CP 6.2 - Equations of a plane

CP 6.2 - Equations of a plane

Two independent directions let you move anywhere on a flat surface. You will construct a plane through three points, convert between its vector and Cartesian descriptions, and recognise when the information describes only a line.

Two independent directions span a plane

Why do three points usually fix a flat sheet, while three points on one straight edge do not?

Start at (1,2,3)(1,2,3). Moving by (1,0,1)(1,0,1) changes one horizontal coordinate and the height together; moving independently by (0,1,2)(0,1,2) supplies a second way across the surface. One parameter controls each movement.

You already know

A line uses one direction: r=a+λb\mathbf r=\mathbf a+\lambda\mathbf b. The displacement from AA to BB is b−a\mathbf b-\mathbf a. Proportional directions describe the same line direction.

The second direction must genuinely add a choice: a scalar multiple of the first cannot open out a plane.

Vector equation of a plane

Through point AA, parallel to two non-zero, non-parallel vectors u,v\mathbf u,\mathbf v:

r=a+λu+μv,λ,μ∈R\mathbf r=\mathbf a+\lambda\mathbf u+\mu\mathbf v, \qquad\lambda,\mu\in\mathbb R

Through three non-collinear points A,B,CA,B,C, choose u=b−a\mathbf u=\mathbf b-\mathbf a, v=c−a\mathbf v=\mathbf c-\mathbf a. Equivalently,

r=(1−λ−μ)a+λb+μc\mathbf r=(1-\lambda-\mu)\mathbf a+\lambda\mathbf b+\mu\mathbf c

In the formula booklet.

The first form is often easier to read: a starting point plus two independent displacements.

[DIAGRAM: asset_name: Two independent directions span a plane; asset_slug: 9fm0-cp40-plane-directions; caption: Two parameters control independent displacements across the same plane; description: An exact spatial diagram shows the plane through A at (1,2,3), with directions u at (1,0,1) and v at (0,1,2). Point B is A plus twice u and point C is A plus v. Arrows from A show the two independent directions and a side panel gives r equals a plus lambda u plus mu v]
Two parameters control independent displacements across the same plane

Worked example 1: a plane through three points

Question. Find a vector equation of the plane through A=(1,2,3)A=(1,2,3), B=(3,2,5)B=(3,2,5), C=(1,3,5)C=(1,3,5).

Plan. Form two displacements from the same point and check that they are not proportional.

AB→=(2,0,2)Tsubtract A from BAC→=(0,1,2)Tsubtract A from Cr=(1,2,3)T+λ(1,0,1)T+μ(0,1,2)Tsimplify the first direction\begin{aligned} \overrightarrow{AB}&=(2,0,2)^{\mathrm T} && \text{subtract }A\text{ from }B \\ \overrightarrow{AC}&=(0,1,2)^{\mathrm T} && \text{subtract }A\text{ from }C \\ \mathbf r&=(1,2,3)^{\mathrm T}+\lambda(1,0,1)^{\mathrm T}+\mu(0,1,2)^{\mathrm T} && \text{simplify the first direction} \end{aligned}

Why this step? The first displacement has a non-zero first component and zero second component; the second has the reverse pattern, so they cannot be parallel.

Answer. The displayed equation, for real λ,μ\lambda,\mu, describes the required plane.

Check. (λ,μ)=(0,0),(2,0),(0,1)(\lambda,\mu)=(0,0),(2,0),(0,1) give A,B,CA,B,C, respectively.

Worked example 2 changes only the height of CC. One of the two spanning directions changes, so the plane tilts.

Worked example 2: one moved point

Question. Keep A,BA,B, but let C=(1,3,6)C=(1,3,6). Find a vector equation of the new plane.

Plan. Retain the first direction and recalculate the second.

AB→=(2,0,2)Tunchanged edgeAC→=(0,1,3)Tnew height differencer=(1,2,3)T+λ(1,0,1)T+μ(0,1,3)Ttwo non-parallel directions\begin{aligned} \overrightarrow{AB}&=(2,0,2)^{\mathrm T} && \text{unchanged edge} \\ \overrightarrow{AC}&=(0,1,3)^{\mathrm T} && \text{new height difference} \\ \mathbf r&=(1,2,3)^{\mathrm T}+\lambda(1,0,1)^{\mathrm T}+\mu(0,1,3)^{\mathrm T} && \text{two non-parallel directions} \end{aligned}

Why this step? Keeping A,BA,B fixes their line, but changing the third point can turn the plane about that line.

Answer. r=(1,2,3)T+λ(1,0,1)T+μ(0,1,3)T\mathbf r=(1,2,3)^{\mathrm T}+\lambda(1,0,1)^{\mathrm T}+\mu(0,1,3)^{\mathrm T}.

Check. λ=0,μ=1\lambda=0,\mu=1 gives (1,3,6)(1,3,6), while μ=0\mu=0 retains the original line ABAB.

This gives a small geometric surprise: two different planes can share infinitely many points, namely every point on their common line ABAB, while remaining different surfaces.

Watch out

Wrong: any three points determine one plane. Three collinear points determine a line, with infinitely many planes through it. Right: check that the two displacements are not proportional.

The next three points vary only the third height once again.

