Vector representation — magnitude, direction, components: Exercises

Here $A = ∣ A ∣$ is the magnitude (the length of the arrow, never negative), $θ$ is the direction angle measured anticlockwise from the +x axis, and $A_{x}, A_{y}$ are the components (the shadows the arrow casts on the x- and y-axes). $\hat{i}, \hat{j}$ are the length-1 arrows pointing along $+ x$ and $+ y$ — see Unit Vectors i, j, k.

Formula The quadrant check, spelled out once and for all

The calculator's $tan^{- 1}$ only ever answers between $- 90°$ and $+ 90°$ — that's the right half-plane ( $A_{x} > 0$ ). To land on the correct $0°$ – $360°$ direction the first time, look at the signs of $A_{x}, A_{y}$ :

Sign of $A_{x}$	Sign of $A_{y}$	Quadrant	What to do with the raw $tan^{- 1}$
$+$	$+$	I	keep it as-is
$-$	$+$	II	add $180°$
$-$	$-$	III	add $180°$
$+$	$-$	IV	add $360°$ (or leave it negative)
$0$	any	—	vertical: $90°$ if $A_{y} > 0$ , $270°$ if $A_{y} < 0$

The one thing to remember: if $A_{x} < 0$ , add $180°$ . That single rule fixes Quadrants II and III, where the calculator points you the exact opposite way.

Before we start, one picture to fix all the vocabulary in one place. In the figure below, the blue arrow is the vector $A$ ; its yellow horizontal shadow is $A_{x} = A cos θ$ , its green vertical shadow is $A_{y} = A sin θ$ , the red arc is the angle $θ$ swept anticlockwise from $+ x$ , and the little white square marks the right angle where the shadows meet. Keep this triangle in your mind's eye for every problem that follows.

Figure — Vector representation — magnitude, direction, components

Level 1 — Recognition

Recall Solution

WHAT: we're just matching each formula to the word it produces.

(i) $A_{x}^{2} + A_{y}^{2}$ is the magnitude $A$ — it's Pythagoras on the two legs of the right triangle, and it gives you a length (a single non-negative number).
(ii) $tan^{- 1} (A_{y} / A_{x})$ is the direction $θ$ — the angle whose tangent (opposite÷adjacent) equals the ratio of the sides.

WHY it works: length comes from the hypotenuse; angle comes from the slope $A_{y} / A_{x}$ of the arrow. Length can't be an angle and vice-versa, so there's no ambiguity.

One caution (don't skip this): the raw $tan^{- 1} (A_{y} / A_{x})$ is only the true direction when $A_{x} > 0$ . Whenever $A_{x} < 0$ you must add $180°$ (the quadrant check in the box above), because $tan$ gives the same ratio for two opposite arrows. We'll do this explicitly from Level 3 onward — but never trust the calculator's angle without first glancing at the sign of $A_{x}$ .

Recall Solution

$A_{x} = A cos θ, A_{y} = A sin θ .$ WHY: drop a perpendicular from the arrow's tip to the x-axis. The x-side lies along the axis you measured $θ$ from — it is the adjacent side, and adjacent÷hypotenuse $= cos θ$ . Multiply by $A$ to free $A_{x}$ . The y-side is opposite, so it uses $sin$ .

Recall Solution

False. $A = (- 8)^{2} + 6^{2} = 64 + 36 = 100 = 10$ . WHY: squaring erases every minus sign, so a magnitude can never come out negative. The "which way" information (that the arrow points up-left) lives in the signs of the components, not in $A$ .

Level 2 — Application

Recall Solution

WHAT: polar → components, so we use $A_{x} = A cos θ$ , $A_{y} = A sin θ$ . $A_{x} = 10 cos 30° = 10 \times 0.8660 = 8.66 m$ $A_{y} = 10 sin 30° = 10 \times 0.5 = 5 m$ WHY these functions: $30°$ is measured from the x-axis, so the x-side is adjacent (cosine), the y-side opposite (sine). CHECK: $8.6 6^{2} + 5^{2} = 75 + 25 = 100 = 10$ ✓ — the magnitude comes back.

Recall Solution

$F = 1 2^{2} + 5^{2} = 144 + 25 = 169 = 13.$ WHY Pythagoras: the components are the two perpendicular legs of a right triangle; the vector itself is the hypotenuse, and the hypotenuse² equals the sum of the leg squares.

