Visighta

Section 1.3 — Systems of Three Linear Equations

We now study how three planes in $\mathbb{R}^3$ can intersect. The possibilities are:

Outcome Geometry Solutions
One point Three planes meet at a single point Unique
A line Three planes share a common line Infinitely many (1 free variable)
A plane All three equations describe the same plane Infinitely many (2 free variables)
Empty No point lies on all three planes None 
from linear_algebra_course.chapter1.section3.scenes import *
from drawsvg_renderer import DrawSVGRenderer

renderer = DrawSVGRenderer(
    width=800,
    height=600,
    frame_width=16.0,
    background_color="#000000",
    output_mode='jupyter',
    fps=30,
    show_progress=False
)

Scene 1 — Three Planes Entering the Scene

We introduce three translucent planes one at a time. Each new equation adds another constraint, progressively restricting the possible solutions:

$$ x + y + z = 4 \qquad 2x - y + z = 3 \qquad x + 2y - z = 2 $$

renderer.display_all(Scene1_ThreePlanesEnter(), display=False)
renderer._finalize_interactivity()
renderer.display_inline()
Cell visualization output

Scene 2 — Unique Solution: Three Planes Meet at One Point

When three independent planes intersect at exactly one point, the system has a unique solution.

$$ \begin{cases} x + y + z = 3 \ x - y + z = 1 \ x + y - z = 1 \end{cases} \qquad \Longrightarrow \qquad (x,y,z) = (1,1,1) $$

renderer.display_all(Scene2_UniqueSolution(), display=False)
renderer._finalize_interactivity()
renderer.display_inline()
Cell visualization output

Scene 3 — Infinitely Many Solutions on a Line

Two planes intersect in a line. If the third plane also contains that line, every point on it satisfies all three equations.

$$ \begin{cases} x + y + z = 3 \ x - y + z = 1 \ x + z = 2 \end{cases} $$

From the first two: $y = 1$, $x + z = 2$. The solution is the line $(t,\,1,\,2-t)$.

renderer.display_all(Scene3_LineOfSolutions(), display=False)
renderer._finalize_interactivity()
renderer.display_inline()
Cell visualization output

Scene 4 — Infinitely Many Solutions on a Plane

When all three equations are scalar multiples of each other, they describe the same plane.

$$ x + y + z = 3 \qquad 2x + 2y + 2z = 6 \qquad 3x + 3y + 3z = 9 $$

The solution set is the entire plane — two free variables.

renderer.display_all(Scene4_PlaneOfSolutions(), display=False)
renderer._finalize_interactivity()
renderer.display_inline()
Cell visualization output

Scene 5 — No Common Solution

When two planes are parallel (same normal, different offset), they never intersect. A third plane may cross each of them individually, but no single point lies on all three.

$$ \begin{cases} x + y + z = 2 \ x + y + z = 5 \ 2x - y + z = 1 \end{cases} $$

Planes 1 and 2 are parallel — the system is inconsistent.

renderer.display_all(Scene5_NoSolution(), display=False)
renderer._finalize_interactivity()
renderer.display_inline()
Cell visualization output

Scene 6 — Constraint Metaphor

Each equation removes a degree of freedom from the solution set:

  1. No equations — any point in 3D space is possible.
  2. One equation — solutions form a plane.
  3. Two equations — solutions narrow to a line.
  4. Three equations — solutions shrink to a point (or nothing).
renderer.display_all(Scene6_ConstraintMetaphor(), display=False)
renderer._finalize_interactivity()
renderer.display_inline()
Cell visualization output

Scene 7 — Classification Table

A summary of the four geometric outcomes for a $3 \times 3$ linear system:

Geometry Solutions Free variables
One point Unique $x$ 0
A line $\infty$ solutions 1
A plane $\infty$ solutions 2
No intersection Inconsistent system — 
renderer.display_all(Scene7_ClassificationTable(), display=False)
renderer._finalize_interactivity()
renderer.display_inline()
Cell visualization output
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