Visual walkthrough — Vector addition — triangle law, parallelogram law

We add two arrows and ask the only question that matters: how long is the straight arrow from where I started to where I stopped, and which way does it point?

---

Step 1 — Two arrows, one shared corner

WHAT. We draw two vectors $\vec{A}$ and $\vec{B}$ starting from the same point $O$. An arrow here means a journey: its length is how far, its direction is which way. The wedge of space between them is the angle $\theta$ (theta, a Greek letter we use as a name for "the angle between them"), swept counter-clockwise from $\vec{A}$ to $\vec{B}$.

WHY start tail-to-tail? Because the only piece of geometry that decides the answer is the opening between the two arrows. If we know $A$ (length of $\vec{A}$), $B$ (length of $\vec{B}$), and $\theta$ (the opening), everything else is forced.

PICTURE. Look at the two arrows leaving $O$ and the shaded wedge $\theta$ between them; the little curved arrow shows the counter-clockwise direction we measure in.

---

Step 2 — Lay $\vec{A}$ flat: build a ruler to measure with

WHAT. We rotate the whole picture so $\vec{A}$ lies flat along a horizontal line (the x-axis) and $O$ sits at the crossing point of the horizontal and vertical lines (the origin). Rotating the whole picture changes nothing about the answer — the arrows keep their lengths and their opening $\theta$.

WHY. A horizontal-and-vertical grid lets us describe any point by two plain numbers: how far right (call it $x$) and how far up (call it $y$). Numbers we can add; slanted arrows we cannot. This is the trick that turns geometry into arithmetic — the same idea behind Resolution of a vector into components. Because $\vec{A}$ now points along $+x$ and $\theta$ is measured counter-clockwise from it, $\vec{B}$ leans up and to the side — into the upper half of the grid.

PICTURE. $\vec{A}$ now runs along the bottom; its tip sits at the point $(A,0)$ — distance $A$ to the right, $0$ up.

---

Step 3 — Break $\vec{B}$ into "right" and "up"

WHAT. From the tip of $\vec{B}$ we drop a straight vertical line onto the x-axis. That makes a right triangle whose slanted side (the hypotenuse) is $\vec{B}$ itself, with the angle $\theta$ sitting at $O$.

WHY these two pieces? We need $\vec{B}$ written as (how far right) and (how far up) so we can add it to $\vec{A}$'s numbers. The tool that reads an angle-plus-hypotenuse into a right/up pair is trigonometry:

• $\cos \theta =\frac{\text{adjacent (the horizontal side)}}{\text{hypotenuse}}$ — so the horizontal side $=B\cos \theta$.
• $\sin \theta =\frac{\text{opposite (the vertical side)}}{\text{hypotenuse}}$ — so the vertical side $=B\sin \theta$.

Why cosine for the "right" part? Cosine measures how much of the arrow points along $\vec{A}$'s direction; sine measures how much points across it. That is exactly the split we want. Because $\theta \in [0^{\circ },180^{\circ }]$, the up-part $B\sin \theta \ge 0$ (the tip of $\vec{B}$ never dips below the base line), while the right-part $B\cos \theta$ may be positive (if $\theta <90^{\circ }$), zero (at $90^{\circ }$), or negative (if $\theta >90^{\circ }$, i.e. $\vec{B}$ leans backward).

PICTURE. The dotted right triangle under $\vec{B}$: base $B\cos \theta$, height $B\sin \theta$.

---

Step 4 — Slide $\vec{B}$ onto the tip of $\vec{A}$ (head-to-tail)

WHAT. Keep $\vec{B}$'s length and direction, but slide it so its tail sits on $\vec{A}$'s head. This is the triangle law: "continue the journey from where the first one ended." Sliding an arrow without turning it does not change it.

WHY. After walking $\vec{A}$ you are at $(A,0)$. Walking $\vec{B}$ from there adds $B\cos \theta$ to your right-position and $B\sin \theta$ to your up-position. So the place you finally land — the head of the resultant — is:

PICTURE. $\vec{A}$ along the base, $\vec{B}$ perched on its tip, and the finish point marked $(R_x,R_y)$.

Here $R_x$ is the total rightward distance from start to finish and $R_y$ the total upward distance. The straight arrow from $O$ to $(R_x,R_y)$ is the resultant $\vec{R}=\vec{A}+\vec{B}$, and its length we call $R=|\vec{R}|$.

