Let the arrow leave the page: a vector that carries size and direction through space.
Picture an arrow drawn on a sheet of paper. It carries a length and a direction, and you may slide it anywhere — that is a plane vector, the star of Stage 27. Now lift that paper off the desk and let the arrow point up out of the page, or down toward the floor, or off into a corner of the room. Nothing about the arrow changed — it still has just a length |a| and a direction — only now the direction is free to point any which way in three dimensions. That single move turns a plane vector into a spatial vector, and the wonderful surprise is that every rule still works. Arrows add tip-to-tail, scale by a number λ, and snap into the collinear test b = λa exactly as before; the dot product a·b = |a||b|cos θ still measures how "in sync" two arrows are. We were never using the flatness of the page — so let it go.
This lesson lifts each plane-vector operation into space one at a time, reasoning out loud and checking the picture against the arithmetic every step. Then it adds the one genuinely new idea — the coplanar test p = xa + yb, which decides whether three arrows can be slid onto a single flat sheet — the seed of the basis we build in 29.2, where three numbers will replace every arrow entirely.
A spatial vector is a directed segment: pin a tail at a point A, point at a head B, and you have AB→. Just like a plane vector, it stores exactly two facts — how long it is, written |a|, and which way it aims. The only difference from Stage 27 is that the "way" now lives in a room, not on a page: an arrow can rise straight up an elevator shaft, or run diagonally from one corner of the room to the opposite one.
A vector is still free: it has no home. Slide it — keeping its length and direction — and it is the very same vector. Two arrows are equal exactly when they have the same length and the same direction, no matter where they sit. So in the box below, the four parallel edges all running "to the right" are one vector drawn four times. The zero vector 0 has length 0 and — this matters — no direction at all.
A spatial vector = a length + a direction in 3-D. It is free (slide it anywhere); equal vectors match in length and direction. The zero vector 0 has length 0 and no direction.
To add two arrows, lay them tip-to-tail: walk along a, then from where you land walk along b; the single arrow from your start to your finish is a + b. That is the triangle rule, and it never asked whether the journey stayed in one plane. Equivalently, draw a and b from a shared tail, complete the parallelogram, and the diagonal from that tail is a + b — the parallelogram rule. Both give the same answer because translating an arrow does not change it.
Take a = (2, 0, 1) and b = (1, 2, 1) — two edges of a slanted box. Tip-to-tail, the resultant runs a + b = (2, 0, 1) + (1, 2, 1) = (3, 2, 2), with length |a + b| = √(3² + 2² + 2²) = √17 ≈ 4.12. The figure below closes that triangle in space; dial b and watch the green diagonal track the sum exactly.
Fix a = (2, 0, 1). Dial the components of b and watch a + b (the green diagonal) follow the triangle and the parallelogram both. The readout is computed, never measured off the picture.
To subtract, ask "a minus b is the arrow that, added to b, gives a." Draw both from a shared tail: a − b is the arrow from the tip of b to the tip of a. With our vectors, a − b = (2, 0, 1) − (1, 2, 1) = (1, −2, 0). And every algebra law from the flat world still holds, for the same reason — nothing flat was ever used:
a + b = b + a (commutative) · (a + b) + c = a + (b + c) (associative) · a + 0 = a · a + (−a) = 0. None of these proofs ever mentioned a plane.
Multiplying an arrow by a number λ — a scalar — does two things and nothing more. It stretches the length by the factor |λ|, and when λ < 0 it also flips the arrow to point the opposite way. So the magnitude obeys |λa| = |λ| · |a| — note the |λ|: a flip does not give a "negative length." Take a = (2, 1, 2), whose length is |a| = √(4 + 1 + 4) = 3. Then:
Drag λ from −2 to 3. The dashed slate ghost is the original a; the green arrow is λa. Watch it stretch by |λ| and flip when λ goes negative — and watch |λa| = |λ||a| stay exact.
Two nonzero arrows are collinear when they lie along one line — same direction or exactly opposite. Scalar multiplication is precisely the tool for that, because λa always points along a's line. So we get a clean, picture-free test:
b ∥ a ⇔ b = λa (for some scalar λ, with a ≠ 0). The sign of λ tells you the direction: λ > 0 same way, λ < 0 opposite.
