2,1,0)B(2,1,0)B(2,1,0), C(0,1,1)C(0,1,1)C(0,1,1)

Step 1: Form two in-plane vectors

Step 2: Cross the vectors to obtain the normal

Step 3: Use a known point and the normal to write the cartesian equation

Intersections

Where does r=(1,0,−1)+t(2,1,3)\mathbf{r} = (1, 0, -1) + t(2, 1, 3)r=(1,0,−1)+t(2,1,3) meet the plane x+3y+2z=5x + 3y + 2z = 5x+3y+2z=5?

Step 1: Write out the parametric coordinates

Step 2: Substitute into the plane equation and solve for ttt

Step 3: Back-substitute to find the intersection point

Angles

Exam-style questions

Practise this

Sources & how we know this

Find the plane through $A(1,0,2)$ , $B(2,1,0)$ , $C(0,1,1)$

Where does $\mathbf{r} = (1, 0, -1) + t(2, 1, 3)$ meet the plane $x + 3y + 2z = 5$ ?

Step 2: Substitute into the plane equation and solve for $t$