Grade by Grade
C. Geometry and Algebra of Vectors · Lines and Planes

Normals and the Scalar Equation of a Plane

C4.3 C4.4 C4.5About 70 minutes
Learning goals
  • I can recognize a normal to a plane geometrically and read one from a scalar equation.
  • I can state and explain properties of normals and planes.
  • I can find the scalar equation of a plane from a point and a normal, or from three points.
  • I can decide whether planes are parallel or perpendicular and find the angle between them.

Words to know

normal (to a plane)
A non-zero vector perpendicular to every vector that lies in the plane. For Ax + By + Cz + D = 0, one normal is [A, B, C].
point-normal form (plane)
n · (r − r₀) = 0, or A(x − x₀) + B(y − y₀) + C(z − z₀) = 0.
non-collinear points
Points that do not all lie on one line.
parallel planes
Planes whose normals are scalar multiples of each other.
angle between planes
The acute angle between their normals.

Which way is "straight out"?

Put a book flat on a table and stand a pencil straight up on it. The pencil is perpendicular to every line you could draw on the book's cover. Tilt the book, and the pencil (if it stays perpendicular to the cover) tilts with it. That pencil is a normal, and it captures the plane's tilt completely. Add one point the plane passes through, and the plane is fixed.

In the last lesson you met planes as solution sets of Ax + By + Cz + D = 0. This lesson explains where that equation comes from, using the dot product and cross product from Dot and Cross Products.

Normals to a plane

A normal to a plane is a non-zero vector perpendicular to the plane, which means perpendicular to every vector lying in the plane.

Geometric properties (try them with the book and pencil):

  • The direction of a normal is the same everywhere on the plane: a flat surface has the same tilt at every point.
  • Every non-zero scalar multiple of a normal is also a normal (2n, −n, ½n, …).
  • Three non-collinear points determine exactly one plane. (Two points, or three on one line, do not: a door swings around its hinge line through infinitely many planes.)
  • If u and v lie in the plane, so does u + v, because n · (u + v) = n · u + n · v = 0 + 0 = 0.
n = [A, B, C]P₀P(x, y, z)r − r₀un · (r − r₀) = 0 for every point P in the plane
A normal n sticks straight out of the plane. It is perpendicular to every vector that lies in the plane, such as u and the vector r − r₀ from P₀ to any point P of the plane.

From a normal and a point to the scalar equation

Let the plane pass through P₀(x₀, y₀, z₀) with normal n = [A, B, C]. A point P(x, y, z) is on the plane exactly when P₀P is perpendicular to n:

n · (r − r₀) = 0, that is, A(x − x₀) + B(y − y₀) + C(z − z₀) = 0.

Expanding gives Ax + By + Cz + D = 0, where D = −(Ax₀ + By₀ + Cz₀).

Reading it backwards: in any scalar equation Ax + By + Cz + D = 0, the coefficients [A, B, C] form a normal. One normal to 3x + 5y − 2z = 6 is [3, 5, −2].

A plane through three points A, B, C: the vectors AB and AC lie in the plane, so their cross product n = AB × AC is perpendicular to both, and therefore normal to the plane. Then substitute one point to find D.

n½n2n−n2x − y + 3z = 1−4x + 2y − 6z = 5
Every normal of a plane points in the same direction (or exactly opposite): n, ½n, 2n and −n are all normals. Planes whose normals are parallel are parallel planes, like these two.

Comparing two planes using normals

  • Parallel planes: normals are scalar multiples. They are the same plane if the whole equations are multiples (including D).
  • Perpendicular planes: n₁ · n₂ = 0.
  • Angle between planes: the acute angle θ between the normals, from cos θ = |n₁ · n₂| / (|n₁||n₂|).
Worked example

Example 1: Equation from a point and a normal

Find a scalar equation of the plane through P₀(2, −1, 4) with normal n = [3, −2, 5].

  1. Point-normal form: 3(x − 2) − 2(y + 1) + 5(z − 4) = 0.
  2. Expand: 3x − 6 − 2y − 2 + 5z − 20 = 0.
  3. Collect: 3x − 2y + 5z − 28 = 0.
  4. Check P₀: 3(2) − 2(−1) + 5(4) − 28 = 6 + 2 + 20 − 28 = 0 ✓.

3x − 2y + 5z − 28 = 0

Worked example

Example 2: The plane through three points

Find a scalar equation of the plane through A(1, 0, 2), B(3, 1, 0) and C(0, 2, 1).

  1. Two vectors in the plane: AB = [2, 1, −2] and AC = [−1, 2, −1].
  2. n = AB × AC = [(1)(−1) − (−2)(2), (−2)(−1) − (2)(−1), (2)(2) − (1)(−1)] = [3, 4, 5].
  3. So 3x + 4y + 5z + D = 0. Substitute A: 3 + 0 + 10 + D = 0, so D = −13.
  4. Check B: 9 + 4 + 0 − 13 = 0 ✓. Check C: 0 + 8 + 5 − 13 = 0 ✓.

3x + 4y + 5z − 13 = 0

Worked example

Example 3: A parallel plane

Find the plane through (1, 2, −3) parallel to 4x − y + 2z = 7.

