Mascot image.
#Math#Vectors#Geometry

More on Planes and Systems

Lesson 2PM attached every plane in R3\mathbb{R}^3 to a single linear expression and identified its normal vector. That description is enough for asking whether a given point lies on the plane, but several natural questions involve the plane only incidentally. A computer rendering a scene has to decide, for every pair of objects, whether a flat surface between them blocks the line of sight; a parametric animation of a ball rolling along a tilted roof has to stay on the surface at every frame. Neither question is answered by finding a point of the plane. The first needs an algebraic test for which side of the plane a point lies on, and the second needs a description that produces points of the plane rather than one that checks them. Both live a short algebraic step from the material we already have, and this lesson walks the step.

Parametric Description of a Plane

The Parametric Equation of a Plane from Lesson 2PM wrote every point of the plane through three non-collinear points AA, BB, CC as a barycentric combination x=ra+sb+tc\mathbf{x} = r\mathbf{a} + s\mathbf{b} + t\mathbf{c} with r+s+t=1r + s + t = 1. Using the constraint to eliminate r=1str = 1 - s - t rearranges that combination into

x=a+s(ba)+t(ca)=p+se+te,\mathbf{x} = \mathbf{a} + s(\mathbf{b} - \mathbf{a}) + t(\mathbf{c} - \mathbf{a}) = \mathbf{p} + s\mathbf{e} + t\mathbf{e}',

with base point p=a\mathbf{p} = \mathbf{a} and direction vectors e=ba\mathbf{e} = \mathbf{b} - \mathbf{a}, e=ca\mathbf{e}' = \mathbf{c} - \mathbf{a} lying inside the plane. The three-point form and this “base point plus two directions” form carry the same information, but the latter is the natural input whenever the plane comes to us through a point on it together with two independent directions it contains, rather than through three named vertices.

Theorem 43 (Parametric Description via Base Point and Directions)

Let pR3\mathbf{p} \in \mathbb{R}^3 and let e,eR3\mathbf{e}, \mathbf{e}' \in \mathbb{R}^3 be linearly independent. The set

Π={p+se+te:s,tR}\Pi = \{ \mathbf{p} + s\mathbf{e} + t\mathbf{e}' : s, t \in \mathbb{R} \}

is a plane in R3\mathbb{R}^3, and every plane arises from some such triple (p,e,e)(\mathbf{p}, \mathbf{e}, \mathbf{e}'). Each point of Π\Pi corresponds to a unique pair (s,t)(s, t).

Proof

The three position vectors p\mathbf{p}, p+e\mathbf{p} + \mathbf{e}, p+e\mathbf{p} + \mathbf{e}' correspond to non-collinear points because e\mathbf{e} and e\mathbf{e}' are linearly independent, so the Parametric Equation of a Plane from Lesson 2PM identifies Π\Pi with the unique plane through them. If p+se+te=p+se+te\mathbf{p} + s\mathbf{e} + t\mathbf{e}' = \mathbf{p} + s'\mathbf{e} + t'\mathbf{e}' then (ss)e+(tt)e=0(s - s')\mathbf{e} + (t - t')\mathbf{e}' = \mathbf{0}, and linear independence forces s=ss = s', t=tt = t', proving uniqueness of the parameters.

Conversely, let Π\Pi' be an arbitrary plane and pick three non-collinear points with position vectors a,b,cΠ\mathbf{a}, \mathbf{b}, \mathbf{c} \in \Pi'. The displacements e=ba\mathbf{e} = \mathbf{b} - \mathbf{a} and e=ca\mathbf{e}' = \mathbf{c} - \mathbf{a} are linearly independent, and the Parametric Equation of a Plane writes every point of Π\Pi' in the required form with p=a\mathbf{p} = \mathbf{a}.

The pair (s,t)(s, t) is a coordinate system inside the plane, with axes pointing along e\mathbf{e} and e\mathbf{e}'. These internal axes need not be perpendicular, and when the plane is viewed from a steep angle the induced grid can look distorted, in the same way that a tilted surface appears foreshortened from the side.

