Choose perpendicular axes and a vector becomes a pair of numbers.
An arrow is a wonderful thing, but it is awkward to compute with. You cannot easily add two pictures, or ask a drawing whether two arrows are parallel. The fix is the oldest trick in mathematics: lay down a grid. Pick two perpendicular directions — call them i pointing east and j pointing north — slide the vector so its tail sits at the origin, and read off where its head lands. Suddenly the arrow a is just an ordered pair (x, y), and every vector operation collapses into ordinary arithmetic on those two numbers. In this lesson we build the dictionary that turns geometry into coordinates: adding, subtracting, scaling, measuring length, testing for parallel, and locating the midpoint.
Suppose I hand you two arrows e1 and e2 that are not parallel, drawn from one shared point. Watch what happens when you try to reach a third arrow a. Slide a copy of e1 and a copy of e2 until they form the two sides of a parallelogram whose diagonal is exactly a. There is one — and only one — parallelogram that works.
The amounts of each arrow you needed are two numbers λ1, λ2, and they are forced — there is no second way to build a from this skeleton. That is the whole content of the theorem.
If e1 and e2 are not parallel, then every vector a in the plane can be written a = λ1e1 + λ2e2 in exactly one way. The pair {e1, e2} is called a basis, and the two numbers (λ1, λ2) are a's coordinates in that basis.
If e1 and e2 pointed the same way (or exactly opposite), every combination λ1e1 + λ2e2 would land on a single line through the origin — you could never reach a vector off that line. Two parallel arrows do not span the plane. You need two genuinely different directions.
Of all the bases you could pick, one is supremely convenient: the perpendicular pair of unit arrows i = (1, 0) pointing east and j = (0, 1) pointing north. With this standard basis, building any vector means walking a horizontal leg and then a vertical leg — exactly the right triangle from the Pythagorean theorem.
Drop a vector's tail at the origin and split it into its horizontal edge and its vertical edge. The horizontal edge is some multiple x of i; the vertical edge is some multiple y of j. Add them head-to-tail and you recover the vector:
a = x i + y j.
This is the orthogonal decomposition. Because the two legs meet at a right angle, the picture is honest: the legs really are perpendicular, and their lengths are exactly x and y. Dial a vector below and watch its two edges rebuild it.
Here is the payoff. Once the standard basis is fixed, a vector is determined by its two amounts (x, y) — and that pair is what we mean by its coordinates:
a = (x, y).
Because a vector is free — you may slide it anywhere without changing it — we always slide its tail to the origin to read these off. But a vector often comes to us as a trip between two named points. If it runs from A(x1, y1) to B(x2, y2), then AB→ is "where you end minus where you start":
AB→ = (x2 − x1, y2 − y1).
From A(1, 4) to B(5, 2): AB→ = (5 − 1, 2 − 4) = (4, −2). You moved 4 right and 2 down — and that is the same free vector wherever you draw it, tail at A or tail at the origin.
Now every operation we built with arrows in 27.2 becomes pure bookkeeping on the coordinates. To add two vectors, add matching components; to scale, multiply both components by the scalar:
| Operation | In coordinates | Example with a=(3,1), b=(−1,2) |
|---|---|---|
| add | (x1, y1) + (x2, y2) = (x1+x2, y1+y2) | a + b = (2, 3) |
| subtract | (x1, y1) − (x2, y2) = (x1−x2, y1−y2) | a − b = (4, −1) |
| scale | λ(x, y) = (λx, λy) | 2a = (6, 2) |
And length? Drop the tail at the origin: the vector is the hypotenuse of a right triangle with legs x and y. The Pythagorean theorem hands you the magnitude directly:
|a| = √(x2 + y2).
For a = (3, 4): |a| = √(32 + 42) = √(9 + 16) = √25 = 5. The famous 3–4–5 right triangle, read straight off the components.
Drive the resultant below. The figure builds a, then carries b on from a's head; the green diagonal is a + b, and every number in the readout is computed component by component.
Two vectors are parallel when one is a scalar multiple of the other — b = λa. In coordinates, b = (x2, y2) is λ times a = (x1, y1) exactly when their components are proportional: x2/x1 = y2/y1. Clearing the fractions (so it works even when a component is zero) gives the clean cross-multiplied test:
a ∥ b ⇔ x1y2 − x2y1 = 0.
That quantity is the 2-D cross of the two vectors; when it is zero the arrows lie along one line. Try to drive b until it lines up with the fixed a — the readout turns green at the instant the cross hits 0.
The parallel test multiplies across: x1y2 − x2y1. A common slip is to write x1x2 − y1y2 — that is something else entirely (it is close to the dot product, which tests perpendicular, coming in 27.4). Pair each x with the other vector's y.