  1. Parallel planes can share the normal [4, −1, 2], so the equation is 4x − y + 2z + D = 0.
  2. Substitute (1, 2, −3): 4 − 2 − 6 + D = 0, so D = 4.

4x − y + 2z + 4 = 0

Worked example

Example 4: The angle between two planes

Find the acute angle between x + y − z = 1 and 2x − y + 3z = 4.

  1. Normals: n₁ = [1, 1, −1], n₂ = [2, −1, 3].
  2. n₁ · n₂ = 2 − 1 − 3 = −2. |n₁| = √3, |n₂| = √14.
  3. cos θ = |−2| / (√3 √14) = 2/√42 ≈ 0.3086.
  4. θ ≈ 72.0°. (Without the absolute value you would get the obtuse angle 108.0° between the normals; the planes meet at 72.0° and 108.0°, and we report the acute one.)

About 72.0°.

Just for fun

A plane walks into a geometry class. The teacher says, "Please stand up straight." The plane replies, "I can't, but my normal can."

Hands-on activity

Book, pencil and three points

You need: a hardcover book, a pencil, three small objects (coins or erasers), and a table.

  1. Lay the book flat and stand the pencil straight up on the cover. Slide the pencil around the cover without tilting it. Its direction never changes: one plane, one normal direction.
  2. Hold the book tilted at some angle. Ask a helper to hold the pencil so it is perpendicular to the cover. Notice the pencil tilts with the book: the normal records the tilt.
  3. Place two coins on the table. Can you rest the book so that its cover touches both coins and is held steady? (No: it can rock around the line through them.) Add a third coin, not in line with the first two. Now the book is held steady: three non-collinear points fix the plane.
  4. Put the three coins in a straight line. Does the book still rock?

What to look for: two points, or three collinear points, leave the book free to rock around a line. Three non-collinear points fix it.

Your turn

1

Give a normal to the plane 2x − 7y + z = 4.

2

Is [−4, 14, −2] also a normal to 2x − 7y + z = 4? Explain.

3

Find a scalar equation of the plane through (1, 1, 1) with normal [2, 3, −1].

4

Find the plane through (0, −2, 5) parallel to x − 3y + 4z = 1.

5

Are 2x − y + 3z = 1 and −4x + 2y − 6z = 5 parallel? Are they the same plane?

6

Are x + 2y − z = 3 and 3x − y + z = 0 perpendicular?

7

Find a scalar equation of the plane through A(1, 2, 3), B(2, 3, 1) and C(0, 1, 4).

More practice: warm-up

1

Give a normal to the plane z = 3.

2

Give a normal to x + y + z = 1.

3

Give a normal to the xz-plane.

4

Is [1, 2, 3] a normal to 2x + 4y + 6z = 5?

5

How many non-collinear points are needed to fix a plane?

More practice: on your own

1

Find the plane through P(3, −1, 2) perpendicular to the line r = t[2, 1, −4].

2

Find the plane through A(2, 0, 0), B(0, 3, 0) and C(0, 0, 4).

3

Find the acute angle between 2x + y + z = 3 and x − y + 2z = 5.

4

Find k so that 3x + ky − z = 1 and x + 2y + 5z = 0 are perpendicular.

5

Explain why A(1, 1, 1), B(2, 3, 4) and C(3, 5, 7) do not determine a single plane.

Challenge

1

Prove that if u and v both lie in a plane with normal n, then every combination au + bv is perpendicular to n.

2

Find a scalar equation of the plane that contains the line r = [1, 0, 2] + t[1, 1, 1] and the point P(2, −1, 3).

3

Find the plane through the origin that is perpendicular to both x + y + z = 1 and x − y = 3.

Check your understanding

Quiz

1.

Which vector is a normal to the plane 3x + 5y − 2z = 6?

2.

Which is the plane through (1, 0, −1) with normal [2, −1, 4]?

3.

Which plane is parallel to 2x − y + 3z = 4?

4.

Which statement about normals to a plane is true?

5.

A, B and C are non-collinear points. Which vector is a normal to the plane through them?

6.

Two planes have normals [1, 0, 0] and [1, 1, 0]. What is the acute angle between the planes?

For parents and teachers

Big idea (C4.3, C4.4, C4.5): a plane is fixed by one point and one normal direction. The scalar equation Ax + By + Cz + D = 0 is just the point-normal condition n · (r − r₀) = 0 written out, so its coefficients [A, B, C] are a normal.

How to help: The book-and-pencil model is excellent: the pencil is the normal. Ask your teen, "If I tilt the book, what happens to the normal? If I slide the book without tilting, does the normal change?" (No: parallel planes share normals.)

Common difficulties:

  • Cross product slips. Encourage the "cover-up" or determinant pattern, and always check the final equation with all three points.
  • Forgetting D: two planes with the same normal can be different (parallel) planes.
  • Reporting the obtuse angle between planes. Use the absolute value of the dot product.
  • Using three collinear points: the cross product comes out as the zero vector, which is a warning sign, not an error in arithmetic.