Visual walkthrough — Position vector, displacement, distance

We will lean on ideas from Scalars and vectors (a number vs a number-with-direction), Coordinate systems and unit vectors (how we pin a point down), Vector addition and subtraction (how arrows combine), and Pythagoras theorem (how we measure a straight length). Everything else we grow from scratch.

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Step 1 — Pin a point down with a reference dot

WHAT. We drop a single dot on the board and call it $O$, the origin — our agreed "home base." Then we place an object at some point $P$ and draw a straight arrow from $O$ to $P$. That arrow is the position vector, written $\vec{r}$.

WHY. A bare location like "the object is at 3" is meaningless — 3 what, measured from where, in which direction? Nothing on a board has a location until we choose a home base to measure from. The arrow $\vec{r}$ bundles two facts into one object: how far (its length) and which way (where it points).

PICTURE. The yellow dot is $O$. The blue arrow is $\vec{r}$, tail nailed at $O$, head resting on $P$. Its horizontal shadow is $x$, its vertical shadow is $y$.

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Step 2 — Two points, two arrows

WHAT. Now the object moves. Mark its start at $A$ and its end at $B$. Draw two position arrows from the same origin $O$: ${\vec{r}}_i$ (to the start) and ${\vec{r}}_f$ (to the end). The subscripts mean $i$ = initial, $f$ = final.

WHY. To talk about motion we need a before and an after. Both arrows share the same tail $O$ so they are measured on the same ruler — that shared origin is the trick that makes the next step collapse so neatly.

PICTURE. Blue arrow ${\vec{r}}_i$ points to $A$; pink arrow ${\vec{r}}_f$ points to $B$. Both grow out of the yellow origin dot.

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Step 3 — The bridging arrow: displacement

WHAT. We ask one question: what single arrow, laid onto the tip of ${\vec{r}}_i$, carries us to the tip of ${\vec{r}}_f$? Call that bridging arrow $\Delta \vec{r}$ (the symbol $\Delta$, "delta," means "the change in").

WHY. This bridging arrow is the physical thing — it points from where you actually started straight to where you actually ended. It ignores the origin entirely, which is exactly what we want: two people using different home-bases must still agree on how the object moved.

PICTURE. The green bridging arrow runs tail-at-$A$, head-at-$B$. Notice it completes a triangle with the two position arrows: ${\vec{r}}_i$, then $\Delta \vec{r}$, arrives where ${\vec{r}}_f$ arrives.

Read the picture as an arrow equation (this is Vector addition and subtraction in action): $\underset{\text{go to start}}{\underbrace{{\vec{r}}_i}}+\underset{\text{bridge across}}{\underbrace{\Delta \vec{r}}}=\underset{\text{arrive at end}}{\underbrace{{\vec{r}}_f}}$

Solve for the bridge by peeling ${\vec{r}}_i$ off both sides: $\boxed{\Delta \vec{r}={\vec{r}}_f-{\vec{r}}_i}$

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Step 4 — How long is that bridge? (Pythagoras enters)

WHAT. The bridge $\Delta \vec{r}$ has a length. We call it $|\Delta \vec{r}|$ — the bars mean "the size of, forget the direction." To find it we drop the horizontal leg $(x_f-x_i)$ and vertical leg $(y_f-y_i)$ and glue them with a right angle.

WHY Pythagoras and not something else? The two legs meet at $90^{\circ }$ (right-step and up-step are perpendicular by construction). The one tool that turns two perpendicular legs into the straight hypotenuse is Pythagoras theorem — that is exactly the question "how long is the diagonal of this right triangle?"

PICTURE. Green hypotenuse $=|\Delta \vec{r}|$; yellow horizontal leg; blue vertical leg; the little square marks the right angle.

$|\Delta \vec{r}|=\sqrt{\underset{{\text{horizontal leg}}^2}{\underbrace{(x_f-x_i{)}^2}}+\underset{{\text{vertical leg}}^2}{\underbrace{(y_f-y_i{)}^2}}}$

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Step 5 — The road the feet actually walked

WHAT. Displacement forgot the journey. Now we remember it. The distance is the total length of the actual path, chopped into little straight segments and added up. It is a scalar (see Scalars and vectors) — just a non-negative number, no direction.

