Add by walking arrows head-to-tail; scale by stretching or flipping.
In Lesson 27.1 a vector earned its keep: it carries both a size and a direction, packed into one arrow. But an idea is only as good as what you can do with it — and arrows turn out to be wonderfully easy to work with. There are exactly three moves. You can lay two arrows head to tail and read off where you end up — that is addition. You can reverse one and add — that is subtraction. And you can stretch or flip a single arrow — that is scaling. Every one of these is something you can literally draw, and once you can draw it you can compute it. By the end you will hold a single equation, b = λa, that captures "these two arrows point along the same line" — the workhorse for proving three points are collinear.
Suppose you walk the displacement a = (3, 1): three steps east, one step north. Then, from wherever you land, you walk b = (1, 3). Where do you end up relative to where you started? Slide the tail of b onto the head of a — line them up head to tail — and the single straight arrow from your starting point to your finishing point is the sum a + b. This is the triangle rule: the two arrows and their sum form a triangle that closes.
The bookkeeping is just component arithmetic — add the easts, add the norths: a + b = (3, 1) + (1, 3) = (4, 4). The arrow really does close, because moving 3 then 1 more east is moving 4 east, and 1 then 3 north is 4 north.
To add a + b, slide b so its tail sits on a's head. The arrow from a's tail to b's head is a + b. In coordinates: (x₁, y₁) + (x₂, y₂) = (x₁ + x₂, y₁ + y₂).
The triangle rule asks you to put the arrows nose to tail. But often two vectors arrive from the same place — two ropes pulling a ring, two forces on a bolt, two velocities at one instant. For that picture there is a twin construction. Draw a and b from a shared tail. Complete the parallelogram by drawing a copy of b off the head of a and a copy of a off the head of b. The diagonal from the shared tail is exactly a + b.
Why is it the same sum? Because the far side of the parallelogram is a slid copy of b — same length, same direction, so it is b as a free vector. Following a and then that copy of b is precisely the triangle rule again. The parallelogram simply draws the triangle rule both ways at once, which is why its diagonal lands on the same green point.
One rope pulls with a = (3, 1), a second with b = (1, 3), both from the same ring. The single pull they amount to — the resultant — is the diagonal a + b = (4, 4), of length |a + b| = √(42 + 42) = √32 ≈ 5.66. One rope pulling (4, 4) would do the same job as both.
Notice that order does not matter: walking a then b lands you in the same place as walking b then a — the parallelogram is symmetric across its diagonal. That is the law a + b = b + a, and we will collect it with its friends in §27.2.6.
Subtraction is addition wearing a disguise: a − b means a + (−b), where −b is just b reversed. But there is a cleaner picture. Draw a and b from the same tail. Then a − b is the arrow that runs from the head of b to the head of a.
Here is the memory hook: a − b is the arrow that "completes the triangle" so that b + (a − b) = a. Start at b's head, travel a − b, and you arrive at a's head — exactly what the equation says. In coordinates it is again component-wise: with a = (4, 1) and b = (1, 3), a − b = (4 − 1, 1 − 3) = (3, −2).
a − b ≠ b − a. They are opposite arrows: b − a = −(a − b). Above, b − a = (−3, 2) points the other way. Pick the right one by asking which arrow you want to point from and point to: a − b points to a, from b.
The third move uses just one arrow and a number. To form λa (read "lambda times a"), keep a's line and rescale its length by |λ|. If λ > 0 the new arrow points the same way; if λ < 0 it flips to the opposite direction; if λ = 0 it collapses to the zero vector. In coordinates, scaling touches every component: λ(x, y) = (λx, λy).
The length obeys a clean law: |λa| = |λ| |a|. Doubling (λ = 2) doubles the length; λ = −1 keeps the length but reverses the arrow; λ = ½ halves it. The sign of λ is the direction switch, the size of λ is the stretch factor.
|a + b| is not generally |a| + |b|. Adding arrows is not adding lengths. With a = (3, 0) and b = (0, 4): |a| + |b| = 3 + 4 = 7, but |a + b| = |(3, 4)| = √25 = 5. The detour through a corner is shorter than the two legs — the famous triangle inequality |a + b| ≤ |a| + |b|. Equality holds only when a and b point the same way.
