Vector representation — magnitude, direction, components: Visual walkthrough

We are answering one question the whole way down:

A slanted arrow is awkward. Can we replace it with two straight, honest numbers — one purely sideways, one purely up — and get back and forth without losing anything?

Step 1 — The arrow, and the two numbers we wish we had

WHAT. Draw an arrow $A$ starting at the origin (the corner where the two axes cross). The horizontal line is the x-axis (rightward is positive). The vertical line is the y-axis (upward is positive).

WHY. Before we compute anything, we must name what we want. We want two clean numbers:

$A_{x}$ = how far the arrow reaches sideways (its shadow on the x-axis),
$A_{y}$ = how far it reaches up (its shadow on the y-axis).

These are called the components. A component is just "how much of the arrow points along one axis."

PICTURE. Look at the arrow and its two dotted shadows below and to the left.

Figure — Vector representation — magnitude, direction, components

Step 2 — Close the shape: a right triangle appears

WHAT. From the tip of the arrow, drop a straight line straight down to the x-axis. That vertical drop, plus the arrow, plus the piece of x-axis underneath, enclose a triangle.

WHY. A slanted arrow alone has no structure we can measure. The moment we drop that perpendicular, we get a right triangle — a triangle with one square (90°) corner. Right triangles are special: their three sides lock into fixed ratios once you fix one angle. That lock is exactly what lets us turn an angle into lengths.

PICTURE. The little square marks the 90° corner. Notice the three named sides.

The hypotenuse (the slanted side, opposite the square corner) is the whole arrow — length $A$ .
The bottom side lies along x — its length is $A_{x}$ .
The vertical side is the drop — its length is $A_{y}$ .
The angle at the origin, measured anticlockwise from the +x axis, is $θ$ .

Step 3 — What "angle" even means as a ratio (defining cosine)

WHAT. Fix the angle $θ$ but imagine sliding the tip further out along the same direction. The triangle grows, but its shape stays identical. So the ratio of any two sides never changes — it depends only on $θ$ .

WHY this tool, cosine, and not another. We want a machine that eats an angle and spits out "how much of the length went sideways." The side touching the angle (the bottom, along x) is called the adjacent side. The ratio

$name of this ratio cos θ = \frac{A _{x} adjacent side}{hypotenuse A}$

is defined to be this fixed number. Reading it in words: cosine of $θ$ is the fraction of the arrow that lies flat along x. When $θ$ is small (nearly flat), almost all the length is sideways, so $cos θ$ is near 1. When $θ = 90°$ (straight up), none is sideways, so $cos θ = 0$ .

PICTURE. Three nested triangles, all the same $θ$ , all the same ratio $A_{x} / A$ .

Step 4 — The "up" fraction (defining sine) and the first golden result

WHAT. Do the same for the vertical side. The side facing the angle (across from it) is the opposite side, of length $A_{y}$ . Its fixed ratio to the hypotenuse gets the name sine:

$name sin θ = \frac{A _{y} opposite side}{hypotenuse A}$

WHY. Cosine captured "how much went sideways." Sine captures "how much went up." Between them they account for the entire arrow.

WHAT WE DO NEXT. Both ratios have $A$ on the bottom. Multiply both sides of each by $A$ to free the component:

$A_{x} = A cos θ A_{y} = A sin θ$

Term by term: $A$ is the arrow's length (a positive number), $cos θ$ is the sideways fraction, so their product $A_{x}$ is the actual sideways distance. Same story for $A_{y}$ with the up-fraction $sin θ$ .

PICTURE. The same triangle, now labelled with the final formulas sitting on their sides.

Step 5 — Going backwards: recover the length (Pythagoras, shown as areas)

WHAT. Now suppose we're given $A_{x}$ and $A_{y}$ and want the length $A$ back.

WHY this tool, Pythagoras. In a right triangle the two legs and the hypotenuse are locked by one law: the square built on the hypotenuse equals the two squares built on the legs, added together. This is the only tool that turns two perpendicular lengths into the one diagonal length — exactly our need.

$square on the arrow A^{2} = square on the flat side A_{x}^{2} + square on the tall side A_{y}^{2} \Rightarrow A = A_{x}^{2} + A_{y}^{2}$

The (square root) just asks "what length, squared, gives that area?" — it undoes the squaring so we land back in length units. Notice $A_{x}^{2}$ and $A_{y}^{2}$ are never negative, so $A$ is always $\geq 0$ : a magnitude can never come out negative.

PICTURE. Literal squares sitting on each side; the two small areas tile into the big one.