The plane x plus 2y plus 3z equals 4 drawn as a translucent surface, with base point P at 4 0 0, direction arrows e and e prime emanating from P into the plane, and a sample point Q equal to P plus e plus e prime at minus 1 1 1 reached by the dashed parameter route

The figure above realises the example x+2y+3z=4x + 2y + 3z = 4 from the next subsection: the base point is P=(4,0,0)P = (4, 0, 0), the direction arrows are e=(2,1,0)\mathbf{e} = (-2, 1, 0) and e=(3,0,1)\mathbf{e}' = (-3, 0, 1), and the sample point Q=P+e+e=(1,1,1)Q = P + \mathbf{e} + \mathbf{e}' = (-1, 1, 1) is reached by the dashed parameter route PP+eP+e+eP \to P + \mathbf{e} \to P + \mathbf{e} + \mathbf{e}'. In the internal (s,t)(s, t)-coordinates of the plane, QQ sits at (s,t)=(1,1)(s, t) = (1, 1).

Translating Between Forms

The equational form ax+c=0\mathbf{a} \cdot \mathbf{x} + c = 0 and the parametric form p+se+te\mathbf{p} + s\mathbf{e} + t\mathbf{e}' answer different questions, and most practical work with planes involves passing from one to the other. The mechanical content of each conversion is short enough to absorb through a worked example.

Example 41 (From Equation to Parametric Form)

Parametrise the plane x+2y+3z=4x + 2y + 3z = 4.

Solve for one coordinate in terms of the others: x=42y3zx = 4 - 2y - 3z. Taking y=sy = s and z=tz = t as free parameters,

[xyz]=[42s3tst]=[400]+s[210]+t[301],\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 4 - 2s - 3t \\ s \\ t \end{bmatrix} = \begin{bmatrix} 4 \\ 0 \\ 0 \end{bmatrix} + s\begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix} + t\begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix},

which places the plane in parametric form with base point P=(4,0,0)P = (4, 0, 0) and direction vectors

e=[210],e=[301].\mathbf{e} = \begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix}, \qquad \mathbf{e}' = \begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix}.

These directions must be orthogonal to the normal

n=[123],\mathbf{n} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix},

and indeed ne=2+2+0=0\mathbf{n} \cdot \mathbf{e} = -2 + 2 + 0 = 0 and ne=3+0+3=0\mathbf{n} \cdot \mathbf{e}' = -3 + 0 + 3 = 0, as required.

Example 42 (From Three Points to Equational Form)

Find an equation for the plane through A=(4,0,0)A = (4, 0, 0), B=(0,2,0)B = (0, 2, 0), C=(1,0,1)C = (1, 0, 1).

The displacements

e=AB=[420],e=AC=[301]\mathbf{e} = \overrightarrow{AB} = \begin{bmatrix} -4 \\ 2 \\ 0 \end{bmatrix}, \qquad \mathbf{e}' = \overrightarrow{AC} = \begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix}

span the directions inside the plane, so a normal vector

a=[a1a2a3]\mathbf{a} = \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix}

is determined by the two orthogonality conditions

ae=4a1+2a2=0,ae=3a1+a3=0.\mathbf{a} \cdot \mathbf{e} = -4 a_1 + 2 a_2 = 0, \qquad \mathbf{a} \cdot \mathbf{e}' = -3 a_1 + a_3 = 0.

The first gives a2=2a1a_2 = 2 a_1 and the second a3=3a1a_3 = 3 a_1, so up to scale

a=[123].\mathbf{a} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}.

The constant cc is pinned down by demanding that AA satisfy the equation: letting pA=[400]\mathbf{p}_A = \begin{bmatrix} 4 \\ 0 \\ 0 \end{bmatrix} denote the position vector of AA,

c=apA=(14+20+30)=4.c = -\mathbf{a} \cdot \mathbf{p}_A = -(1 \cdot 4 + 2 \cdot 0 + 3 \cdot 0) = -4.

The resulting plane x+2y+3z=4x + 2y + 3z = 4 coincides with the one parametrised in the previous example, as intended.

The construction tacitly assumes that A,B,CA, B, C are non-collinear, and that hypothesis is easiest to check directly on the displacement vectors: three points A,B,CA, B, C lie on a common line precisely when one of AB,AC\overrightarrow{AB}, \overrightarrow{AC} is a scalar multiple of the other, with a positive multiple putting BB and CC on the same side of AA and a negative multiple putting them on opposite sides. Otherwise the two displacements span a genuine plane through AA, and the procedure of the previous example goes through.

Example 43 (Collinearity Test via Difference Vectors)

For A=(0,1,1)A = (0, 1, 1), B=(0,2,3)B = (0, 2, 3), C=(1,3,2)C = (1, 3, 2),

AB=[012],AC=[121].\overrightarrow{AB} = \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}, \qquad \overrightarrow{AC} = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}.

