Introduction to Equations of Planes

Welcome to the study of planes! In our previous chapters, we looked at vectors and lines in three-dimensional space. While a line is a one-dimensional object stretching forever in a single direction, a plane is a flat, two-dimensional surface that extends infinitely in 3D space. Think of a sheet of paper that never ends, or the surface of a giant table.

In this chapter, we will learn three different ways to describe a plane using mathematics. Don't worry if it feels a bit abstract at first—once you see the connection between vectors and these equations, it will click! We will focus on the Vector Form, the Normal Form, and the Cartesian Form.

1. The Vector Equation of a Plane

To define a line, we needed one point and one direction. To define a plane, we need one point and two directions (as long as those directions aren't pointing the same way!).

Imagine standing on a specific spot on a giant flat floor. To describe any other spot on that floor, you could say: "Start at my spot, walk some distance in direction A, and then walk some distance in direction B."

The vector equation is written as:
\( \mathbf{r} = \mathbf{a} + \lambda \mathbf{b} + \mu \mathbf{c} \)

Where:

  • \( \mathbf{r} \) is the position vector of any point \( (x, y, z) \) on the plane.
  • \( \mathbf{a} \) is the position vector of a fixed point known to be on the plane.
  • \( \mathbf{b} \) and \( \mathbf{c} \) are two direction vectors that lie parallel to the plane. These must not be parallel to each other!
  • \( \lambda \) (lambda) and \( \mu \) (mu) are scalars (parameters) that can be any real number. By changing these values, you can "reach" any point on the infinite surface.

Quick Review: This looks very similar to the equation of a line (\( \mathbf{r} = \mathbf{a} + \lambda \mathbf{b} \)), but we added a second parameter (\( \mu \)) and a second direction (\( \mathbf{c} \)) to "spread" the line out into a flat surface.

2. The Normal Form (Scalar Product Form)

While the vector equation uses vectors inside the plane, the Normal Form uses a vector that is perpendicular (at a 90-degree angle) to the plane. This perpendicular vector is called the Normal Vector, usually denoted as \( \mathbf{n} \).

If you take any vector lying on the plane and find its scalar product with the normal vector \( \mathbf{n} \), the result will always be zero because they are perpendicular. This leads to the beautiful and simple equation:

\( \mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n} \)

Where:

  • \( \mathbf{r} \) is the position vector of a general point \( (x, y, z) \).
  • \( \mathbf{n} \) is the normal vector \( \begin{pmatrix} a \\ b \\ c \end{pmatrix} \).
  • \( \mathbf{a} \) is the position vector of a known point on the plane.

Note: The right side of the equation (\( \mathbf{a} \cdot \mathbf{n} \)) results in a single number (a scalar), which we often call \( d \).

Did you know? This form is often the most useful for solving IB exam problems because it's compact and links directly to the Cartesian form.

3. The Cartesian Equation of a Plane

This is the most common algebraic way to see a plane. If you expand the Normal Form \( \mathbf{r} \cdot \mathbf{n} = d \) using components, you get the Cartesian Equation:

\( ax + by + cz = d \)

The Golden Rule of Planes: In the equation \( ax + by + cz = d \), the coefficients \( a, b, \) and \( c \) are exactly the components of the normal vector \( \mathbf{n} = \begin{pmatrix} a \\ b \\ c \end{pmatrix} \). This is a massive shortcut! If you see the equation \( 2x - 3y + 5z = 10 \), you immediately know that the vector \( \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix} \) is perpendicular to that plane.

Wait, how do I find \( d \)?

To find the value of \( d \), simply substitute the coordinates of any point \( (x, y, z) \) that you know is on the plane into the equation.

Example: Find the Cartesian equation of a plane with normal vector \( \mathbf{n} = \begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix} \) passing through the point \( (3, 0, 1) \).
1. Start with the template: \( 1x + 4y - 2z = d \).
2. Plug in the point: \( 1(3) + 4(0) - 2(1) = d \).
3. Calculate: \( 3 + 0 - 2 = 1 \). So, \( d = 1 \).
4. Final Answer: \( x + 4y - 2z = 1 \).

4. Converting Between Forms

In the IB exams, you will often need to switch between these formats. Here is a simple guide on how to handle the most common conversion: Vector Form to Cartesian Form.

If you have the vector equation \( \mathbf{r} = \mathbf{a} + \lambda \mathbf{b} + \mu \mathbf{c} \), you need to find the normal vector \( \mathbf{n} \). Since \( \mathbf{n} \) must be perpendicular to both direction vectors \( \mathbf{b} \) and \( \mathbf{c} \), you can find it using the Vector Product (Cross Product) which you learned in AHL 3.16.

Step-by-Step:

  1. Calculate \( \mathbf{n} = \mathbf{b} \times \mathbf{c} \).
  2. The components of \( \mathbf{n} \) give you the \( a, b, c \) for the Cartesian equation \( ax + by + cz = d \).
  3. Substitute the point \( \mathbf{a} \) into the equation to find \( d \).

Summary of Key Takeaways

1. Vector Form: \( \mathbf{r} = \mathbf{a} + \lambda \mathbf{b} + \mu \mathbf{c} \). Uses a point and two directions.
2. Normal Form: \( \mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n} \). Uses the scalar product and a perpendicular vector.
3. Cartesian Form: \( ax + by + cz = d \). The coefficients \( a, b, c \) are the components of the normal vector.
4. Important Connection: To find the normal vector from two direction vectors, use the vector product: \( \mathbf{n} = \mathbf{b} \times \mathbf{c} \).

Common Mistake to Avoid: When writing the vector equation, make sure your two direction vectors (\( \mathbf{b} \) and \( \mathbf{c} \)) are not multiples of each other. If they are, you've just described a line, not a plane!

Don't worry if this seems tricky at first. Practice finding the normal vector—it is the "key" that unlocks almost every problem involving planes. In the next chapter, we will look at how planes intersect with lines and other planes!