Solving Linear Equations: The Geometry of Space (FP2.14)
Hello! Welcome to one of the most practical and geometrically satisfying topics in Further Maths. Here, we tackle systems of linear equations—not just two lines in 2D, but up to three planes in 3D space!
This chapter is all about finding where these planes intersect. The method we use is fundamental to computer science, engineering, and physics. Don't worry if the algebra seems long; the real fun is in understanding the geometry!
Quick Review: What is a Linear Equation in 3D?
In standard A-Level Maths, you know that the equation \(y = mx + c\) represents a straight line in 2D.
In 3D, a linear equation involving \(x, y,\) and \(z\) defines a plane (a perfectly flat, infinitely large surface).
The general form is:
\(ax + by + cz = d\)
Did you know? The coefficients \(a, b, c\) form the normal vector \(\mathbf{n} = \begin{pmatrix} a \\ b \\ c \end{pmatrix}\), which is perpendicular to the plane. Knowing this helps us spot parallel planes quickly!
Section 1: The Goal – Intersecting Three Planes
When we are asked to solve a system of three linear equations:
\(E_1: a_1x + b_1y + c_1z = d_1\)
\(E_2: a_2x + b_2y + c_2z = d_2\)
\(E_3: a_3x + b_3y + c_3z = d_3\)
We are simply asking: What point \((x, y, z)\) lies on all three planes simultaneously?
Analogy: The Walls of a Room
Imagine you are standing in a standard rectangular room.
- Plane 1: The floor.
- Plane 2: A side wall.
- Plane 3: The back wall.
But what if the planes aren't arranged so nicely? That's where the different configurations come in!
Section 2: Solving the System Using Gaussian Elimination
While substitution works, it gets messy quickly. The most reliable and efficient method taught in Further Maths for determining the nature of the solution is Gaussian Elimination (also known as row reduction).
Gaussian elimination systematically transforms the system of equations into an equivalent, simpler system called Row Echelon Form (a triangle shape), from which we can easily find the solution using back-substitution.
Step-by-Step Process
Let's look at the process. We often use an augmented matrix \((A | \mathbf{d})\) to keep the numbers tidy.
Step 1: Get a leading '1' (or a simple number) in the top left.
Choose the equation that makes elimination easiest. Swapping equations is allowed.
Step 2: Eliminate \(x\) from the second and third equations.
We perform row operations (adding or subtracting a multiple of the top equation) to make the first term in \(E_2\) and \(E_3\) equal to zero.
(e.g., New \(E_2\) = \(E_2 - 2E_1\))
Step 3: Eliminate \(y\) from the new third equation.
Now, use the new second equation to eliminate \(y\) from the new third equation. This gives you an equation solely in terms of \(z\). The system is now in triangle form.
Step 4: Back-Substitute.
Solve the final equation for \(z\). Substitute \(z\) into the second equation to find \(y\). Substitute both \(y\) and \(z\) into the first equation to find \(x\).
Common Mistake Alert!
Arithmetic errors are the biggest challenge here. Be meticulous when multiplying and subtracting entire equations. Always check your final answer by substituting the values back into the *original* three equations.
Section 3: The Three Possible Outcomes (The Algebraic Test)
After the elimination process, the third equation is the diagnostic tool. There are three things that can happen to the final line of algebra (or the bottom row of your matrix):
1. Unique Solution
This is the standard case where you find a single point of intersection.
Algebraic Result: The final equation gives you a clear value for \(z\).
\(0x + 0y + Cz = D\) where \(C \neq 0\).
Example: \(5z = 10 \implies z = 2\). You can then easily find \(x\) and \(y\).
2. Infinitely Many Solutions (Dependent System)
This occurs when the three planes meet along a common line or are the same plane.
Algebraic Result: The final equation is true, but gives no new information.