This is how you prove three points lie on one line without drawing anything. Points A, B, C are collinear exactly when AB→ and AC→ are collinear — i.e. AC→ = λ AB→. Take A(1, 0, 1), B(3, 1, 2), C(7, 3, 4). Then AB→ = B − A = (2, 1, 1) and AC→ = C − A = (6, 3, 3). Since (6, 3, 3) = 3 · (2, 1, 1), every component shares the same multiplier λ = 3, so AC→ = 3 AB→ — the three points are collinear, and C is three times as far from A as B is.
Here is the one idea that is genuinely new in space. Two arrows always lie in some plane. Three arrows might or might not. Three vectors are coplanar when you can slide them — keeping length and direction — so that all three lie flat on a single sheet. The test mirrors the collinear one, but with two "ingredient" vectors instead of one:
Let a and b be not collinear. A vector p is coplanar with them ⇔ p = xa + yb for a unique pair of scalars x, y. (Two non-collinear arrows span a plane; "coplanar" means p can be reached by combining them.)
For instance, with a = (1, 0, 0) and b = (0, 1, 1), the vector p = 2a − b = (2, −1, −1) is coplanar with a and b — it is built entirely out of them. But the basis vector k = (0, 0, 1) is not a combination of those two: no x, y make x(1,0,0) + y(0,1,1) = (0,0,1) (the second and third components would force y = 0 and y = 1 at once). Three arrows like i, j, k that genuinely fill space — that no flat sheet can hold — are called non-coplanar, and they are exactly the skeleton we will build coordinates on in 29.2.
A blue a and amber b span a flat patch. Dial p = x·a + y·b + lift·k: while lift = 0, p stays on the patch (coplanar). Add any lift and p rises off it — no x, y can reach it any more.
The last tool carries up untouched. The dot product of two arrows is the single number
a·b = |a| · |b| · cos θ, where θ is the angle between them (0° ≤ θ ≤ 180°).
It measures how "in sync" the arrows are, and its sign reads off the angle at a glance — because |a| and |b| are positive, the sign of a·b is the sign of cos θ:
| a·b | cos θ | the angle θ |
|---|---|---|
| > 0 | positive | acute (arrows lean together) |
| = 0 | 0 | right — so a·b = 0 ⇔ a ⊥ b |
| < 0 | negative | obtuse (arrows lean apart) |
That middle row is the workhorse of all of Stage 29: two arrows are perpendicular exactly when their dot product is zero. Set up a cube of side 2 and put your corner at A. The edges AB→ = (2, 0, 0) and AD→ = (2 in the depth direction) meet at a square corner, and indeed AB→ · AD→ = 0 — the figure's right angle and the number agree. Compare the edge AB→ with the space diagonal AC₁→ = (2, 2, 2): there a·b = 4 > 0, so the angle is acute, and we can even read it: cos θ = 4 / (2 · √12) ≈ 0.58, so θ ≈ 54.7°. (In 29.2 we will see this same number drop out of coordinates, with no cosine table in sight.)
Both arrows start at corner A of a cube (side 2). Pick a second arrow and read a·b, the lengths, cos θ, and the angle — all computed by the vector helpers, with the honest angle arc drawn at A.
• The zero vector has length 0 and no direction — never speak of "the direction of 0."
• "Any three vectors lie in a plane" is false. That is exactly what non-coplanar means — arrows like i, j, k fill space and no sheet can hold them.
• a·b is a number, not an arrow. Writing "a·b = (…, …, …)" is a category error. (The thing that returns a vector is the cross product a×b, a different operation we meet in 29.3.) And there is no such thing as dividing by a vector.
