Three non-coplanar arrows turn every vector and point into three numbers.
In the plane you learned a quiet magic trick: pin down two perpendicular unit arrows, and every arrow on the page collapses into a pair of numbers — and once that happens, adding arrows, measuring them, and even finding the angle between them all turn into arithmetic. Space is the same trick with one more arrow. Pick three arrows that do not all lie in one plane — a little skeleton — and every vector in space is built from them in exactly one way. Choose the skeleton to be perpendicular unit arrows i, j, k, and a vector becomes a triple (x, y, z). From there, length, distance, the dot product, and the cosine of an angle are all just sums and square roots. That is the whole engine of three-dimensional geometry, and this lesson installs it.
Back in 29.1 we found the coplanar test: if a and b are two arrows that are not parallel, then any arrow p lying in their plane is some combination p = x a + y b — and that combination is unique. But that only reaches arrows that lie flat in the plane of a and b. A combination x a + y b can never leave that plane. To reach off the plane we need a third arrow c that genuinely points out of it.
Three arrows a, b, c are called non-coplanar when you cannot slide them all onto one flat sheet — c sticks up out of the plane of a and b. Such a trio is a basis, a skeleton for all of space. Here is the payoff.
If a, b, c are non-coplanar, then every vector p in space can be written
p = x a + y b + z c
in exactly one way. The three scalars (x, y, z) are the coordinates of p in this basis.
Why does it work? Build the box on a, b, c. Starting at the tail of p, slide along a until you are directly "below" the head (you can, because a and b span a whole plane), then slide along b to land under the head, then rise along c to reach it. Each slide is some multiple — x of a, y of b, z of c — and the three add up to p. Why exactly one way? If x a + y b + z c and x′a + y′b + z′c both equalled p, subtracting gives (x−x′)a + (y−y′)b + (z−z′)c = 0. Non-coplanar arrows can only combine to the zero vector when every coefficient is zero, so x = x′, y = y′, z = z′. One vector, one triple.
If the three arrows are coplanar — all lying in one plane — they are not a basis. Every combination x a + y b + z c stays trapped in that plane, so the arrows that point out of it are unreachable. Three coplanar vectors cannot span space. "Non-coplanar" is exactly the condition that the skeleton fills all three directions.
Any non-coplanar trio works, but one choice makes the arithmetic cleanest: take three arrows that are mutually perpendicular and each of length 1. Such a basis is called orthonormal (ortho = perpendicular, normal = unit length). Name them
i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1).
With this skeleton, the basis theorem reads
a = x i + y j + z k,
and the three coordinates are literally how far you walk east (along i), then how far back into the page (along j), then how far up (along k). Because the three hops meet at right angles, the picture is a perfect staircase: a horizontal blue leg, a horizontal amber leg pointing into the page, and a vertical violet leg — and the green resultant is the diagonal that closes it. Try the staircase below; the readout reassembles the vector from its legs and computes its length.
To pin a basis to actual points, plant the three axes at one corner of a box. Choose a square corner A, run the x-axis along one edge, the y-axis along the second edge, and the z-axis straight up the third. Because a box has right angles at every corner, these three axes are automatically perpendicular — an orthonormal frame for free. Now every corner of the box gets a triple of coordinates, read straight off the three edges.
Take the cube with edge length 2, corner A at the origin. Walk its eight corners:
| corner | A | B | C | D | A₁ | B₁ | C₁ | D₁ |
|---|---|---|---|---|---|---|---|---|
| (x, y, z) | (0,0,0) | (2,0,0) | (2,2,0) | (0,2,0) | (0,0,2) | (2,0,2) | (2,2,2) | (0,2,2) |
The bottom face ABCD all sit at z = 0; the top face A₁B₁C₁D₁ is the same square lifted to z = 2. The far top corner C₁ = (2, 2, 2) is "2 right, 2 back, 2 up" — the end of the space diagonal from A. We will lean on this labelled cube for the rest of the lesson, so every number stays honest.