\(0x + 0y + 0z = 0\)
How to proceed: Since you have two meaningful equations and three variables, you must introduce a parameter (like \(\lambda\), \(t\), or \(k\)) for one variable (often \(z\)). Express \(x\) and \(y\) in terms of this parameter. The solution will be a line:
\(\mathbf{r} = \mathbf{a} + t\mathbf{d}\)
3. No Solution (Inconsistent System)
This means the three planes never all meet at a single point.
Algebraic Result: The final equation is a clear contradiction (a mathematical lie).
\(0x + 0y + 0z = K\) where \(K \neq 0\).
Example: \(0 = 7\). This is impossible! Therefore, the system is inconsistent, and there are no solutions.
If the bottom row is:
\(0 = 0\) → Happy math, infinite solutions (set up a parameter).
\(0 = K\) (K non-zero) → Contradiction, NO solution.
Section 4: Geometrical Interpretation of Planes
The crucial part of the FP2 syllabus is linking these algebraic results to the physical arrangement of the planes in 3D space.
Case A: Unique Solution
Geometry: The three planes intersect at a single point.
This is the 'corner of the room' scenario. The planes are non-parallel and they meet cleanly.
Visual check: None of the normal vectors are parallel.
Case B: Infinitely Many Solutions
When the system is dependent (\(0=0\)), there are two geometric configurations:
B1: Intersection in a Line
Geometry: The three planes meet along a single straight line.
Analogy: Think of a half-open book. The pages are two planes, and they meet along the spine (a line). If a third plane (a piece of paper) also lies exactly along that spine, all three share that line.
Visual check: The normal vectors are not parallel, but the system of equations is dependent.
B2: Coincident Planes
Geometry: Two or all three planes are identical.
Example: \(x+y+z=5\) and \(2x+2y+2z=10\). These are the same plane. If the third plane is also the same (or intersects along a line), you get infinite solutions. If all three are coincident, the entire plane is the solution set.
Visual check: The equation of one plane is a direct multiple of another.
Case C: No Solution (Inconsistent Systems)
When the system is inconsistent (\(0=K\)), the planes never all meet:
C1: At Least Two Planes are Parallel and Distinct
Geometry: If Plane 1 is parallel to Plane 2, they will never intersect. No matter where the third plane goes, it can't solve the problem of the first two not meeting.
Visual check: The normal vectors of two planes are proportional, but the constants \(d\) are not (e.g., \(x+y+z=1\) and \(x+y+z=5\)).
C2: Sheaf/Triangular Prism (Planes Intersect in Parallel Pairs)
Geometry: The planes intersect in pairs, forming three parallel lines. They enclose an empty space (like a triangular tunnel).
Analogy: Imagine holding three rulers on a table so that they cross over each other, but they never meet at a central point. Or think of a Toblerone box that extends infinitely. Each face is a plane, and the intersection lines are parallel.
Visual check: No two planes are parallel, but the system is still inconsistent (this is where students often get confused!). This occurs when the three normal vectors are coplanar (lie on the same plane) but the planes themselves do not contain a common line or point.
To check if two planes \(P_1\) and \(P_2\) are parallel, check if their normal vectors \(\mathbf{n}_1\) and \(\mathbf{n}_2\) are multiples of each other.
Summary Table of Geometric Configurations
This table summarizes the possible outcomes you must be able to interpret:
Type 1: Unique Solution (1 Solution)
Configuration: Three planes intersecting at a single point (corner).
Algebraic Result: Final equation gives \(z = k\).
Type 2: Consistent (Infinite Solutions)
Configuration: Intersecting in a common line (pages of a book) OR Coincident planes.
Algebraic Result: Final row is \(0 = 0\).
Type 3: Inconsistent (No Solution)
Configuration: Parallel and distinct planes OR Sheaf/Triangular Prism (pairwise intersection).
Algebraic Result: Final row is \(0 = K\) (\(K \neq 0\)).
Don't worry if the Sheaf configuration seems hard to picture! The key is that if you find No Solution (\(0=K\)) but no two planes are parallel, you must have the triangular prism arrangement.
Keep practicing the Gaussian Elimination, and pay close attention to that very last equation. It tells the whole story! Good luck!