Turn a geometry problem into an equation, and a force diagram into arithmetic.
For four lessons we have been collecting tools: an arrow that carries size and direction, the rules for adding and scaling, coordinates that turn a vector into a pair of numbers, and the dot product that reads off an angle. Now we put them to work. A vector argument is short because it hides the picture inside arithmetic — to prove two segments are parallel you produce a single number λ; to prove they are perpendicular you check that one number, a·b, is zero. The same arrows that prove a clean geometry theorem also describe a force on a bridge or a boat steering across a current. By the end you will derive the law of cosines in two lines, balance three forces, split a velocity into components — and add one more coordinate to step from the plane into space.
Two questions come up again and again in geometry, and vectors answer both with one short test each.
Parallel. Segments PQ and RS point the same way (or exactly opposite) when their vectors are scalar multiples: PQ→ = λ RS→ for some number λ. Finding such a λ is the proof. In coordinates, a ∥ b exactly when the cross is zero, x1y2 − x2y1 = 0.
Perpendicular. Segments meet at a right angle when their vectors have dot product 0: a ⊥ b ⇔ a·b = 0 ⇔ x1x2 + y1y2 = 0. No protractor, no slope formula with its awkward "negative reciprocal" — just multiply and add.
Parallel ⇒ find a λ. Perpendicular ⇒ make the dot zero. Choose coordinates, write the two vectors, compute. The figure below proves a classic theorem this way.
Let a triangle have one vertex at the origin O and the other two at the tips of b and c. Let M be the midpoint of side O→(b) and N the midpoint of O→(c). Then MN→ = N − M = ½c − ½b = ½(c − b). But the third side is exactly c − b. So MN→ = ½ × (third side): the midsegment is parallel to the base and half as long — Stage 16's theorem, in two lines. (See triangle midsegment.)
Once a shape sits in coordinates, a side length is just a magnitude and an angle falls out of the dot-product formula. Nothing new to memorize — the tools from 27.3 and 27.4 do the work.
A length is |a| = √(x2 + y2) — the Pythagorean theorem wearing vector clothes. An angle at a vertex is the angle between the two side-vectors leaving it: cos θ = a·b|a| |b|.
Triangle with A(1, 1), B(5, 2), C(2, 5). At A the two side-vectors are AB→ = (4, 1) and AC→ = (1, 4). Then AB·AC = 4·1 + 1·4 = 8, and |AB| = |AC| = √17. So cos A = 8 / (√17·√17) = 8/17 ≈ 0.47, giving A ≈ 61.9°. Every step is multiply-and-add.
Here vectors earn their keep. Put the angle C at the origin, with sides a and b as the two vectors leaving it. The third side, opposite C, runs from the tip of b to the tip of a — that is the vector c = a − b. Now just square it with the dot product:
|c|2 = (a − b)·(a − b) = a·a − 2a·b + b·b = |a|2 + |b|2 − 2a·b.
And a·b = |a||b| cos C, so writing a = |a|, b = |b|, c = |c|:
c2 = a2 + b2 − 2ab cos C.
That is the law of cosines (Stage 25's law of cosines), proved in two lines with no diagram of dropped altitudes. When C = 90°, cos C = 0 and it collapses to c2 = a2 + b2 — the Pythagorean theorem is just the right-angle case.
The 2-D cross a × b = axby − aybx measures the signed area of the parallelogram on a and b; in fact |a × b| = |a||b| sin θ. For a triangle with sides leaving a vertex, its area is ½|a × b|.
Compute a triangle's area three ways — once from each pair of sides — and you always get the same number. Writing the area as ½ab sin C = ½bc sin A = ½ca sin B and dividing through gives the law of sines:
asin A = bsin B = csin C.
So the same vector picture delivers both laws: the dot product (which knows cos θ) gives the law of cosines, and the cross product (which knows sin θ) gives the law of sines.
a·b sees the cosine of the angle (and a length); the cross a × b sees the sine (and an area). Together they pin the angle down completely.
A force has a size and a direction — it is a vector. When several forces act at one point, they combine by the same parallelogram rule as displacements: the single force that would do the same job is their resultant, the vector sum. A body is in equilibrium — not accelerating — exactly when all the forces on it cancel:
F1 + F2 + F3 + … = 0.
Geometrically, three forces in balance form a closed triangle: lay them head-to-tail and you return to the start. If the triangle doesn't close, the gap is the leftover resultant that makes the body move.
A force diagram balances only when the resultant is exactly 0 — not "small," not "nearly closed." And never add forces that point different ways as plain numbers: |F1| + |F2| is generally wrong. A 3 N force east and a 4 N force north give a 5 N resultant, not 7 N — add the vectors, then take the magnitude.
