Visual walkthrough — Unit vectors — î, ĵ, k̂; constructing unit vector

This page is the picture-version of the parent note, Unit vectors. If a word here feels unfamiliar, it gets built in the very step where it first shows up.

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Step 1 — Draw an arrow. Notice it does two jobs at once.

WHAT. We draw one arrow on gridded paper and call it $\vec{A}$ (the little arrow-hat over the letter, $\stackrel{⃗}{}$, just means "this letter is an arrow, not an ordinary number").

WHY. Before we can split a vector into "how much" and "which way", we have to see that it secretly bundles both. A single arrow already answers two questions: how long is it? and where does it point?

PICTURE. Look at the amber arrow. Its length (how far the tip is from the tail) answers "how much". Its tilt answers "which way". These are two independent facts glued into one object.

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Step 2 — Lay the arrow on the axes and read off its components.

WHAT. We slide the tail of $\vec{A}$ to the origin (the corner where the axes cross) and drop two dashed lines from its tip: one straight down to the horizontal axis, one straight across to the vertical axis. This gives two numbers, $A_x$ and $A_y$.

WHY. Grid paper can only measure things along its own lines — rightward and upward. So we ask: how far right does the arrow reach, and how far up? Those two answers, $A_x$ and $A_y$, are the components. They rebuild the arrow completely, because "go right $A_x$, then up $A_y$" lands exactly on the tip.

PICTURE. The cyan horizontal leg has length $A_x$; the cyan vertical leg has length $A_y$; the amber arrow is the slanted line closing the corner.

$\vec{A}=\underset{\text{how far right}}{\underbrace{A_x}}\underset{\text{right-pointer}}{\underbrace{\hat{i}}}+\underset{\text{how far up}}{\underbrace{A_y}}\underset{\text{up-pointer}}{\underbrace{\hat{j}}}$

Here $\hat{i}$ is a length-1 arrow pointing right and $\hat{j}$ is a length-1 arrow pointing up — the tiny fixed rulers of the grid. (Full detail lives in Vectors — components and resolution.)

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Step 3 — See the right triangle hiding inside the arrow.

WHAT. We notice the two legs $A_x$, $A_y$ and the arrow itself form a right triangle: the two legs meet at a perfect $90^{\circ }$ corner, and the arrow is the slanted side (the hypotenuse).

WHY. A right triangle is the one shape where length obeys a clean rule (Pythagoras theorem). Since we want the arrow's length, we must find the shape whose length we know how to compute — and here it is, for free.

PICTURE. The small square in the corner marks the $90^{\circ }$ angle. The horizontal leg $A_x$ and the vertical leg $A_y$ are perpendicular; the amber hypotenuse is $\vec{A}$.

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Step 4 — Turn the triangle into the magnitude formula.

WHAT. We apply Pythagoras to the triangle from Step 3.

WHY. Pythagoras is the tool that converts "two perpendicular legs" into "length of the slanted side". We reach for it — and not, say, plain addition — precisely because the legs are perpendicular. (Adding $A_x+A_y$ would be the length only if they lay on the same line, which they don't.)

PICTURE. The squares built on each side visualise the theorem: the area on the hypotenuse equals the two leg-areas combined.

$\underset{{\text{(arrow length)}}^2}{\underbrace{|\vec{A}{|}^2}}=\underset{{\text{right-leg}}^2}{\underbrace{A_x^2}}+\underset{{\text{up-leg}}^2}{\underbrace{A_y^2}}$

Take the square root of both sides. Length can't be negative, so we keep the positive root:

For 3D just do it twice: first the flat diagonal $r=\sqrt{A_x^2+A_y^2}$, then combine with the height $A_z$ to get $|\vec{A}|=\sqrt{A_x^2+A_y^2+A_z^2}$. Same right-triangle move, stacked. See Magnitude and direction of a vector.

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Step 5 — The key idea: shrink the arrow by its own length.

WHAT. We divide the whole arrow by the single positive number $|\vec{A}|$ we just computed: $\hat{A}=\frac{\vec{A}}{|\vec{A}|}.$

WHY. We want a pointer of length exactly $1$ that still aims the same way. Multiplying a vector by a positive number never rotates it — it only stretches or squashes it (a negative number would flip it; zero would destroy it). The magic multiplier is $\frac{1}{|\vec{A}|}$: it is positive, so direction survives, and it is sized to cancel the length.

