Solids, Planes, and Systems in Space

In Lesson 2AM we built the vector space $\mathbb{R}^3$, recovered the Euclidean metric through iterated Pythagoras, and worked out which collections of space vectors are linearly independent. What was still missing was any genuine use of that machinery against solid figures: tetrahedra, planes fixed by three non-aligned points, cevians viewed as segments inside a triangular slice of space. We now put the algebra to work. Each definition that follows is chosen so that the object it introduces earns its keep in the proof it enables, and each proof settles a question that space geometry had previously left open.

Displacement Vectors in Space

The first thing space geometry asks of us is a way to describe the relative position of two points without forcing the origin into every calculation. In Lesson 1PM we did exactly this in the plane by attaching the displacement $\mathbf{b} - \mathbf{a}$ to the directed segment $\overrightarrow{AB}$. The construction transplants into $\mathbb{R}^3$ component by component.

Because the algebraic laws of addition and scalar multiplication are identical in $\mathbb{R}^2$ and $\mathbb{R}^3$, every theorem from Lesson 1PM whose proof used only those laws transfers without a single altered line. Three pieces will matter for the rest of this lesson:

• Midpoint Formula: the midpoint $M$ of $\overline{AB}$ has position vector $\mathbf{m} = \tfrac{1}{2}(\mathbf{a} + \mathbf{b})$.
• Collinearity: three points $A$, $B$, $C$ are collinear if and only if $\overrightarrow{AC} = k\,\overrightarrow{AB}$ for some scalar $k$.
• Parametric Line Equation: the line through distinct $A$ and $B$ consists of the points $X$ with $\mathbf{x} = (1-t)\mathbf{a} + t\mathbf{b}$, and the interpretation of $t$ fixed in Lesson 1PM is unchanged.

A triangle in space still lies in a unique plane, so any statement about a triangle proved purely from position vectors is inherited at once. In particular, the concurrency of medians at the centroid holds for a triangle placed anywhere in $\mathbb{R}^3$: the proof in Lesson 1PM never touched a second coordinate.

The Tetrahedron

With displacements and lines available, we can turn to a figure that cannot be drawn in the plane at all. A triangle is the smallest closed figure the plane admits; in space the smallest closed solid requires a fourth vertex lifted out of the plane of the first three.

The requirement that the four points not be coplanar is the spatial analogue of the non-collinearity demanded of the vertices of a triangle, and Lesson 2AM already supplies the exact test: the displacements $\overrightarrow{AB}$, $\overrightarrow{AC}$, $\overrightarrow{AD}$ must be linearly independent. Every tetrahedron therefore comes with a built-in certificate of its own three-dimensionality.

A triangle has three medians. A tetrahedron carries two families of distinguished internal segments.

There are four medians (one per vertex) and three bimedians (one per pair of opposite edges). Seven internal segments is a lot to keep track of, and the next theorem is precisely what keeps the picture tractable.

The figure above fixes the labels used in the proof: $U$ is the midpoint of $BC$, $V$ the midpoint of $AD$, and $E$ the centroid of face $BCD$. The point $M$ is where the bimedian $UV$ and the median $AE$ intersect, and the same intersection holds for every other pair of median and bimedian by symmetry.

Planes, Barycentric Coordinates, and Ceva’s Theorem

The proof above quietly used a fact that deserves to be isolated: the average of three vertex vectors is the centroid of the triangular face they span, and more generally every point of that face is a weighted average of the three vertices. This is the geometric motivation behind the parametric equation of a plane, which plays the same structural role in $\mathbb{R}^3$ as the parametric line equation did in $\mathbb{R}^2$.

The tetrahedron proof has already shown the first instance of this idea. When we wrote the centroid of face $BCD$ as

we were using a weighted average of the three vertices of that face. That is not a special trick reserved for centroids. More generally, every point in the plane determined by three non-collinear points can be described by weighting those three vertices, provided the weights add to $1$. This is the planar analogue of the parametric line equation: on a line through $A$ and $B$, the coefficients of $\mathbf{a}$ and $\mathbf{b}$ add to $1$; on a plane through $A$, $B$, and $C$, the coefficients of $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$ do the same.

The triple $(r, s, t)$ is called the set of barycentric coordinates of $X$ relative to $\triangle ABC$. Physically, they are exactly the weights one would place at $A$, $B$, $C$ so that the centre of mass of the weighted system sits at $X$. When $r = s = t = 1/3$ the centre of mass is the centroid, which recovers the formula we already used for the face centroid in the tetrahedron proof. This is not a coincidence: the tetrahedron centroid was constructed out of face centroids, and the face centroid is the simplest barycentric point.

Once points in a triangle are written in this weighted form, ratios along cevians become algebraic relations among the coefficients. With that perspective, the next theorem becomes almost purely algebraic, even though it is one of the oldest and most elegant results about triangles.

The Dot Product in Space

Everything to this point has been affine: positions, displacements, ratios along lines, concurrence. To say anything about lengths and angles inside a spatial figure, we need the dot product again, and we need it for three-component vectors. The extension is dictated by the component structure set up in Lesson 2AM.

The only change from Lesson 1PM is the presence of a third summand. As in the planar case, the self dot product recovers the square of the magnitude fixed in Lesson 2AM:

so the Euclidean distance between space points can be written as $|AB| = \|\mathbf{b} - \mathbf{a}\|$, bridging the dot product and the distance formula from Lesson 2AM.

