Chapter 07 — Vectors (Part 5): Planes in Three Dimensions

CHAPTER 07: VECTORS

PART 5 — PLANES IN THREE DIMENSIONS

A-Level Mathematics HipHop | Advanced 3D Vector Geometry

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Learning Objectives

By the end of this lesson, you should be able to:

• Understand the meaning of a plane in three-dimensional space.
• Identify a normal vector to a plane.
• Find the Cartesian equation of a plane.
• Use the vector equation of a plane.
• Convert between different forms of plane equations.
• Find the point where a line intersects a plane.
• Determine whether a line is parallel to, or lies in, a plane.

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1. Understanding a Plane in 3D

A plane is a flat surface extending indefinitely in two directions within three-dimensional space.

Unlike a straight line, which has one direction, a plane contains two independent directions.

A plane can be described using:

• A point lying on the plane.
• A normal vector perpendicular to the plane.
• Two independent direction vectors lying in the plane.

Important distinction

A direction vector lies along a line or within a plane. A normal vector is perpendicular to the plane.

2. The Cartesian Equation of a Plane

The general Cartesian equation of a plane is:

ax + by + cz = d

Here, a, b and c are constants, and the vector

n = (a, b, c)

is a normal vector to the plane.

The constant d determines the position of the plane.

3. Finding the Equation of a Plane

If a plane passes through a point A(x₁, y₁, z₁) and has normal vector n = (a, b, c), its equation is:

a(x − x₁) + b(y − y₁) + c(z − z₁) = 0

Worked Example 1

Find the equation of the plane passing through A(2, −1, 3) with normal vector n = (2, 1, −2).

Solution

Using the point-normal equation:

2(x − 2) + 1(y + 1) − 2(z − 3) = 0

Expanding:

2x − 4 + y + 1 − 2z + 6 = 0

Collecting terms:

2x + y − 2z + 3 = 0

This is the required equation.

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4. The Vector Equation of a Plane

A plane can also be represented using a position vector and two independent direction vectors.

r = a + λb + μc

Where:

• r is the position vector of any point on the plane.
• a is the position vector of a fixed point on the plane.
• b and c are two non-parallel direction vectors in the plane.
• λ and μ are real parameters.

The two direction vectors must be independent: one must not be a scalar multiple of the other.

Worked Example 2

A plane passes through A(1, 2, 0) and contains the two direction vectors:

b = (1, 1, 0),   c = (0, 1, 2).

Write its vector equation.

Solution

The position vector of A is a = (1, 2, 0).

Therefore:

r = (1, 2, 0) + λ(1, 1, 0) + μ(0, 1, 2)

This is the vector equation of the plane.

5. Converting a Vector Plane Equation into Cartesian Form

Suppose:

r = (1, 2, 0) + λ(1, 1, 0) + μ(0, 1, 2).

Comparing components gives:

• x = 1 + λ
• y = 2 + λ + μ
• z = 2μ

From the first equation:

λ = x − 1

From the third equation:

μ = z/2

Substitute into the second equation:

y = 2 + x − 1 + z/2

y = x + 1 + z/2

Multiplying by 2 and rearranging:

2x − 2y + z + 2 = 0

This is the Cartesian equation of the plane.

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6. Finding the Point of Intersection of a Line and a Plane

To find where a line meets a plane:

1. Write the line in parametric form.
2. Substitute its x, y and z expressions into the plane equation.
3. Solve for the parameter.
4. Substitute the parameter back into the line to obtain the intersection point.

Worked Example 3

Find the point of intersection of the line

r = (1, 0, 2) + t(2, 1, −1)

and the plane

x + 2y + z = 8.

Solution

From the line equation:

• x = 1 + 2t
• y = t
• z = 2 − t

Substitute into the plane:

(1 + 2t) + 2t + (2 − t) = 8

3 + 3t = 8

3t = 5

t = 5/3

Now substitute into the line:

x = 1 + 10/3 = 13/3

y = 5/3

z = 2 − 5/3 = 1/3

Intersection point = (13/3, 5/3, 1/3)

7. The Relationship Between a Line and a Plane

A line and a plane can have three possible relationships.

Case 1: The line intersects the plane

There is exactly one point common to the line and the plane.

Case 2: The line is parallel to the plane

The line has no point in common with the plane.

Case 3: The line lies in the plane

Every point on the line also lies on the plane.

If the plane has normal vector n and the line has direction vector d:

n · d = 0

This means the line's direction is parallel to the plane. The line may either lie in the plane or be strictly parallel to it. Substitute a point from the line into the plane equation to distinguish the two cases.

Worked Example 4

Consider the plane:

2x + y − z = 5

and the line:

r = (1, 0, 0) + t(1, −1, 1).

Determine their relationship.

Solution

The normal vector of the plane is n = (2, 1, −1).

The direction vector of the line is d = (1, −1, 1).

Calculate the scalar product:

n · d = 2(1) + 1(−1) − 1(1)

= 2 − 1 − 1 = 0.

Therefore, the line is parallel to the plane or lies in it.

Test the point (1, 0, 0) in the plane:

2(1) + 0 − 0 = 2 ≠ 5.

Conclusion: The line is parallel to the plane and does not lie in it.

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8. Summary of Important Formulae

Cartesian plane: ax + by + cz = d

Point-normal form: a(x − x₁) + b(y − y₁) + c(z − z₁) = 0

Vector plane: r = a + λb + μc

Normal vector: n = (a, b, c)

Line parallel to plane: n · d = 0

9. Practice Exercises

Attempt the following questions before opening the answers.

1. Find the equation of the plane passing through (1, 2, 3) with normal vector (2, −1, 4).
2. Find the Cartesian equation of the plane represented by r = (2, 0, 1) + λ(1, 1, 0) + μ(0, 2, 1).
3. Find the point where the line r = (0, 1, 2) + t(1, 2, −1) meets the plane x + y + z = 6.
4. Determine whether the line r = (1, 1, 1) + t(1, 0, 1) is parallel to the plane 2x + y − 2z = 4. If it is parallel, determine whether it lies in the plane.
5. A plane has equation 3x − 2y + z = 7. State a normal vector to the plane.

Open Answers and Worked Solutions

1.

2(x − 1) − (y − 2) + 4(z − 3) = 0

Answer: 2x − y + 4z = 12.

2.

The direction vectors are b = (1, 1, 0) and c = (0, 2, 1).

A normal vector is perpendicular to both. Their vector product gives n = (1, −1, 2).

Using the point (2, 0, 1):

(x − 2) − y + 2(z − 1) = 0

Answer: x − y + 2z = 4.

3.

x = t, y = 1 + 2t, z = 2 − t.

Substitute: t + 1 + 2t + 2 − t = 6.

2t + 3 = 6, so t = 3/2.

Answer: (3/2, 4, 1/2).

4.

Normal vector n = (2, 1, −2), direction vector d = (1, 0, 1).

n · d = 2 + 0 − 2 = 0.

The line is parallel to the plane or lies in it. Test (1, 1, 1):

2(1) + 1 − 2(1) = 1 ≠ 4.

Answer: The line is parallel to, but does not lie in, the plane.

5.

Answer: n = (3, −2, 1).

10. Final Learning Check

Before moving to the next part, make sure you can:

• Recognise the normal vector from a Cartesian plane equation.
• Construct a plane equation from a point and a normal vector.
• Work with a plane in vector form.
• Find a line-plane intersection accurately.
• Distinguish a line lying in a plane from one parallel to it.

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