Crossing a river, a boat points straight across at some speed, but the current carries it downstream. The boat's true velocity over the ground is the vector sum of the two. Conversely, any velocity splits cleanly into a downstream (x) part and an across-river (y) part — the very i, j decomposition from 27.3 — and you handle each part on its own before recombining.
The across-river component decides the crossing time (it does the real work of getting to the far bank); the downstream component decides how far down you land. Compute each separately, then put them back together with |resultant| and its direction.
Everything here lived in the flat plane, where a vector is a pair a = (x, y). Reality has one more direction — up. Add a third coordinate z and the pair becomes a triple, a = (x, y, z). The rules survive untouched: add and scale component by component, the length grows to |a| = √(x2 + y2 + z2) (Pythagoras, twice), and the dot product becomes a·b = x1x2 + y1y2 + z1z2 — still a number, still zero exactly when the vectors are perpendicular.
That single extra coordinate is the doorway to Stage 28 · Solid Geometry: lines and planes in space, distances, and the angles between them — all run by the same arrows you already command. Same vectors, one more number.
The whole strand collapses into one short toolkit — read the question, pick the row, compute.
| To show… | Use… | Test (in coordinates) |
|---|---|---|
| parallel a ∥ b | find a scalar λ with a = λb | x1y2 − x2y1 = 0 |
| perpendicular a ⊥ b | make the dot product zero | a·b = x1x2 + y1y2 = 0 |
| a length | take a magnitude | |a| = √(x2 + y2) |
| an angle θ | the dot-product formula | cos θ = (a·b)/(|a||b|) |
| an area | the cross product | area = ½|x1y2 − x2y1| |
| forces balance | the resultant is zero | ΣF = 0 (the triangle closes) |
And the headline proof to remember: writing the third side as c = a − b and squaring it, |c|2 = |a|2 + |b|2 − 2a·b = a2 + b2 − 2ab cos C — the law of cosines in two lines.
Are a = (6, −4) and b = (−3, 2) parallel? If so, find the λ with a = λb.
Cross-test: x1y2 − x2y1 = 6·2 − (−3)·(−4) = 12 − 12 = 0, so yes, a ∥ b. To find λ, match a component: 6 = λ·(−3) gives λ = −2. Check the other: −2·2 = −4 ✓. So a = −2b (they point in opposite directions).
Show that u = (2, 3) and v = (6, −4) are perpendicular.
u·v = 2·6 + 3·(−4) = 12 − 12 = 0. A dot product of zero means the angle between them is 90°, so u ⊥ v. No slopes needed.
A triangle has vertices A(0, 0), B(4, 0), C(1, 3). Find the length of side AC and the angle at A.
AC→ = (1, 3), so |AC| = √(12 + 32) = √10 ≈ 3.16. The angle at A is between AB→ = (4, 0) and AC→ = (1, 3): cos A = (4·1 + 0·3)/(4·√10) = 4/(4√10) = 1/√10 ≈ 0.316, so A ≈ 71.6°.
Use the vector method to find the third side c of a triangle with a = 5, b = 8, and included angle C = 60°.
Law of cosines (the two-line vector proof): c2 = a2 + b2 − 2ab cos C = 25 + 64 − 2·5·8·cos 60° = 89 − 80·(½) = 89 − 40 = 49. So c = √49 = 7.
A boat heads straight across a 120 m river at 4 m/s; the current flows downstream at 3 m/s. Find the boat's speed over the ground, the crossing time, and how far downstream it lands.
Velocity = (downstream, across) = (3, 4). Ground speed = |(3, 4)| = √(9 + 16) = 5 m/s. The crossing depends only on the across component, 4 m/s: time = 120/4 = 30 s. In that time the current carries it 3·30 = 90 m downstream.
Two forces act at a point: F1 = (5, 1) N and F2 = (−2, 3) N. What third force F3 holds the point in equilibrium, and what is its magnitude?
Equilibrium means the sum is the zero vector, so F3 = −(F1 + F2). First F1 + F2 = (3, 4), so F3 = (−3, −4) N. Its magnitude is √(9 + 16) = 5 N — equal and opposite to the resultant of the other two.
Six questions to lock it in. Tap the answer you think is right.
This capstone lesson applies plane vectors as a proof tool and a physics model, aligning with the Common Core vector standards HSN-VM.A (represent and model with vector quantities — magnitude, direction, components), HSN-VM.B (perform operations on vectors: add, subtract, scale, and find the resultant), and HSN-VM.C (work with vectors and their algebra). It also revisits HSG-SRT.D.10–11 (proving and applying the laws of sines and cosines) by deriving the law of cosines directly from the dot product, and connects to HSN-VM.A.3 (solving problems involving velocities and forces). The "find a λ for parallel, make the dot zero for perpendicular" toolkit is the high-school analytic-geometry workhorse and the bridge into vectors in space (Stage 28).