PICTURE. The long amber arrow $\vec{A}$ and the short cyan arrow $\hat{A}$ lie exactly along the same line; only the length changed, from $|\vec{A}|$ down to $1$.

Check the new length by pulling the constant $\frac{1}{|\vec{A}|}$ out of the length bars: $|\hat{A}|=\left|\frac{\vec{A}}{|\vec{A}|}\right|=\frac{|\vec{A}|}{|\vec{A}|}=1.✓$

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Step 6 — Walk one numeric case all the way through.

WHAT. Take $\vec{A}=3\hat{i}+4\hat{j}$ and grind the recipe.

WHY. Numbers pin down that the abstract steps actually land on length $1$.

PICTURE. The $3$-$4$-$5$ triangle: right-leg $3$, up-leg $4$, hypotenuse $5$; the unit vector is the same slant shrunk to length $1$.

$|\vec{A}|=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$ $\hat{A}=\frac{3\hat{i}+4\hat{j}}{5}=0.6\hat{i}+0.8\hat{j}$ $|\hat{A}|=\sqrt{0.6^2+0.8^2}=\sqrt{0.36+0.64}=\sqrt{1}=1✓$

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Step 7 — Edge and degenerate cases (never leave a scenario unshown).

WHAT / WHY / PICTURE, three cases side by side:

• Negative components — $\vec{C}=-3\hat{i}-4\hat{j}$ points down-left. Its length is still $\sqrt{(-3{)}^2+(-4{)}^2}=5$ (squares kill the signs), so $\hat{C}=-0.6\hat{i}-0.8\hat{j}$. The minus signs stay — they are the "which-way", and stripping them would silently rotate the pointer $180^{\circ }$.
• An axis vector — $\vec{E}=5\hat{j}$ has $A_x=0$. Then $|\vec{E}|=\sqrt{0^2+5^2}=5$ and $\hat{E}=\hat{j}$. A basis vector like $\hat{j}$ is already a unit vector; the recipe just confirms it.
• The zero vector — $\vec{0}=0\hat{i}+0\hat{j}$ has length $0$. The recipe demands division by $0$, which is undefined. Geometrically there is no arrow to point along, so the zero vector has no direction and no unit vector. This is the one input the construction refuses.

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The one-picture summary

Read it left to right: an arrow carries length × direction; drop its shadows to get components; those components are the legs of a right triangle whose hypotenuse-length is $\sqrt{A_x^2+A_y^2}$ by Pythagoras; divide the whole arrow by that length and you keep the tilt but reset the size to $1$ — a pure pointer $\hat{A}$.

Recall Feynman: the whole walkthrough in plain words

I drew a slanted arrow on grid paper. It was doing two jobs — telling me how far and which way. To separate them, I measured how far it went rightward and how far upward; those two numbers are its shadows on the two rulers. The shadows meet at a square corner, so the arrow, the right-shadow, and the up-shadow make a right triangle. For right triangles I know one magic fact — square the two legs, add, square-root — and out pops the arrow's true length. Now the finishing move: I shrink the entire arrow by exactly that length. Shrinking by a positive number never spins an arrow, it only changes its size, so my little arrow still points the same way but is now exactly 1 long. That length-1 arrow is the unit vector: a stick that only knows which way, never how far. The one arrow it can't be built from is the arrow of length zero — there's nothing to point along, and you'd be dividing by nothing.

Recall Quick self-test

Unit vector of $-3\hat{i}-4\hat{j}$? ::: $-0.6\hat{i}-0.8\hat{j}$ (length is still $5$; keep the signs). Why divide by $|\vec{A}|$ and not some other number? ::: Only dividing by the arrow's own length forces the result to length exactly $1$ while a positive divisor keeps the direction. Which vector has no unit vector? ::: The zero vector $\vec{0}$ — dividing by $0$ is undefined and it points nowhere.

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Connections

• Unit vectors — î, ĵ, k̂; constructing unit vector (parent)
• Vectors — components and resolution
• Magnitude and direction of a vector
• Pythagoras theorem
• Vector addition — triangle and parallelogram laws
• Dot product
• Cross product
• Position vector and displacement