Eliminating the two parameters

A plane's Cartesian equation is one linear relation shared by all its points; eliminating both parameters reveals that relation.

The form is ax+by+cz=dax+by+cz=d, where a,b,ca,b,c are not all zero. A zero coefficient just means that coordinate is absent from the constraint. For example, z=3z=3 is a horizontal plane, not an incomplete equation. The Cartesian plane form is in the formula booklet.

Worked example 3: a Cartesian equation by substitution

Question. Find a Cartesian equation for r=(1,2,3)T+λ(1,0,1)T+μ(0,1,2)T\mathbf r=(1,2,3)^{\mathrm T}+\lambda(1,0,1)^{\mathrm T}+\mu(0,1,2)^{\mathrm T}.

Plan. Use x,yx,y to isolate the parameters, then substitute them into zz.

x=1+λ,y=2+μ,z=3+λ+2μcoordinate equationsλ=x−1,μ=y−2isolate parametersz=3+(x−1)+2(y−2)=x+2y−2substitutex+2y−z=2collect coordinates\begin{aligned} x&=1+\lambda,&y&=2+\mu,&z&=3+\lambda+2\mu && \text{coordinate equations} \\ \lambda&=x-1,&\mu&=y-2 && \text{isolate parameters} \\ z&=3+(x-1)+2(y-2)=x+2y-2 && \text{substitute} \\ x+2y-z&=2 && \text{collect coordinates} \end{aligned}

Why this step? Substitution leaves one relation with no parameter, so it applies to every point of the plane.

Answer. x+2y−z=2x+2y-z=2.

Check. The anchor gives 1+4−3=21+4-3=2; substitution of the full parameter expressions also cancels both parameters.

Worked example 4 changes only the last component of the second direction, from 22 to 33.

Worked example 4: the changed tilt

Question. Find a Cartesian equation for r=(1,2,3)T+λ(1,0,1)T+μ(0,1,3)T\mathbf r=(1,2,3)^{\mathrm T}+\lambda(1,0,1)^{\mathrm T}+\mu(0,1,3)^{\mathrm T}.

Plan. Keep the first two substitutions and update the height equation.

λ=x−1,μ=y−2unchanged first coordinatesz=3+(x−1)+3(y−2)new height equation=x+3y−4expand and collectx+3y−z=4Cartesian form\begin{aligned} \lambda&=x-1,&\mu&=y-2 && \text{unchanged first coordinates} \\ z&=3+(x-1)+3(y-2) && \text{new height equation} \\ &=x+3y-4 && \text{expand and collect} \\ x+3y-z&=4 && \text{Cartesian form} \end{aligned}

Why this step? The changed direction alters a coefficient and the constant; keeping the old right-hand side would lose the anchor point.

Answer. x+3y−z=4x+3y-z=4.

Check. (1,3,6)(1,3,6) gives 1+9−6=41+9-6=4.

You can reverse the process too. From x+3y−z=4x+3y-z=4, choose x=1+s,y=2+tx=1+s,y=2+t; then z=3+s+3tz=3+s+3t, recovering the vector equation. There is no need for a new operation on vectors.

Watch out

Wrong: use different values of the same parameter in different coordinates. Right: substitute the same λ,μ\lambda,\mu consistently. A candidate Cartesian equation must hold for the entire vector expression, not just one point.

The next plane changes the same coefficient once more.

What the plane equation includes

A valid plane equation describes all of a surface, even when some coordinates are fixed or some parameters look quite different.

For x+2y−z=2x+2y-z=2, any choice of x,yx,y gives exactly one z=x+2y−2z=x+2y-2. Thus it has two independent choices, like the two parameters. The point (1,2,3)(1,2,3) lies on it, while (1,2,4)(1,2,4) does not. Substitution is a test for membership, not a search for a special parameter.

Non-collinear spanning points

Three points are non-collinear when they do not lie on one straight line. Their two displacements from one chosen point are not scalar multiples. This is the condition needed for the three-point plane formula; repeated points or proportional directions cannot determine a unique plane.

Different equations can describe the same surface. Multiplying x+2y−z=2x+2y-z=2 by 33 gives 3x+6y−3z=63x+6y-3z=6, with exactly the same points. But changing only the right side to 55 produces a different, parallel surface. These equivalence and membership checks follow from the linear equation and are not separate formula-booklet results.

The opening question now has a precise answer. Three non-collinear points supply two independent directions. Collinear points supply only one, leaving the sheet free to turn about their line.

Key idea

A plane needs one point and two independent directions, or one non-trivial linear Cartesian constraint. Its parametrisation and its coefficients need not be unique.

Check independence before trusting a three-point construction.

Quick Recap

  • r=a+λu+μv\mathbf r=\mathbf a+\lambda\mathbf u+\mu\mathbf v describes a plane when the two directions are non-parallel.
  • Three non-collinear points provide two such displacements.
  • Eliminate parameters to obtain ax+by+cz=dax+by+cz=d, and reverse this by choosing two free coordinates.
  • Test points by substitution; non-zero scalar multiples of a Cartesian equation describe the same plane.

Three points fix the sheet only when they are non-collinear; three points along one edge leave infinitely many possible planes.