Scaling has a beautiful payoff. Take any nonzero a. Every scalar multiple λa lies on the same line through the origin as a — stretching and flipping never leave that line. Run it backwards: if b lies along a's line, there is a number λ that rescales a onto b. That gives the whole theorem.
For a ≠ 0: b ∥ a ⇔ there is exactly one number λ with b = λa. The λ is the stretch-and-flip factor: λ > 0 means same direction, λ < 0 opposite, |λ| the length ratio.
This is the standard tool for proving three points lie on a line. To show A, B, C are collinear, form the vectors AB→ and AC→ from the common point A; if you can write AC→ = λ AB→ for some single λ, the two arrows share a line, and since they share the point A, all three points are on one line.
Let A = (1, 1), B = (3, 2), C = (7, 4). Then AB→ = (2, 1) and AC→ = (6, 3). Is AC→ a multiple of AB→? Try λ = 3: 3·(2, 1) = (6, 3) = AC→. ✓ One λ works, so AC→ = 3 AB→ and A, B, C are collinear — with C three times as far from A as B is.
Here is the quiet miracle that makes everything later possible: vector addition and scaling obey the same rules as ordinary numbers. So you may rearrange a vector expression exactly the way you rearrange algebra — collect terms, factor out a scalar, swap an order — and never go wrong.
| Rule | What it says | Name |
|---|---|---|
| a + b = b + a | order of adding does not matter | commutative |
| (a + b) + c = a + (b + c) | grouping does not matter | associative |
| a + 0 = a | the zero vector changes nothing | identity |
| a + (−a) = 0 | every arrow has an opposite | inverse |
| λ(a + b) = λa + λb | scaling spreads over a sum | distributive |
| (λ + μ)a = λa + μa | scalars add the way you'd hope | distributive |
For instance, 2(a + b) − 2b = 2a + 2b − 2b = 2a — pure schoolbook algebra, now with arrows. These rules are exactly what Lesson 27.3 will cash in when it turns every vector into a coordinate pair and collapses all of this into arithmetic.
Three moves, one theorem, and the algebra that ties them together:
| Operation | The picture | In coordinates |
|---|---|---|
| a + b (triangle) | lay b head-to-tail on a; arrow start→finish | (x₁+x₂, y₁+y₂) |
| a + b (parallelogram) | shared tail; the diagonal | same sum |
| a − b | shared tail; arrow from b's head to a's head | (x₁−x₂, y₁−y₂) |
| λa | stretch by |λ|, flip if λ<0 | (λx, λy), |λa| = |λ||a| |
| b ∥ a (a≠0) | same line through the tail | ⇔ b = λa for one λ |
And the algebra is ordinary: a + b = b + a, (a+b)+c = a+(b+c), λ(a+b) = λa+λb. Two cautions to keep: a − b ≠ b − a, and |a + b| ≤ |a| + |b| (the triangle inequality), with equality only when a, b point the same way.
Six questions to lock it in. Tap the answer you think is right.
This lesson develops the linear operations on plane vectors and their geometry. It aligns with the Common Core vector standards HSN-VM.A (recognize vectors as quantities with magnitude and direction; find components) and especially HSN-VM.B — adding vectors end-to-end and by the parallelogram rule (HSN-VM.B.4a), understanding that the magnitude of a sum is not generally the sum of magnitudes (the triangle inequality), vector subtraction a − b = a + (−b) (HSN-VM.B.4c), and scalar multiplication with its effect on magnitude and direction, |λa| = |λ||a| (HSN-VM.B.5). The collinear theorem b = λa is the standard bridge to coordinate proofs in the next lesson. The figures are drawn from a shared vector engine, so every arrow and length is computed, never eyeballed.