Metric Identities in Space

Because the algebraic laws of the dot product are the same in $\mathbb{R}^2$ and $\mathbb{R}^3$, the entire chain of metric inequalities already proved in Lesson 1PM rebroadcasts into space with no further work.

The absence of any new work here is itself the punchline. Metric geometry in space inherits metric geometry in the plane once the algebraic framework is in place; and the reason is simple enough to state precisely. Any triangle drawn inside a spatial figure lies in a plane of its own, and every metric inequality above is a statement about pairs of vectors that collectively span at most a single plane.

Angles, Orthogonality, and Projection

Because the Cauchy-Schwarz inequality still holds, the quantity $\dfrac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\|\,\|\mathbf{b}\|}$ remains in $[-1, 1]$ for all non-zero $\mathbf{a}, \mathbf{b} \in \mathbb{R}^3$. The definition of angle used in Lesson 1PM lifts to space without modification.

This is consistent with the geometric angle $\angle AOB$ formed by the segments $OA$ and $OB$: the triangle $OAB$ lies in the unique plane determined by $\mathbf{a}$ and $\mathbf{b}$, and inside that plane the angle reduces to the planar one. The ambient third dimension plays no role.

Orthogonal Projection

Resolving a vector into a component along a reference direction and a component perpendicular to it is another theorem that needs no new proof: the argument in Lesson 1PM depended only on bilinearity and positive definiteness. In space, the decomposition still takes place inside a plane, namely the plane spanned by the two vectors involved.

The triangle with vertices $O$, $\mathrm{proj}_{\mathbf{a}}\mathbf{b}$, $B$ lies in the plane spanned by $\mathbf{a}$ and $\mathbf{b}$ and is right-angled at the tip of the projection. Its hypotenuse has length $\|\mathbf{b}\|$ and the leg along $\mathbf{a}$ has length

Reading the inequality hypotenuse $\ge$ leg out of this triangle gives a purely geometric proof of Cauchy-Schwarz, this time without a discriminant in sight. The equality cases of the metric inequalities now fall out immediately.

Planes in Space

The orthogonal projection just developed turns the dot product from an angle-measuring device into a way of writing down loci. A plane in $\mathbb{R}^3$ is the simplest such locus: fix a direction, and demand that every displacement from a chosen point be orthogonal to it. Unpacking that single sentence delivers the whole elementary theory of the plane as a set of points, and its results will recover in space everything Lesson 1PM established for the line in the plane.

Begin with the homogeneous linear equation

which the dot product collapses to $\mathbf{a} \cdot \mathbf{x} = 0$, where

This is the set of position vectors orthogonal to the fixed vector $\mathbf{a}$, which is precisely the plane through the origin with $\mathbf{a}$ as its normal. A general plane need not pass through the origin, but the inhomogeneous form

handles that case with no structural change. If $Q$ is any point on the plane, with position vector $\mathbf{q}$, then $\mathbf{a} \cdot \mathbf{q} + c = 0$, so $c = -\mathbf{a} \cdot \mathbf{q}$, and substituting back rewrites the defining equation as

Read geometrically, the plane is the set of points $X$ whose displacement $\overrightarrow{QX}$ is orthogonal to $\mathbf{a}$. Every choice of $Q$ produces the same locus; only the algebraic label $c$ shifts.

Distance from a Point to a Plane

The point-line distance problem in the plane was solved in Lesson 1PM by projecting a displacement onto the normal of the line. The same strategy works in space, and this is the first place the orthogonal projection theorem earns its keep against a problem with no honest two-dimensional version.

Dividing the defining equation through by $\|\mathbf{a}\|$ recasts the plane into its normal form

and in this form the distance formula collapses to $d(P, E) = |f(\mathbf{p})|$, where $f$ denotes the linear polynomial on the left. The value of $f$ at any point is literally its signed distance from the plane, which is a small but extremely useful fact: wherever we have a plane written in normal form, the function $f$ doubles as a ruler.

Systems of Planes

A single plane is one linear condition. Two or more planes turn solid geometry into linear algebra in miniature: whether they meet, how they meet, and which further planes are generated by their intersection. The remainder of the lesson treats these questions with the tools already in hand.

Parallel Planes

Write two planes $E$ and $F$ as

with respective normals $\mathbf{a}$ and $\mathbf{b}$. The planes are parallel precisely when their normals are collinear, i.e. $\mathbf{b} = r\mathbf{a}$ for some non-zero scalar $r$. Dividing the equation for $F$ through by $r$ replaces its normal by $\mathbf{a}$ without altering the locus, so no information is lost by assuming from the outset that two parallel planes are presented with a common normal vector.

Intersecting Planes and Sheaves

Two planes that are not parallel share an entire line. Their normals $\mathbf{a}$ and $\mathbf{b}$ are linearly independent and together span the plane orthogonal to the intersection line $L$, while the direction of $L$ itself is the unique space direction (up to sign) orthogonal to both. The natural numerical quantity attached to such a pair is the angle between them, defined through their normals:

The absolute value pins the result to the dihedral angle along $L$, ignoring the irrelevant orientation of the two normals.

The more interesting question is algebraic: which linear equations describe the remaining planes that contain $L$? The answer is remarkably clean, and hinges on the observation that any linear combination of $f$ and $g$ still vanishes on $L$.

The sheaf equation is, at bottom, the observation that containing $L$ is itself a linear condition on the space of linear equations. Its real utility is that it converts statements about the geometry of the sheaf into statements about pairs $(r, s)$, and in doing so exposes an invariant that is not at all apparent from the geometry alone. For that invariant we need one more classical notion, which is new to this course but built entirely out of quantities we have already met.