Coordinates also locate points along a segment. The midpoint of A(x1, y1) and B(x2, y2) is just the average of the endpoints:
M = ((x1+x2)/2, (y1+y2)/2).
More generally, the point P that divides AB in the ratio AP : PB = λ is a weighted average that leans toward whichever endpoint carries more weight:
P = (A + λB) / (1 + λ) (λ = 1 gives the midpoint).
Move B and slide the ratio λ below. The green dot M is always the midpoint; the green dot P divides AB in your chosen ratio, and the readout shows the averaged coordinates.
Once perpendicular axes are chosen, geometry becomes arithmetic. Keep this dictionary:
| Idea | In coordinates |
|---|---|
| a vector | a = x i + y j = (x, y) |
| from A to B | AB→ = (x2−x1, y2−y1) |
| add / subtract | (x1,y1) ± (x2,y2) = (x1±x2, y1±y2) |
| scale | λ(x, y) = (λx, λy) |
| length | |a| = √(x2 + y2) |
| parallel test | x1y2 − x2y1 = 0 |
| midpoint | ((x1+x2)/2, (y1+y2)/2) |
| section AP:PB = λ | (A + λB)/(1 + λ) |
Next, in 27.4 The dot product, we multiply two vectors into a single number, a·b = x1x2 + y1y2, and read the angle between them straight off their coordinates.
Write the vector AB→ in coordinates, where A(2, −1) and B(6, 5). Then find |AB→|.
AB→ = (6 − 2, 5 − (−1)) = (4, 6). Length: |AB→| = √(4² + 6²) = √(16 + 36) = √52 = 2√13 ≈ 7.21. The components are "end minus start," and the magnitude is the hypotenuse of the 4-by-6 right triangle.
Given a = (3, −2) and b = (−1, 4), compute a + b, a − b, and 2a − b.
Work component by component. a + b = (3 + (−1), −2 + 4) = (2, 2). a − b = (3 − (−1), −2 − 4) = (4, −6). For 2a − b: first 2a = (6, −4), then subtract b = (−1, 4): (6 − (−1), −4 − 4) = (7, −8).
Are a = (2, 3) and b = (6, 9) parallel? Are c = (2, 3) and d = (6, 8)? Use the cross test.
For a, b: x1y2 − x2y1 = 2·9 − 6·3 = 18 − 18 = 0, so a ∥ b (indeed b = 3a). For c, d: 2·8 − 6·3 = 16 − 18 = −2 ≠ 0, so c and d are not parallel. The cross multiplies across; only the first pair lines up.
Find the value of t that makes a = (t, 6) parallel to b = (2, 3).
Set the cross to zero: x1y2 − x2y1 = t·3 − 2·6 = 3t − 12 = 0, so t = 4. Check: a = (4, 6) = 2·(2, 3) = 2b. ✓ Proportional components, as required.
A(−2, 5) and B(4, −1). Find the midpoint M of AB, and the point P dividing AB with AP : PB = 2.
Midpoint: M = ((−2 + 4)/2, (5 + (−1))/2) = (1, 2) = (1, 2). Section with λ = 2: P = (A + 2B)/(1 + 2) = ((−2 + 8)/3, (5 + (−2))/3) = (6/3, 3/3) = (2, 1). Since AP : PB = 2 : 1, P sits two-thirds of the way from A to B — past the midpoint, leaning toward B.
Explain why the two parallel arrows e1 = (1, 2) and e2 = (2, 4) cannot serve as a basis for the plane.
They are parallel: the cross is 1·4 − 2·2 = 0, so e2 = 2e1. Any combination λ1e1 + λ2e2 = λ1(1,2) + λ2(2,4) = (λ1 + 2λ2)(1, 2) is always a multiple of (1, 2) — every reachable vector lies on the single line through the origin in direction (1, 2). A vector like (1, 0) off that line is unreachable, so the pair does not span the plane. A basis needs two non-parallel directions.
Six questions to lock it in. Tap the answer you think is right.
This lesson develops the coordinate (component) representation of plane vectors and the operations on it, aligned with the Common Core vector and matrix standards HSN-VM.A.2 (find the components of a vector as the difference of the coordinates of an initial and terminal point), HSN-VM.A.3 (solve problems with vectors), HSN-VM.B.4a (add vectors and find magnitude and direction from components), and HSN-VM.B.5 (multiply a vector by a scalar component-wise). It rests on the Pythagorean theorem (8.G.B / Stage 15) for the magnitude formula and on the coordinate plane and slope (Stage 20–21). A productive home conversation: a vector is a single arrow you may slide anywhere; choosing perpendicular axes is just a bookkeeping choice that turns that arrow into two numbers, after which adding, scaling, and measuring are ordinary arithmetic.