An equation AC=μAB\overrightarrow{AC} = \mu \overrightarrow{AB} would force 1=μ01 = \mu \cdot 0 in the first coordinate, which has no solution, so the two displacements are not scalar multiples of each other. The three points are therefore non-collinear and determine a plane; running the recipe of the previous example on them gives normal n=[321]\mathbf{n} = \begin{bmatrix} 3 \\ -2 \\ 1 \end{bmatrix} and equation 3x2y+z=13x - 2y + z = -1.

Starting from a Point and a Normal

A third natural input packages a plane as a point PP lying on it together with a vector n\mathbf{n} perpendicular to it. The equational form is immediate from the perpendicularity of n\mathbf{n} to every displacement inside the plane:

n(xp)=0.\mathbf{n} \cdot (\mathbf{x} - \mathbf{p}) = 0.

Passing to a parametric form requires two linearly independent directions orthogonal to n\mathbf{n}. A reliable recipe, which sidesteps the cross product and so generalises verbatim to Rn\mathbb{R}^n, is to read off two auxiliary points of the plane by fixing two of the three coordinates at convenient values (typically 00 and 11) and solving the equation for the third. Repeating with a different coordinate held constant gives a second auxiliary point, and a quick collinearity check on the two displacements from PP certifies that they are independent.

Example 44 (From a Point and a Normal to Both Forms)

Give equational and parametric descriptions of the plane through P=(1,2,3)P = (1, 2, 3) with normal n=[345]\mathbf{n} = \begin{bmatrix} 3 \\ 4 \\ 5 \end{bmatrix}.

The perpendicularity condition expands to

3(x1)+4(y2)+5(z3)=0,i.e.3x+4y+5z=26.3(x - 1) + 4(y - 2) + 5(z - 3) = 0, \qquad \text{i.e.} \qquad 3x + 4y + 5z = 26.

For the parametric form, set y=0,z=1y = 0, z = 1: 3x+5=263x + 5 = 26 gives x=7x = 7, so Q=(7,0,1)Q = (7, 0, 1) lies on the plane. Set z=0,x=1z = 0, x = 1: 3+4y=263 + 4y = 26 gives y=23/4y = 23/4, so R=(1,23/4,0)R = (1, 23/4, 0) lies on the plane. The displacements

PQ=[622],PR=[015/43]\overrightarrow{PQ} = \begin{bmatrix} 6 \\ -2 \\ -2 \end{bmatrix}, \qquad \overrightarrow{PR} = \begin{bmatrix} 0 \\ 15/4 \\ -3 \end{bmatrix}

disagree in first coordinate (66 versus 00), so they are not scalar multiples, and

[xyz]=[123]+s[622]+t[015/43]\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} + s \begin{bmatrix} 6 \\ -2 \\ -2 \end{bmatrix} + t \begin{bmatrix} 0 \\ 15/4 \\ -3 \end{bmatrix}

parametrises the plane.

When the plane instead passes through the origin, the recipe shortens, because P=0P = \mathbf{0} supplies the base point and forces the constant term to vanish. For the plane through 0\mathbf{0} with normal [218]\begin{bmatrix} 2 \\ 1 \\ 8 \end{bmatrix}, the equation is 2x+y+8z=02x + y + 8z = 0; setting y=1,z=0y = 1, z = 0 gives Q=(1/2,1,0)Q = (-1/2, 1, 0), and setting x=0,z=1x = 0, z = 1 gives S=(0,8,1)S = (0, -8, 1). The displacements from 0\mathbf{0} are QQ and SS themselves, not scalar multiples (their zz-coordinates are 00 and 11), and

[xyz]=s[1/210]+t[081]\begin{bmatrix} x \\ y \\ z \end{bmatrix} = s\begin{bmatrix} -1/2 \\ 1 \\ 0 \end{bmatrix} + t\begin{bmatrix} 0 \\ -8 \\ 1 \end{bmatrix}

parametrises the plane.