WHY. Real feet don't teleport along the bridge; they follow roads that bend. Distance is what an odometer reads. To measure a bendy path we approximate it by many tiny straight pieces and sum their lengths.

PICTURE. The wandering pink road from $A$ to $B$, cut into short chalk segments $s_1,s_2,s_3,\dots$; the straight green bridge $\Delta \vec{r}$ cuts underneath for comparison.

$\text{distance}=\underset{\text{add every little step}}{\underbrace{s_1+s_2+s_3+\cdots }}\ge 0$

Each $s_k$ is a length, so it can never be negative, so the total can never be negative.

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Step 6 — Why the bridge always loses to the road

WHAT. Take just two road pieces, $\vec{a}$ then $\vec{b}$, that together go from $A$ to $B$. The bridge is $\vec{a}+\vec{b}$. The triangle inequality says the direct side is never longer than the two-side detour: $|\vec{a}+\vec{b}|\le |\vec{a}|+|\vec{b}|$

WHY. In any triangle, one side alone cannot out-reach the other two summed — otherwise the triangle wouldn't close. Chain this fact across all the little road pieces and the whole road wins every time. Geometrically: the straight line is the shortest route between two points.

PICTURE. A triangle with green direct side $\vec{a}+\vec{b}$ and two pink detour sides $\vec{a}$, $\vec{b}$ — the detour is visibly longer.

Chaining it over every segment gives the headline result: $\boxed{|\Delta \vec{r}|\le \text{distance}}$

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Step 7 — Every case, including the sneaky ones

WHAT / WHY / PICTURE — the full case sweep. The inequality has one equality case and several degenerate ones. The reader must never meet a scenario we skipped.

• Case A — straight, one direction (equality). Road is the bridge, so $|\Delta \vec{r}|=\text{distance}$. This is the only equality case (the Motion in a straight line special case).
• Case B — the path bends. Road longer than bridge: $|\Delta \vec{r}|<\text{distance}$.
• Case C — forward then back (partial reversal). Bridge shrinks while road keeps growing; the gap widens.
• Case D — closed loop, $A=B$. Here ${\vec{r}}_f={\vec{r}}_i$, so $\Delta \vec{r}=\vec{0}$ and $|\Delta \vec{r}|=0$, yet distance $>0$. Bird flew nowhere; feet still got tired.
• Case E — didn't move at all. Path length $0$ and displacement $\vec{0}$: both sides $0$, equality holds trivially.

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The one-picture summary

Everything above compressed into a single board: origin $O$, the two position arrows, the green straight bridge $\Delta \vec{r}$, the wandering pink road, the right triangle that measures the bridge, and the inequality stamped across the bottom.

Recall Feynman retelling — the whole walk in plain words

Drop a home dot on the ground; that's the origin. An arrow from home to you is your position — it says how far and which way. Now walk somewhere: draw a home-arrow to your start and another to your end. The single arrow bridging start-to-end is your displacement — and because both home-arrows are measured from the same home, the home-dot cancels out, so displacement is the same no matter where anyone puts their home. To measure that bridge's length, drop its sideways and upward shadows, make a right triangle, and use Pythagoras. Meanwhile, chop your actual winding road into tiny steps and add them all up — that's distance, and since each step is a length it's never negative. Finally, in any little triangle the straight side can't beat the two bent sides, and stacking that fact over your whole road proves the straight bridge is never longer than the road: $|\Delta \vec{r}|\le \text{distance}$. They tie only when you march straight without ever turning back — and if you loop right back home, the bridge is zero while your feet are exhausted.

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Connections

• Position vector, displacement, distance — the parent this walkthrough visualises.
• Scalars and vectors — displacement is a vector, distance a scalar.
• Vector addition and subtraction — Step 3 solves ${\vec{r}}_i+\Delta \vec{r}={\vec{r}}_f$.
• Coordinate systems and unit vectors — $\hat{i},\hat{j}$ and coordinates in Steps 1–2.
• Pythagoras theorem — Step 4 measures the bridge length.
• Motion in a straight line — Case A, the only equality case.
• Average velocity and average speed — velocity uses the bridge, speed uses the road.