What is line-plane angle formula slip?

The dot product of direction and normal gives the complement; the line-plane angle is 90^ minus that.

§-Syllabus dot point

SASpecialist MathematicsSyllabus dot point

How do vector and cartesian equations describe lines and planes, and how do we find where they meet?

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

Vector, parametric and cartesian forms for lines and planes in three dimensions, the role of direction and normal vectors, and finding intersections and angles between lines and planes.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

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

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What this dot point is asking
Equations of a line
Equations of a plane
Intersections
Angles

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.

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.

Exam-style practice questions

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

SACE 20233 marksCalculator-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 20232 marksCalculator-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 20212 marksCalculator-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]