The plane x plus 2y plus 3z equals 4 drawn with a blue lattice of internal coordinate lines s equals constant and t equals constant overlaid on it, with base point P at 4 0 0, red direction arrows e and e prime, and the lattice point at s t equals 1 1 sitting at minus 1 1 1

The lattice in the figure is the image of the integer (s,t)(s, t)-grid under the parametrisation x(s,t)=p+se+te\mathbf{x}(s, t) = \mathbf{p} + s\mathbf{e} + t\mathbf{e}' of the plane x+2y+3z=4x + 2y + 3z = 4 from Example 41. The two families of gridlines are the fibres {s=const}\{s = \text{const}\} and {t=const}\{t = \text{const}\}, each itself a line inside the plane, and they meet at the lattice points.

Traces Inside a Plane

One payoff of the parametric form is that it constrains motion to the plane for free. For any continuous curve τ(s(τ),t(τ))\tau \mapsto (s(\tau), t(\tau)) in the (s,t)(s, t)-parameter space, the composition

x(τ)=p+s(τ)e+t(τ)e\mathbf{x}(\tau) = \mathbf{p} + s(\tau)\mathbf{e} + t(\tau)\mathbf{e}'

is automatically a continuous curve in R3\mathbb{R}^3 lying entirely on the plane. The equational form ax+c=0\mathbf{a} \cdot \mathbf{x} + c = 0 offers no such guarantee, so whenever the task is to generate points on a plane, for example when rendering a path along a surface, the parametric form is the right tool, whereas whenever the task is to test points already in hand, the equational form is.

Sides of a Plane

A plane partitions R3\mathbb{R}^3 into three disjoint pieces: the plane itself and the two open regions on either side. The linear expression whose zero set defines the plane already contains the information needed to tell these regions apart: the only thing that changes between them is the sign of that expression.

Definition 35 (Half-Spaces of a Plane)

Let f(x)=ax+cf(\mathbf{x}) = \mathbf{a} \cdot \mathbf{x} + c with a0\mathbf{a} \neq \mathbf{0}, and let E={f=0}E = \{f = 0\} be the associated plane. The open half-spaces of EE are

H+={xR3:f(x)>0},H={xR3:f(x)<0}.H^+ = \{\mathbf{x} \in \mathbb{R}^3 : f(\mathbf{x}) > 0\}, \qquad H^- = \{\mathbf{x} \in \mathbb{R}^3 : f(\mathbf{x}) < 0\}.

Two points P,QEP, Q \notin E lie on the same side of EE when f(p)f(\mathbf{p}) and f(q)f(\mathbf{q}) share a sign, and on opposite sides when the signs differ.

Every point of R3\mathbb{R}^3 belongs to exactly one of EE, H+H^+, HH^-, since ff takes a unique real value at each point and the three cases f>0f > 0, f=0f = 0, f<0f < 0 exhaust the possibilities. The geometric question of which side of EE a point lies on therefore reduces to a single sign comparison, and the algebraic shortcut extends to the related question of when two points are separated by the plane.

Theorem 44 (Sign Test for Sides of a Plane)

Let f(x)=ax+cf(\mathbf{x}) = \mathbf{a} \cdot \mathbf{x} + c define a plane EE, and let P,QP, Q be points with P,QEP, Q \notin E. The segment PQ\overline{PQ} meets EE if and only if f(p)f(\mathbf{p}) and f(q)f(\mathbf{q}) have opposite signs, and in that case the intersection point is unique.

Proof

The Ratio of Division by a Plane from Lesson 2PM fixes the location of the intersection of the line PQPQ with EE: a point RR with directed ratio λ=dr(P,Q;R)\lambda = \mathrm{dr}(P, Q;\, R) lies on EE precisely when

λ=f(p)f(q).\lambda = -\frac{f(\mathbf{p})}{f(\mathbf{q})}.

Because QEQ \notin E, the denominator is non-zero, so λ\lambda is a well-defined real number. The intersection RR lies on the closed segment PQ\overline{PQ} rather than on its extension exactly when λ0\lambda \geq 0, and this inequality holds if and only if f(p)f(\mathbf{p}) and f(q)f(\mathbf{q}) have opposite signs (the case λ=0\lambda = 0 corresponds to f(p)=0f(\mathbf{p}) = 0, which the hypothesis PEP \notin E excludes). Uniqueness of λ\lambda gives uniqueness of RR.

The test turns the geometric question of whether a plane separates two points into one arithmetic evaluation per point. That is the mechanism behind one of the basic primitives in three-dimensional rendering, and it also sits at the algebraic heart of a much larger family of questions across dimensions.