---

Step 5 — Pythagoras: the length of the straight-home arrow

WHAT. The resultant reaches $R_x$ to the right and $R_y$ up. Those two form the legs of a right triangle; $\vec{R}$ is the hypotenuse, and its length is $R=|\vec{R}|$.

WHY Pythagoras and not something else? Because $R_x$ (horizontal) and $R_y$ (vertical) meet at a right angle by construction — the grid is square. The theorem that turns two perpendicular legs into a straight-line distance is exactly $(\text{leg}{)}^2+(\text{leg}{)}^2=(\text{hypotenuse}{)}^2$.

PICTURE. The right triangle $O$–(foot)–(head), with legs $R_x$, $R_y$ and hypotenuse $R$.

$R=\sqrt{R_x^2+R_y^2}=\sqrt{\underset{R_x^2}{\underbrace{(A+B\cos \theta {)}^2}}+\underset{R_y^2}{\underbrace{(B\sin \theta {)}^2}}}$

---

Step 6 — Multiply out, and watch a term vanish

WHAT. We expand the square. Expanding $(A+B\cos \theta {)}^2$ gives three pieces; the $(B\sin \theta {)}^2$ adds one more:

$R^2=\underset{\text{from }{A}^2}{\underbrace{A^2}}+\underset{\text{the cross term}}{\underbrace{2AB\cos \theta }}+\underbrace{B^2{\cos }^2\theta }+\underbrace{B^2{\sin }^2\theta }$

WHY it simplifies. The last two share $B^2$: $B^2({\cos }^2\theta +{\sin }^2\theta )$. And ${\cos }^2\theta +{\sin }^2\theta =1$ always — it is Pythagoras applied to the little unit triangle inside every angle. So those two collapse into a single $B^2$.

PICTURE. A colour-coded map of the four terms: two "clean" squares $A^2$, $B^2$, and the orange cross term $2AB\cos \theta$ that carries all the angle information.

This is the Law of cosines in disguise; the $\cos \theta$ inside is exactly the quantity a Scalar (dot) product would hand you.

---

Step 7 — Which way does $\vec{R}$ point?

WHAT. We want $\alpha$ (alpha), the angle $\vec{R}$ makes with the base arrow $\vec{A}$, measured counter-clockwise from $\vec{A}$ just like $\theta$. In the right triangle of Step 5, the side opposite $\alpha$ is $R_y$ and the side adjacent to $\alpha$ is $R_x$.

WHY tangent? We know both legs but no hypotenuse-angle yet. The one trig ratio built from opposite over adjacent — needing neither the hypotenuse nor anything else — is the tangent. It answers precisely "how steep is this arrow above the base?"

PICTURE. The angle $\alpha$ tucked at $O$ between $\vec{A}$ and $\vec{R}$, with $R_y$ (opposite) and $R_x$ (adjacent) labelled.

Reading $\alpha$ back out — and why one ${\tan }^{-1}$ is not enough. The ordinary inverse-tangent ${\tan }^{-1}$ (also written $\arctan$) answers "which angle has this tangent?", but it can only ever return an answer in its principal range $(-90^{\circ },+90^{\circ })$ — angles that point into the right half of the plane. Our resultant, though, can point up-and-left (when $R_x<0$), and a bare ${\tan }^{-1}$ would fold that back into the wrong half. The clean fix is a single function that looks at both $R_y$ and $R_x$ separately (their individual signs, not just their ratio) and returns the correct wedge in one shot. That function is called $atan2$:

The last two lines are the degenerate "flat resultant" outcomes; the middle picture below shows why they happen.

---

Step 8 — Every edge case, drawn

WHAT. We sweep $\theta$ from $0^{\circ }$ to $180^{\circ }$ and watch $R$ change. Only the cross term $2AB\cos \theta$ moves, because $\cos \theta$ slides from $+1$ down to $-1$.

WHY check these? A formula you cannot stress-test is a formula you do not trust. Each limit must match common sense.

PICTURE. Three stacked diagrams — aligned, perpendicular, opposed — with the resultant redrawn each time.