Every plane-vector operation lifts into space without a single new rule — plus one fresh idea, the coplanar test, that sets up coordinates next.
| Operation | What it does | The rule / test |
|---|---|---|
| Add a + b | tip-to-tail, or the parallelogram diagonal | a + b = b + a; (a+b)+c = a+(b+c) |
| Subtract a − b | from b's tip to a's tip (shared tail) | a − b = a + (−b) |
| Scale λa | stretch by |λ|, flip if λ < 0 | |λa| = |λ| · |a| |
| Collinear? | two arrows on one line | b ∥ a ⇔ b = λa (a ≠ 0) |
| Coplanar? | three arrows fit one flat sheet | p = xa + yb (a, b not collinear) |
| Dot product a·b | a number: how in-sync the arrows are | a·b = |a||b|cos θ; a⊥b ⇔ a·b = 0 |
Next, in 29.2 · Coordinates in space, we pick three non-coplanar arrows as a skeleton and let every vector become three numbers (x, y, z) — and watch addition, length, and even the dot product collapse into plain arithmetic.
Let a = (2, 0, 1) and b = (1, 2, 1). Find a + b, a − b, and 2a − b.
Work component by component. a + b = (2+1, 0+2, 1+1) = (3, 2, 2). a − b = (2−1, 0−2, 1−1) = (1, −2, 0). And 2a = (4, 0, 2), so 2a − b = (4−1, 0−2, 2−1) = (3, −2, 1).
a = (1, −2, 2). Find |a|, then −3a and |−3a|. Does the magnitude rule |λa| = |λ||a| check out?
|a| = √(1² + (−2)² + 2²) = √9 = 3. −3a = (−3, 6, −6), which is the reverse arrow stretched 3×. Its length is √(9 + 36 + 36) = √81 = 9, and |−3| · |a| = 3 · 3 = 9 ✓ — the flip never makes the length negative.
Show the points A(2, 1, 0), B(4, 4, 2), C(8, 10, 6) are collinear, and say how the distances compare.
AB→ = B − A = (2, 3, 2) and AC→ = C − A = (6, 9, 6). Since (6, 9, 6) = 3 · (2, 3, 2), the same multiplier λ = 3 works for all three components, so AC→ = 3 AB→ — the points are collinear. C is three times as far from A as B is, on the same side.
Let a = (1, 0, 1) and b = (0, 1, 1). Is p = (2, 3, 5) coplanar with a and b? If so, write it as xa + yb.
Try p = xa + yb = (x, y, x + y). Matching components: x = 2 and y = 3, and then the third must satisfy x + y = 5 — and indeed 2 + 3 = 5 ✓. So p = 2a + 3b, and yes, p is coplanar with a and b. (Had the third equation failed, p would lift off the a-b sheet and be non-coplanar.)
Compute a·b for a = (3, −1, 2), b = (1, 4, 1). Is the angle between them acute, right, or obtuse? Then find a nonzero vector perpendicular to u = (2, 2, −1) of the form (1, 1, z).
a·b = 3·1 + (−1)·4 + 2·1 = 3 − 4 + 2 = 1. It is positive, so cos θ > 0 and the angle is acute. For the perpendicular: we need u·(1,1,z) = 0, i.e. 2·1 + 2·1 + (−1)·z = 4 − z = 0, so z = 4 and (1, 1, 4) works — check 2 + 2 − 4 = 0 ✓.
A classmate writes "a·b = (2, 0, 2)" and "since a·b ≠ 0, the arrows must be parallel." Find both mistakes.
First, the dot product is a number, not a vector — "(2, 0, 2)" can never be a value of a·b. (A vector answer would come from the cross product a×b instead.) Second, a·b ≠ 0 only tells you the arrows are not perpendicular; it says nothing about parallel. Parallel is the collinear test, b = λa. Two different errors, both fixed by remembering what each operation returns.
Six questions to lock it in. Tap the answer you think is right.
This lesson opens Stage 29 by extending plane-vector arithmetic into three dimensions: addition and subtraction (triangle and parallelogram rules), scalar multiplication with |λa| = |λ||a|, the collinearity test b = λa, the coplanarity theorem p = xa + yb, and the geometric dot product a·b = |a||b|cos θ with its perpendicularity test. It addresses the Common Core vector standards HSN-VM (Vector & Matrix Quantities — representing vectors, adding and scaling them, and using them to model) and HSG-GMD / G-MG (geometric measurement and modeling in space — reasoning about lengths, directions, and angles inside solids). The recurring habit to reinforce: a vector stores only a length and a direction, so its operations are independent of the page — and the dot product returns a number, the key to turning every later angle and distance question into a calculation.