Example 45 (Line of Sight Across a Tilted Roof)

A video game places a viewer at V=(1,1,2)V = (1, 1, 2) at the bottom of a driveway, and a tree at T=(2,1,3)T = (-2, 1, 3) in the backyard. The relevant section of roof lies in the plane E:x+2y+3z=4E: x + 2y + 3z = 4. Does the roof block the viewer’s sight of the tree?

Set f(x)=x+2y+3z4f(\mathbf{x}) = x + 2y + 3z - 4 and evaluate:

f(v)=1+2+64=5,f(t)=2+2+94=5.f(\mathbf{v}) = 1 + 2 + 6 - 4 = 5, \qquad f(\mathbf{t}) = -2 + 2 + 9 - 4 = 5.

Both values are positive, so V,TH+V, T \in H^+; by the Sign Test the segment VT\overline{VT} does not meet EE and the sight line is unobstructed. If the tree were instead at T=(3,1,2)T' = (3, 1, -2), the evaluation would give f(t)=3+264=5<0f(\mathbf{t}') = 3 + 2 - 6 - 4 = -5 < 0, placing THT' \in H^-. The signs at VV and TT' would then disagree, the segment VT\overline{VT'} would cross EE, and the Ratio of Division by a Plane would place the obstruction at directed ratio

λ=f(v)f(t)=55=1,\lambda = -\frac{f(\mathbf{v})}{f(\mathbf{t}')} = -\frac{5}{-5} = 1,

i.e. at the midpoint of the sight line.

A tilted plane E partitioning space into two open half-spaces, with two points V and T on the positive side joined by an unobstructed green segment, and a third point T prime on the negative side joined to V by a dashed red segment that crosses E at a point R
Remark

The entire discussion above uses only the sign of a single linear expression and so transplants verbatim to Rn\mathbb{R}^n: a linear form f(x)=ax+cf(\mathbf{x}) = \mathbf{a} \cdot \mathbf{x} + c on Rn\mathbb{R}^n cuts space into a hyperplane {f=0}\{f = 0\} and two open half-spaces, and the same sign test decides when a segment between two points is split by the hyperplane. In large nn this is exactly the linear-algebraic machinery behind separating hyperplanes in convex optimisation and support vector machines, but the proof ingredients are the ones already used in R3\mathbb{R}^3.

Problem 28

Let EE be the plane 2xy+z=32x - y + z = 3, and let P1=(3,1,0)P_1 = (3, 1, 0), P2=(0,1,4)P_2 = (0, -1, 4), P3=(1,3,2)P_3 = (1, 3, 2). Determine which pairs among P1,P2,P3P_1, P_2, P_3 lie on the same side of EE and which lie on opposite sides. For each separated pair, use the Ratio of Division by a Plane from Lesson 2PM to compute the directed ratio at which the connecting segment meets EE.

Lines in Three-Dimensional Space

A line in R3\mathbb{R}^3 is not the zero set of a single linear expression, since that zero set cuts out a plane. The parametric form, which carried two direction vectors alongside a base point when describing a plane, collapses to the natural description of a line once a single direction is kept.

Theorem 45 (Parametric Description of a Line)

Let pR3\mathbf{p} \in \mathbb{R}^3 and let dR3\mathbf{d} \in \mathbb{R}^3 be non-zero. The set

L={p+td:tR}L = \{ \mathbf{p} + t\mathbf{d} : t \in \mathbb{R} \}

is the unique line through p\mathbf{p} with direction d\mathbf{d}, and each point of LL corresponds to a unique tRt \in \mathbb{R}. Every line in R3\mathbb{R}^3 arises from some such pair (p,d)(\mathbf{p}, \mathbf{d}).

When the line is presented through two distinct points P,QP, Q, the displacement d=qp\mathbf{d} = \mathbf{q} - \mathbf{p} supplies the direction, and the parametric form becomes

x(t)=p+t(qp)=(1t)p+tq,\mathbf{x}(t) = \mathbf{p} + t(\mathbf{q} - \mathbf{p}) = (1 - t)\mathbf{p} + t\mathbf{q},

which recovers PP at t=0t = 0, QQ at t=1t = 1, and the midpoint of PQ\overline{PQ} at t=1/2t = 1/2. For t1t \neq 1, the Directed Ratio from Lesson 2PM is

dr(P,Q;X(t))=t1t,\mathrm{dr}(P, Q;\, X(t)) = \frac{t}{1-t},

so the parameter tt determines the directed ratio, but is not itself the directed ratio.