$\theta$	$\cos \theta$	$R=\sqrt{A^2+B^2+2AB\cos \theta }$	$\alpha =atan2(R_y,R_x)$	meaning
$0^{\circ }$	$+1$	$\sqrt{(A+B{)}^2}=A+B$	$0^{\circ }$ ($R_y=0,R_x>0$)	arrows aligned → longest
$90^{\circ }$	$0$	$\sqrt{A^2+B^2}$	quadrant I ($R_x=A>0$)	perpendicular → plain Pythagoras
$180^{\circ }$	$-1$	$\sqrt{(A-B)^2}=	A-B	$

At $\theta =0^{\circ }$ and $\theta =180^{\circ }$ we have $R_y=B\sin \theta =0$ — the resultant lies flat along the base line, so $atan2$ returns $0^{\circ }$ (head along $+\vec{A}$, when $R_x>0$) or $180^{\circ }$ (head along $-\vec{A}$, when $R_x<0$), the two "$R_y=0$" outcomes listed in Step 7.

So the resultant is trapped in $|A-B|\le R\le A+B$. If additionally $A=B$, then at $\theta =180^{\circ }$ the two arrows cancel and $R=0$ — the degenerate zero vector, a journey that ends where it began (and $\alpha$ is undefined, because a zero-length arrow points nowhere).

---

The one-picture summary

Everything above lives in a single diagram: lay $\vec{A}$ flat, drop $\vec{B}$'s components, close the right triangle, read $R$ off Pythagoras and $\alpha$ off $atan2$.

Recall Feynman: tell it to a 12-year-old

Put two arrows tip-tails at the same dot. Turn the page so the first arrow lies flat like a ruler, and always measure angles going anti-clockwise from that ruler. Now the second arrow leans off it at some slant — split that slant arrow into "how far sideways" (that's the cosine part) and "how far up" (that's the sine part, which stays positive because the arrow leans up). Stack the second arrow onto the end of the first: you've now walked sideways $A+B\cos \theta$ and up $B\sin \theta$. Draw the straight line home. Its length $R$ is found by the flat-triangle rule $a^2+b^2=c^2$ — that gives $R=\sqrt{A^2+B^2+2AB\cos \theta }$ — and its tilt is "up-part over sideways-part." If the home-arrow points straight up, its tilt is exactly $90^{\circ }$; if it leans back to the left, it is between $90^{\circ }$ and $180^{\circ }$ — the $atan2$ tool sorts out which just by looking at whether "sideways" is positive or negative. Swing the second arrow from lined-up to opposite and the home-arrow shrinks from $A+B$ all the way down to $|A-B|$. That's the whole story.

Recall Quick self-test

Why does the cross term carry all the angle info? ::: $A^2$ and $B^2$ never change; only $2AB\cos \theta$ depends on $\theta$. Where does ${\cos }^2\theta +{\sin }^2\theta =1$ get used? ::: In Step 6, to collapse $B^2{\cos }^2\theta +B^2{\sin }^2\theta$ into $B^2$. Why tangent for the direction and not sine? ::: We know both legs $R_x,R_y$ but not the hypotenuse; tangent = opposite/adjacent uses exactly those. Why does only one cross-term sign ever appear? ::: Because $\theta \in [0^{\circ },180^{\circ }]$ keeps $\sin \theta \ge 0$, so $R_y$ stays upward; all sign changes hide inside $\cos \theta$. What does $R$ (no arrow) mean versus $\vec{R}$? ::: $\vec{R}$ is the resultant vector; $R=|\vec{R}|$ is just its length. Why is a bare ${\tan }^{-1}$ not enough, and what replaces it? ::: ${\tan }^{-1}$ only returns $(-90^{\circ },90^{\circ })$ so it mis-places leftward ($R_x<0$) arrows and divides by zero at $R_x=0$; $atan2(R_y,R_x)$ uses both signs to give the correct wedge.

---

Connections

• Resolution of a vector into components — Steps 2–3 are resolution.
• Vectors — components and unit vectors — the $(R_x,R_y)$ bookkeeping in $\hat{i},\hat{j}$ form.
• Law of cosines — the Step 6 result rewritten.
• Scalar (dot) product — supplies the $\cos \theta$ inside the cross term.
• Subtraction of vectors and the difference vector — the $\theta =180^{\circ }$ edge case.
• Relative velocity — the same triangle for boats and rivers.
• Vector addition — triangle law, parallelogram law — parent topic.