Step 6 — Going backwards: recover the angle (why tangent, why divide)

WHAT. For the direction, divide the two component formulas from Step 4:

$\frac{A _{y}}{A _{x}} = \frac{A s i n θ}{A c o s θ} = \frac{s i n θ}{c o s θ} = tan θ$

WHY divide. Dividing kills the unknown $A$ (it cancels top and bottom). What survives is a pure "steepness" number: rise $A_{y}$ over run $A_{x}$ — that is the tangent. Tangent is the right tool because it depends only on the angle, not the length, which is exactly what "direction" should mean.

To get $θ$ itself we ask the reverse question — "which angle has this tangent?" — written $tan^{- 1}$ (arctan). It undoes tangent:

$θ = tan^{- 1} (\frac{A _{y}}{A _{x}})$

PICTURE. Two triangles of different sizes, same slope, showing $A_{y} / A_{x}$ is a steepness that ignores size.

Step 7 — The degenerate & every-quadrant cases (never skip these)

WHAT. The calculator's $tan^{- 1}$ only ever answers between $- 90°$ and $+ 90°$ — it always points rightward. But arrows can point left, and $tan$ repeats every $180°$ , so an up-left arrow and a down-right arrow share the same ratio $A_{y} / A_{x}$ . We must repair the answer using the signs of $A_{x}, A_{y}$ .

WHY. Two opposite directions ( $θ$ and $θ + 180°$ ) give identical tangents. The signs of the components are the only clue to which of the two is real.

PICTURE. Four arrows, one per quadrant, each with its raw angle and its fix.

$A_{x}$	$A_{y}$	Quadrant	Fix to raw $tan^{- 1}$
$+$	$+$	I	none
$-$	$+$	II	add $180°$
$-$	$-$	III	add $180°$
$+$	$-$	IV	leave (negative) or add $360°$

Degenerate inputs (must be handled by hand — $tan^{- 1}$ chokes):

The one-picture summary

One triangle carries the entire round trip: the hypotenuse is the arrow $A$ at angle $θ$ ; multiply by $cos θ, sin θ$ to go down into components; use Pythagoras and arctan (with a sign check) to climb back up to polar.

Recall Feynman retelling — the whole walkthrough in plain words

You've got a slanted arrow and it's annoying to work with. So you drop a straight line from its tip to the floor. That splits the arrow's journey into "walk this far East, then this far North" — two honest numbers instead of one awkward slant. The triangle you just drew has fixed shape for its angle, so the fraction of the arrow lying flat (call it cosine) and the fraction standing up (call it sine) depend only on the tilt. Multiply the arrow's length by those fractions and out pop the East and North numbers. To go back, you build squares on the two legs — their areas add up to the square on the arrow (that's Pythagoras), and taking the square root gives the length back. For the direction you divide North by East; the length cancels, leaving a pure steepness, and "arctan" asks which angle has that steepness. Last, you glance at the plus/minus signs to know if you're pointing left or down — because the calculator only ever points right. Straight-up, straight-down, and the lonely zero-length dot you just read off by eye. That's it: one arrow, two numbers, and a clean way home.

Connections

Vector Representation — magnitude, direction, components — the parent this walkthrough expands.
Scalars vs Vectors — why one number can't hold both facts.
Vector Addition — Triangle & Parallelogram Law — components make adding arrows trivial.
Resolution of Forces — Steps 1–4 applied to forces.
Dot Product and Cross Product — built directly on components.
Projectile Motion — split velocity into $v_{x}, v_{y}$ using Step 4.
Unit Vectors i, j, k — the axes $\hat{i}, \hat{j}$ our components ride on.

Active recall

Why drop a perpendicular from the arrow's tip?

It creates a right triangle whose fixed side-ratios convert the angle into lengths.

In the triangle, which side is adjacent to

θ

The x-side (bottom), because

θ

is measured from the x-axis.

Define

cos θ

geometrically.

adjacent ÷ hypotenuse — the "sideways fraction" of the arrow.

Why does dividing

A_{y}

A_{x}

remove the length

A

Both equal

A sin θ

and

A cos θ

; the

A

's cancel, leaving

tan θ

What does

tan^{- 1}

(arctan) actually ask?

"Which angle has this tangent?" — it undoes tangent.

When must you add

180°

to the raw arctan?

When

A_{x} < 0

(quadrants II and III).

Direction of

A = (0, - 5)

270°

(or

- 90°

) — straight down, arctan fails so read it by eye.

Direction of the zero vector?

Undefined — a point has no direction.

Why is magnitude never negative?

A = A_{x}^{2} + A_{y}^{2}

; squares can't be negative.