Example 46 (Lines in Point-Direction and Two-Point Forms)

For base point p=(1,5,2)\mathbf{p} = (1, -5, 2) and direction d=(3,2,7)\mathbf{d} = (-3, 2, 7), the line through PP along d\mathbf{d} is

x(t)=[152]+t[327]=[13t5+2t2+7t].\mathbf{x}(t) = \begin{bmatrix} 1 \\ -5 \\ 2 \end{bmatrix} + t\begin{bmatrix} -3 \\ 2 \\ 7 \end{bmatrix} = \begin{bmatrix} 1 - 3t \\ -5 + 2t \\ 2 + 7t \end{bmatrix}.

For the line through the distinct points p=(3,1,1)\mathbf{p} = (3, 1, -1) and q=(2,4,5)\mathbf{q} = (2, 4, 5), the direction is d=qp=(1,3,6)\mathbf{d} = \mathbf{q} - \mathbf{p} = (-1, 3, 6), and

x(t)=(1t)[311]+t[245]=[3t1+3t1+6t].\mathbf{x}(t) = (1 - t)\begin{bmatrix} 3 \\ 1 \\ -1 \end{bmatrix} + t\begin{bmatrix} 2 \\ 4 \\ 5 \end{bmatrix} = \begin{bmatrix} 3 - t \\ 1 + 3t \\ -1 + 6t \end{bmatrix}.

The single-parameter description of a line and the two-parameter description of a plane are algebraic shadows of their geometric dimensions, and the number of free parameters needed to sweep out a figure will be our first handle on the notion of dimension in later lessons, where it extends far beyond the R3\mathbb{R}^3 of daily experience.

Assembling a Plane from Parallel Slices

Throughout this lesson and the last, every equation ax+by+cz=dax + by + cz = d with (a,b,c)0(a, b, c) \neq \mathbf{0} has been treated as the zero set of a plane, without a derivation of that fact from anything more primitive. One way to see why the claim is true is to assemble the surface from its slices by vertical planes, building on the familiar fact that a linear equation in two variables cuts out a line.

Relabelling axes if necessary, arrange that c0c \neq 0 and divide through to put the equation in the form

z=αx+βy+γ.z = \alpha x + \beta y + \gamma.

Fixing x=hx = h collapses the equation to z=βy+(αh+γ)z = \beta y + (\alpha h + \gamma), a line of slope β\beta in the (y,z)(y, z)-plane sitting at {x=h}\{x = h\}, with vertical intercept f(h)=αh+γf(h) = \alpha h + \gamma. As hh varies, these parallel-sloped lines stack to form the full surface, and the intercept f(h)f(h) is linear in hh, so its graph is a straight ruling and the stacked slices fit together flat.

Two side-by-side three-dimensional plots: on the left a blue plane z equals minus 2x minus y plus 3 sliced by three vertical planes x equals minus 1, 0, 1, each slice a parallel line of slope minus 1 whose vertical intercepts space evenly so the surface is flat; on the right a wavy blue surface sliced by the same three vertical planes, each slice again a line of slope minus 1 but with intercepts that no longer space evenly, producing a curved ruled surface instead of a plane

The contrast with the right-hand surface is instructive. Both pictures show a surface whose every slice {x=h}\{x = h\} is a line of slope 1-1 in the (y,z)(y, z)-plane. On the left the intercept f(h)=2h+3f(h) = -2h + 3 depends linearly on hh and the stacked slices produce the plane z=2xy+3z = -2x - y + 3; on the right f(h)=212h3+16h2+76hf(h) = 2 - \tfrac{1}{2}h^3 + \tfrac{1}{6}h^2 + \tfrac{7}{6}h is a cubic, and the same parallel-slope lines assemble instead into a wavy ruled surface. Linearity of ff in hh is the structural ingredient that turns a family of parallel slices into a flat plane, and its absence is what allows an assembly of parallel-slope lines to buckle.

Problem 29

Let LL be the line through A=(1,0,2)A = (1, 0, 2) with direction d=(2,2,3)\mathbf{d} = (2, 2, -3), and let EE be the plane xy+2z=4x - y + 2z = 4. Find the value of tt at which the parametrisation x(t)=a+td\mathbf{x}(t) = \mathbf{a} + t\mathbf{d} meets EE, give the coordinates of the intersection point, and determine whether that intersection lies between AA and A+dA + \mathbf{d} on the line or outside the segment.