A vector is free: slide its tail to the origin O, and the head lands on a point whose triple is the vector's coordinates. So a point and a position vector carry the same three numbers. For two points A and B, the arrow between them is "head minus tail," component by component:
AB→ = B − A = (x₂−x₁, y₂−y₁, z₂−z₁).
For the cube, the edge AB→ = B − A = (2,0,0) − (0,0,0) = (2, 0, 0) — pure x, as expected. The space diagonal AC₁→ = C₁ − A = (2,2,2) − (0,0,0) = (2, 2, 2). And the face diagonal DB₁→ = B₁ − D = (2,0,2) − (0,2,0) = (2, −2, 2) — note the real minus in the y-slot, because the arrow runs forward in depth.
Adding, subtracting, and scaling are all done slot by slot — the plane rules with a z-term tacked on:
With a = (2, 0, 1) and b = (1, 2, 1):
a + b = (2+1, 0+2, 1+1) = (3, 2, 2).
a − b = (2−1, 0−2, 1−1) = (1, −2, 0).
3a = (3·2, 3·0, 3·1) = (6, 0, 3).
In Stage 27 the dot product was geometric: a·b = |a||b|cos θ, a single number that measures how much two arrows agree. In coordinates it becomes a one-line sum. Because i, j, k are perpendicular unit arrows, i·i = j·j = k·k = 1 and every different pair gives 0; multiplying out a·b with those facts kills the cross-terms and leaves
a·b = x₁x₂ + y₁y₂ + z₁z₂.
It is still a number, not an arrow. For a = (2,0,1) and b = (1,2,1): a·b = 2·1 + 0·2 + 1·1 = 2 + 0 + 1 = 3. And whenever the dot product is 0, the arrows are perpendicular — the cube's two edges AB→ = (2,0,0) and AD→ = (0,2,0) give 2·0 + 0·2 + 0·0 = 0, confirming the right angle at corner A without a protractor.
The single most common slip in space is computing x₁x₂ + y₁y₂ and forgetting the third product. The dot product in space is x₁x₂ + y₁y₂ + z₁z₂ — three products, not two. Drop the z-term and a perpendicular pair can look slanted, or a clean angle comes out wrong. Always add all three.
Everything geometric now falls out of the same three numbers. Length is Pythagoras applied twice. In the base plane the diagonal of the x-by-y rectangle has length √(x²+y²); raise by z and Pythagoras again gives
|a| = √(x² + y² + z²).
The space diagonal AC₁→ = (2,2,2) has length √(2²+2²+2²) = √12 = 2√3 ≈ 3.46. Distance between two points is just the length of the arrow between them:
|AB→| = √((x₂−x₁)² + (y₂−y₁)² + (z₂−z₁)²).
And the angle between two arrows comes straight from the dot product, rearranged from a·b = |a||b|cos θ:
cos θ = a·b|a| |b| = x₁x₂ + y₁y₂ + z₁z₂√(x₁²+y₁²+z₁²) · √(x₂²+y₂²+z₂²).
Try the cube calculator. Pick any two arrows that spring from corner A — an edge, a face diagonal, the space diagonal — and watch the dot product, the two lengths, and the cosine assemble themselves; the sign of a·b alone tells you acute, right, or obtuse.
Notice the patterns the cube hands you: two edges give a·b = 0 and a 90° corner; an edge and a face diagonal give 45°; two face diagonals meeting at A give exactly 60°; and the space diagonal leans on an edge at ≈ 54.7° (cos θ = 1/√3) — a famous number you have now computed from nothing but counting and a square root.
Three non-coplanar arrows are a basis: every vector is x a + y b + z c in exactly one way. Choose the orthonormal trio i, j, k, and a vector becomes a triple — after which every geometric question is arithmetic on those three numbers.
