Plane equations and normal vectors | Year 12

01 / LEARN

Understand the idea.

A plane and its normal

A plane's normal vector is perpendicular to every displacement within it. If the plane passes through $a$ with nonzero normal $n$ , its points satisfy $n \cdot (r - a) = 0$ . This directly converts geometric perpendicularity into an equation.

Expanding gives $n_{x} x + n_{y} y + n_{z} z = d$ , where $d = n \cdot a$ . Multiplying the entire equation by one nonzero scalar gives the same plane. Its normal uses the three coordinate coefficients; the constant sets the plane's offset.

Use geometric information

Three non-collinear points supply two nonparallel side vectors from a common point. Their cross product is a nonzero normal. If that product is zero, the points are collinear and do not determine a unique plane.

For a parametric plane $r = a + s u + t v$ , cross its independent directions to obtain a normal, then use the anchor to find $d$ . Substitute each supplied point into the final Cartesian equation as a check.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example 1

Find the plane through (1,2,3) with normal (2,-1,4).

Worked solution

1
Use the point-normal condition.
$(2, - 1, 4) \cdot (x - 1, y - 2, z - 3) = 0$
Every displacement from the given point to the plane is perpendicular to its normal.
2
Expand the dot product.
$2 (x - 1) - (y - 2) + 4 (z - 3) = 0$
The dot product multiplies corresponding components and adds.
3
Collect the constant terms.
$2 x - y + 4 z = 12$
The constant is $2 (1) - 2 + 4 (3) = 12$ .
4
Check the supplied point.
$2 (1) - 2 + 4 (3) = 12$
The original point satisfies the completed plane equation.

Answer

$2 x - y + 4 z = 12$

EXAMPLE 02

Worked example 2

Find the plane through (0,0,1), (1,0,1), (0,2,1).

Worked solution

1
Form two side vectors from the first point.
$\begin{aligned} u & = (1, 0, 1) - (0, 0, 1) = (1, 0, 0) \\ v & = (0, 2, 1) - (0, 0, 1) = (0, 2, 0) \end{aligned}$
Their common origin ensures both displacements lie within the required plane.
2
Calculate a normal.
$n = u \times v = (0, 0, 2)$
The nonzero cross product also confirms that the points are non-collinear.
3
Apply the normal through the anchor.
$(0, 0, 2) \cdot (x, y, z - 1) = 0 ⟹ 2 (z - 1) = 0$
The plane passes through $(0, 0, 1)$ and is perpendicular to its normal.
4
Simplify the plane equation.
$z = 1$
All three original points have height one, so they satisfy this horizontal plane.

Answer

$z = 1$

03 / YOUR TURN

Try it yourself.

Check your understanding

Is (1,1,2) on 2x+y-z=1?

Show answer & reasoning

Yes.

Substitute the point into the original equation: $2 (1) + 1 - 2 = 1$ . It equals the right-hand constant, so $(1, 1, 2)$ lies on the plane.

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Give a normal to $3 x - 2 y + z = 7$ .
2 marks
Find the plane through $(2, 0, - 1)$ with normal $(1, 2, 3)$ .
2 marks
Find the plane through $O$ spanned by $(1, 0, 1)$ and $(0, 1, 1)$ .
2 marks
Are $2 x - 4 y + 6 z = 8$ and $x - 2 y + 3 z = 4$ equivalent planes?
2 marks
Do $O$ , $(1, 2, 3)$ , $(2, 4, 6)$ determine a unique plane?
2 marks
Parametrise the plane through $(1, 0, 2)$ with directions $(1, 1, 0)$ , $(0, 1, 1)$ .
2 marks

Read the worked solutions

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

Give a normal to $3 x - 2 y + z = 7$ .
2 marks
Read the coefficients of $x$ , $y$ and $z$ .
$(3, - 2, 1)$
Find the plane through $(2, 0, - 1)$ with normal $(1, 2, 3)$ .
2 marks
The constant is $2 + 0 - 3 = - 1$ .
Use the normal coefficients.
$x + 2 y + 3 z = - 1$
Find the plane through $O$ spanned by $(1, 0, 1)$ and $(0, 1, 1)$ .
2 marks
A normal is $(- 1, - 1, 1)$ .
Use zero constant and multiply by $- 1$ .
$x + y - z = 0$
Are $2 x - 4 y + 6 z = 8$ and $x - 2 y + 3 z = 4$ equivalent planes?
2 marks
Divide every coefficient and the constant of the first by 2.
Yes.
Do $O$ , $(1, 2, 3)$ , $(2, 4, 6)$ determine a unique plane?
2 marks
The points are collinear.
Infinitely many planes contain their common line.
No.
Parametrise the plane through $(1, 0, 2)$ with directions $(1, 1, 0)$ , $(0, 1, 1)$ .
2 marks
Use both independent direction parameters.
The two directions are not parallel.
$r = (1, 0, 2) + s (1, 1, 0) + t (0, 1, 1)$ .

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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.

A plane and its normal

Use geometric information

See the method in action.

Worked example 1

Use the point-normal condition.

Expand the dot product.

Collect the constant terms.

Check the supplied point.

Worked example 2

Form two side vectors from the first point.

Calculate a normal.

Apply the normal through the anchor.

Simplify the plane equation.

Try it yourself.

Check your understanding

Keep practising