Toward Systems of Linear Equations

Every equation a1x1+a2x2+a3x3=ba_1 x_1 + a_2 x_2 + a_3 x_3 = b encountered in this lesson is a single linear equation in three unknowns, cutting out a plane in R3\mathbb{R}^3. Problem 29 already nudged past that one-equation setting: finding where a line meets a plane is the problem of jointly satisfying the line’s three parametric equations and the plane’s single equation, four scalar constraints tying together four unknowns (x,y,z,t)(x, y, z, t). The intersection of two non-parallel planes is the joint solution set of their two defining equations in three unknowns, and the intersection of three generic planes is the joint solution set of three. Each configuration is a small instance of one general object: a system of linear equations.

The general system of mm linear equations in nn unknowns has the form

{a11x1+a12x2++a1nxn=b1,a21x1+a22x2++a2nxn=b2,am1x1+am2x2++amnxn=bm,\begin{cases} a_{11} x_1 + a_{12} x_2 + \cdots + a_{1n} x_n = b_1, \\ a_{21} x_1 + a_{22} x_2 + \cdots + a_{2n} x_n = b_2, \\ \qquad\qquad\qquad\quad\vdots \\ a_{m1} x_1 + a_{m2} x_2 + \cdots + a_{mn} x_n = b_m, \end{cases}

with coefficients aija_{ij} and right-hand sides bib_i. Equivalently, the iith equation may be written in the compact summation form

j=1naijxj=bi.\sum_{j=1}^n a_{ij} x_j = b_i.

When every bi=0b_i = 0 the system is called homogeneous, and otherwise inhomogeneous. Both the aija_{ij} and the bib_i are drawn from a chosen field, and we write F\mathbb{F} for such a generic field: a set carrying addition, subtraction, multiplication, and division subject to the nine field axioms (A1)-(A9) laid out for the real numbers in the MCEA lesson “Predicates, Numbers, and Sets”. Those nine axioms, and only those, are what every manipulation in the chapters ahead is ultimately allowed to invoke, so any statement proved from them applies uniformly to every choice of F\mathbb{F}; the familiar ones are Q\mathbb{Q}, R\mathbb{R}, and C\mathbb{C}, and unless stated otherwise F\mathbb{F} will mean one of these three with the default reading F=R\mathbb{F} = \mathbb{R}.

Systems of this form have been studied since the beginnings of algebra in its modern sense. Newton’s Arithmetica Universalis, compiled from his Cambridge lectures of the 1670s and published in 1707, already set out systematic elimination for small linear systems; and the matrix notation that packages the coefficients aija_{ij} into a rectangular array was introduced by Arthur Cayley in his 1858 Memoir on the Theory of Matrices, building on the determinantal calculus developed a few years earlier by his Cambridge colleague James Joseph Sylvester. The linear-algebraic investigation of such systems has since grown into the hub of the whole subject, dragging along with it the theories of linear spaces, linear transformations, and matrices.

Three basic questions organise everything that follows.

  1. Existence. Does the system have any solution at all?
  2. Uniqueness. When a solution exists, is it the only one?
  3. Structure. When there are infinitely many solutions, what geometric shape does the solution set have, and how is it parametrised?

For the tiny cases visited in this lesson and the last, geometry in R2\mathbb{R}^2 and R3\mathbb{R}^3 already answers all three at a glance: two non-parallel lines in the plane R2\mathbb{R}^2 meet in a single point; two non-parallel planes in R3\mathbb{R}^3 meet in a line, a one-parameter family of solutions; three generic planes in R3\mathbb{R}^3 meet in a point; and two parallel but distinct planes meet in the empty set. In higher dimensions, and with more than three equations, that intuition runs out, and we will need a systematic algebraic procedure together with the language of matrices and ranks to reduce existence, uniqueness, and structure to bookkeeping. That procedure, and the vocabulary that goes with it, is the subject of the next few lessons, where we work out row reduction for arbitrary m×nm \times n systems and see how it settles all three questions uniformly.

Problem 30

Consider the system

{x+y+z=1,xy+2z=0.\begin{cases} x + y + z = 1, \\ x - y + 2z = 0. \end{cases}

How many equations and unknowns does it have? Is it homogeneous or inhomogeneous? What geometric object does each equation define in R3\mathbb{R}^3, and what shape should you expect for the full solution set before carrying out any elimination?