| idea | in coordinates | cube example |
|---|---|---|
| a vector | a = x i + y j + z k = (x, y, z) | AC₁→ = (2, 2, 2) |
| arrow between points | AB→ = B − A = (x₂−x₁, y₂−y₁, z₂−z₁) | DB₁→ = (2, −2, 2) |
| add / subtract / scale | slot by slot: (x₁±x₂, …), λ(x, y, z) | (2,0,1)+(1,2,1) = (3, 2, 2) |
| length | |a| = √(x² + y² + z²) | |AC₁→| = √12 = 2√3 |
| dot product | x₁x₂ + y₁y₂ + z₁z₂ (a number) | AB→·AD→ = 0 → ⊥ |
| angle | cos θ = (a·b) / (|a| |b|) | AC₁→ & AB→: cos = 1/√3 |
Two traps to remember. A basis must be non-coplanar — three flat arrows can't reach off their plane. And the dot product carries the + z₁z₂ term; never stop at x₁x₂ + y₁y₂. Next, in 29.3, we hand a line its direction vector and a plane its normal, so "parallel" and "perpendicular" become one dot-product check.
Let a = (1, −2, 2) and b = (3, 0, −1). Find a + b, 2a − b, and the length |a|.
Work slot by slot. a + b = (1+3, −2+0, 2−1) = (4, −2, 1). For 2a − b, first 2a = (2, −4, 4), then subtract: (2−3, −4−0, 4−(−1)) = (−1, −4, 5). Length |a| = √(1² + (−2)² + 2²) = √(1+4+4) = √9 = 3.
For the points A(1, 0, 2) and B(3, 4, 2), write the vector AB→ and find the distance from A to B.
AB→ = B − A = (3−1, 4−0, 2−2) = (2, 4, 0) — the z-slot is 0 because A and B sit at the same height. Distance = |AB→| = √(2² + 4² + 0²) = √20 = 2√5 ≈ 4.47.
Find the dot product and the angle between u = (1, 1, 0) and v = (0, 1, 1).
u·v = 1·0 + 1·1 + 0·1 = 1 (don't forget all three products). Lengths: |u| = √2, |v| = √2. So cos θ = 1 / (√2·√2) = 1/2, giving θ = 60°. These are two face diagonals of a cube meeting at a corner — the classic 60° angle.
For which value of t are the vectors (t, 2, −1) and (3, t, 1) perpendicular?
Perpendicular means the dot product is zero: (t)(3) + (2)(t) + (−1)(1) = 3t + 2t − 1 = 5t − 1. Set 5t − 1 = 0, so t = 1/5 = 0.2. (Check: (0.2, 2, −1)·(3, 0.2, 1) = 0.6 + 0.4 − 1 = 0. ✓)
On the cube of edge 2 with A at the origin, the corners are A(0,0,0), B(2,0,0), C(2,2,0), D(0,2,0) and the top set A₁, B₁, C₁, D₁ at height 2. Find the distance from B to D₁.
D₁ = (0, 2, 2), so BD₁→ = D₁ − B = (0−2, 2−0, 2−0) = (−2, 2, 2). Distance = √((−2)² + 2² + 2²) = √12 = 2√3 ≈ 3.46 — a space diagonal, equal in length to AC₁.
On the same cube, find the angle between the space diagonal AC₁→ and the face diagonal AC→ that share corner A.
AC₁→ = (2,2,2) and AC→ = (2,2,0). Dot product: 2·2 + 2·2 + 2·0 = 8. Lengths: |AC₁→| = 2√3, |AC→| = 2√2. So cos θ = 8 / (2√3 · 2√2) = 8 / (4√6) = 2/√6 ≈ 0.816, giving θ = ≈ 35.3°. (This is the angle a space diagonal makes with the base — a number you would dread without coordinates.)
Six questions to lock it in. Tap the answer you think is right.
This lesson is the three-dimensional companion to the plane vectors of Stage 27, and it lands squarely in HSN-VM (Vector & Matrix Quantities): writing a vector in component form, adding and scaling componentwise, computing magnitude, and using the dot product to find lengths and angles — extended from two coordinates to three. The basis theorem (every vector is a unique combination of three non-coplanar arrows) is the structural idea behind a coordinate system, and setting axes at a corner of a solid to assign coordinates and compute distances is geometric modeling in space, touching HSG-GMD and HSG-MG (Modeling with Geometry). A good home check: ask your student to set coordinates on a shoebox corner, read off two corner-to-corner vectors, and decide whether they are perpendicular by testing whether the dot product (all three products) is zero.