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.

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 4.79 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
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 $A(1,0,2)$ , $B(2,1,0)$ , $C(0,1,1)$

Form two in-plane vectors:

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

A normal is their cross product:

\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)

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

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.

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.

2023 SACE Stage 23 marksConsider the plane P: x + 2y - z = 4 and the line l defined by x = 1 + t, y = 2 - 2t, z = 3 - t, where t is a real parameter. Show that the point of intersection of P and l is A(0, 4, 4).

Show worked answer →

To find where the line meets the plane, substitute the parametric coordinates into the plane equation and solve for t.

x + 2y - z = (1 + t) + 2(2 - 2t) - (3 - t) = 1 + t + 4 - 4t - 3 + t = 2 - 2t. [1 mark]

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

Substitute t = -1 back into the line: x = 1 + (-1) = 0, y = 2 - 2(-1) = 4, z = 3 - (-1) = 4. So the point of intersection is A(0, 4, 4), as required. [1 mark]

2023 SACE Stage 22 marksFor the plane P: x + 2y - z = 4 and line l with x = 1 + t, y = 2 - 2t, z = 3 - t, find the equation of the plane which passes through A(0, 4, 4) and is perpendicular to l.

Show worked answer →

A plane perpendicular to the line l has the line's direction vector as its normal. From the parametric equations, the direction of l is d = (1, -2, -1) (the coefficients of t). [1 mark]

The plane through A(0, 4, 4) with normal (1, -2, -1) has equation 1(x - 0) - 2(y - 4) - 1(z - 4) = 0, i.e. x - 2y + 8 - z + 4 = 0, which simplifies to x - 2y - z + 12 = 0, or x - 2y - z = -12. [1 mark]

2017 SACE Stage 22 marksThree planes are P1: x + 2y + 2z = 4, P2: 2x + y - 2z = 5 and P3: 3x + 2y - 2z = 8. Show that P1 and P2 are perpendicular.

Show worked answer →

Two planes are perpendicular when their normal vectors are perpendicular, that is, when the dot product of the normals is zero.

The normal to a plane is given by the coefficients of x, y and z. So the normal to P1 is n1 = (1, 2, 2) and the normal to P2 is n2 = (2, 1, -2). [1 mark]

Compute the dot product: n1 . n2 = (1)(2) + (2)(1) + (2)(-2) = 2 + 2 - 4 = 0.

Since n1 . n2 = 0 and neither normal is the zero vector, the normals are at right angles, hence the planes P1 and P2 are perpendicular. [1 mark]