Line/plane intersections and angles | Year 12

01 / LEARN

Understand the idea.

Substitute before solving

To intersect a line and plane, substitute all three line components into the plane equation. The resulting coefficient of $t$ is $n \cdot d$ . A nonzero coefficient gives exactly one intersection parameter and therefore one point.

If that coefficient is zero, the direction is parallel to the plane. A true constant equation means the whole line lies in the plane; a false one means no intersection. Substitute any final point into both original objects.

Match the angle to the geometry

Angles depend on which objects are compared. The smaller angle between planes equals the smaller angle between their normals. For a line and plane, the required angle is complementary to the angle between the line direction and the plane normal.

Consequently, use a cosine ratio for two plane normals and a sine ratio for a line-plane angle. Absolute values select the smaller, non-obtuse geometric angle; do not use them automatically for an angle between directed vectors.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example 1

Intersect $(1, 0, 2) + t (1, 2, - 1)$ with x+y+z=6.

Worked solution

1
Substitute the line into the plane.
$(1 + t) + 2 t + (2 - t) = 6$
Every intersection point must satisfy the plane equation and all line components.
2
Solve for the common parameter.
$3 + 2 t = 6 ⟹ t = \frac{3}{2}$
The nonzero parameter coefficient gives a unique intersection.
3
Find all coordinates at that parameter.
$r = (1 + \frac{3}{2}, 2 \cdot \frac{3}{2}, 2 - \frac{3}{2}) = (\frac{5}{2}, 3, \frac{1}{2})$
Use the same parameter in every coordinate.
4
Check the plane condition.
$\frac{5}{2} + 3 + \frac{1}{2} = 6$
The point satisfies the original plane as well as its line parametrisation.

Answer

$t = 3 / 2$ , point $(5 / 2, 3, 1 / 2)$ .

EXAMPLE 02

Worked example 2

Find the smaller angle of direction (1,0,1) with z=0.

Worked solution

1
Choose a normal to the plane.
$n = (0, 0, 1), d = (1, 0, 1)$
The plane $z = 0$ is perpendicular to the $z$ axis.
2
Compute the line-plane sine ratio.
$sin θ = \frac{∣ d \cdot n ∣}{∣ d ∣ ∣ n ∣} = \frac{1}{\sqrt{2}}$
The line-plane angle is complementary to the direction-normal angle.
3
Take the smaller geometric angle.
$θ = {sin}^{- 1} (\frac{1}{\sqrt{2}}) = 45^{\circ}$
The requested angle lies between $0^{\circ}$ and $90^{\circ}$ .

Answer

$45^{\circ}$

03 / YOUR TURN

Try it yourself.

Check your understanding

Describe $(1, 0, 0) + t (0, 1, 0)$ relative to z=0.

Show answer & reasoning

The line lies in the plane.

Every point of the line is $(1, t, 0)$ , whose $z$ coordinate is zero for every real $t$ . Hence the entire line is contained in the plane $z = 0$ .

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Intersect $t (1, 1, 1)$ with $x + y + z = 6$ .
2 marks
Does $(0, 0, 1) + t (1, 0, 0)$ meet $z = 0$ ?
2 marks
Find the smaller angle between $x = 0$ and $x + y = 0$ .
2 marks
Find the smaller angle of direction $(0, 0, 1)$ with $z = 2$ .
2 marks
Find the intersection line of $x + y = 1$ and $z = 2$ .
2 marks
Find $k$ so $(0, 0, k) + t (1, - 1, 0)$ lies in $x + y + z = 3$ .
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Intersect $t (1, 1, 1)$ with $x + y + z = 6$ .
2 marks
The plane equation gives $3 t = 6$ .
Substitute $t = 2$ .
$(2, 2, 2)$
Does $(0, 0, 1) + t (1, 0, 0)$ meet $z = 0$ ?
2 marks
Its $z$ coordinate remains 1.
It is a distinct parallel line.
No.
Find the smaller angle between $x = 0$ and $x + y = 0$ .
2 marks
Use normals $(1, 0, 0)$ and $(1, 1, 0)$ .
Their smaller-angle cosine is $\frac{1}{\sqrt{2}}$ .
$45^{\circ}$
Find the smaller angle of direction $(0, 0, 1)$ with $z = 2$ .
2 marks
The line direction is parallel to the plane normal.
It is perpendicular to the plane.
$90^{\circ}$
Find the intersection line of $x + y = 1$ and $z = 2$ .
2 marks
Set $y = t$ .
Then $x = 1 - t$ and $z = 2$ .
$r = (1, 0, 2) + t (- 1, 1, 0)$ .
Find $k$ so $(0, 0, k) + t (1, - 1, 0)$ lies in $x + y + z = 3$ .
2 marks
The direction contributes zero to $x + y + z$ .
The initial point must satisfy the plane.
$k = 3$

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist Mathematics vector algebra, vector/Cartesian equations and linear systems. Original explanations and tasks.

Official curriculum source ↗

In this lesson

Understand the idea.

Substitute before solving

Match the angle to the geometry

See the method in action.

Worked example 1

Substitute the line into the plane.

Solve for the common parameter.

Find all coordinates at that parameter.

Check the plane condition.

Worked example 2

Choose a normal to the plane.

Compute the line-plane sine ratio.

Take the smaller geometric angle.

Try it yourself.

Check your understanding

Keep practising