Plane equations and normal vectors | Year 12

01 / LEARN

The idea, step by step

A plane and its normal

A plane through a with nonzero normal n satisfies $n \cdot (r - a) = 0$ . Expanding gives $n_{x} x + n_{y} y + n_{z} z = d$ , where $d = n \cdot a$ . Its normal is perpendicular to every in-plane displacement.

Multiplying all coefficients and the constant by the same nonzero number gives an equivalent plane equation. The constant controls the offset, not the normal direction.

Use geometric information

Three non-collinear points give two nonparallel side directions; their cross product is a normal. A zero cross product shows that the points do not determine a unique plane.

A parametric plane is $r = a + s u + t v$ for nonparallel u,v. Cross the directions to convert to normal form. Check proposed points by substitution in the original plane equation.

A common trap

The normal to ax+by+cz=d is (a,b,d).

Instead: The normal is (a,b,c); d is the equation constant.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example 1

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

Compute $d = 2 - 2 + 12 = 12$ .
Substitute into the normal equation.

2 x - y + 4 z = 12

Example 2

Worked example 2

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

The two side vectors are $(1, 0, 0)$ and $(0, 2, 0)$ .
Their cross product is normal to a horizontal plane with z=1.

z = 1

Check your understanding

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

Show answer

Yes.

Substitution gives 2+1-2=1.

03 / WORKSHEET

Your turn

6 questions · 12 marks · Work at your own pace.

Worksheet PDF 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.
$r = (1, 0, 2) + s (1, 1, 0) + t (0, 1, 1)$ .

Checking saved progress…

Continue learning

Line/plane intersections and angles

Next lesson

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 ↗

Back to Vectors in three dimensions