Recall Solution

WHAT: a unit vector is the arrow shrunk to length 1: $\overset{v}{^} = v /∣ v ∣$ . First the length: $∣ v ∣ = 6^{2} + 8^{2} = 36 + 64 = 100 = 10$ . Divide each component by $10$ : $\overset{v}{^} = \frac{6}{10} \hat{i} + \frac{8}{10} \hat{j} = 0.6 \hat{i} + 0.8 \hat{j} .$ WHY dividing keeps the direction: both components shrink by the same factor $1/10$ , so their ratio $A_{y} / A_{x}$ — which sets the angle — is untouched. Only the length changes. CHECK: $0. 6^{2} + 0. 8^{2} = 0.36 + 0.64 = 1 = 1$ ✓.

Level 3 — Analysis

Recall Solution

Magnitude: $A = (- 4)^{2} + 3^{2} = 16 + 9 = 25 = 5$ . Raw angle: $tan^{- 1} (3/ (- 4)) = tan^{- 1} (- 0.75) = - 36.87°$ . Quadrant check: $A_{x} < 0$ and $A_{y} > 0$ ⇒ the arrow points up and to the left ⇒ Quadrant II. The calculator's $- 36.87°$ points down-right (Quadrant IV) — the wrong direction. Add $180°$ : $θ = - 36.87° + 180° = 143.13°.$ WHY the calculator lies: $tan$ repeats every $180°$ , so $tan (143.13°) = tan (- 36.87°)$ — two opposite arrows share one ratio. Only the signs of the components tell them apart.

In the figure below, the green arrow is the true vector $A = (- 4, 3)$ sitting in Quadrant II; the dashed red arrow is what the calculator's $- 36.87°$ actually points to — $(4, - 3)$ , the exact opposite corner (Quadrant IV). The white dotted line through both shows they lie on one straight line, which is precisely why they share the same $tan$ ratio. The yellow label marks the $+ 180°$ jump that carries you from the wrong arrow to the right one.

Recall Solution

Raw angle: $tan^{- 1} ((- 2) / (- 2)) = tan^{- 1} (1) = 45°$ . Quadrant check: both components negative ⇒ arrow points down-left ⇒ Quadrant III. Add $180°$ : $θ = 45° + 180° = 225°.$ WHY: the ratio $A_{y} / A_{x} = (- 2) / (- 2) = + 1$ is the same as for $(2, 2)$ in Quadrant I, so the calculator gives $45°$ . The two minus signs mark the true direction as the diagonally-opposite one, $225°$ .

Recall Solution

"Below the x-axis" means the angle is measured clockwise, so $θ = - 40°$ . $F_{x} = 50 cos (- 40°) = 50 \times 0.7660 = 38.30 N$ $F_{y} = 50 sin (- 40°) = 50 \times (- 0.6428) = - 32.14 N$ WHY $F_{y}$ is negative: the arrow dips below the axis, so its vertical shadow points downward ( $- y$ ). $cos$ is even ( $cos (- 40°) = cos 40°$ ) so $F_{x}$ stays positive; $sin$ is odd ( $sin (- 40°) = - sin 40°$ ) so $F_{y}$ flips sign. The maths automatically records "downward" as a minus.

Level 4 — Synthesis

Recall Solution

Step 1 — components (WHY: you add components, never magnitudes). $F_{1 x} = 30 cos 0° = 30, F_{1 y} = 30 sin 0° = 0.$ $F_{2 x} = 40 cos 90° = 0, F_{2 y} = 40 sin 90° = 40.$ Step 2 — add like with like: $R_{x} = 30 + 0 = 30, R_{y} = 0 + 40 = 40.$ Step 3 — back to polar: $R = 3 0^{2} + 4 0^{2} = 900 + 1600 = 2500 = 50 N .$ $θ = tan^{- 1} (40/30) = tan^{- 1} (1.333) = 53.13°.$ Both components positive ⇒ Quadrant I ⇒ no correction. Resultant: $50 N$ at $53.13°$ . See Vector Addition — Triangle & Parallelogram Law for the geometric picture.

In the figure below, the yellow arrow is $F_{1}$ (30 N along $+ x$ ) and the green arrow is $F_{2}$ (40 N straight up), drawn tip-to-tail: start $F_{2}$ where $F_{1}$ ends. The blue arrow closing the triangle from origin to final tip is the resultant $R$ — the familiar 3-4-5 triangle scaled up ten times, so its hypotenuse is exactly 50. The red arc at the origin is the $53.13°$ direction we computed.

Recall Solution

Sum of vectors: $A + B = (5 - 5) \hat{i} + 0 \hat{j} = 0$ , so $∣ A + B ∣ = 0$ . Sum of magnitudes: $∣ A ∣ + ∣ B ∣ = 5 + 5 = 10$ . WHY they clash: the two arrows point exactly opposite, so they cancel — the resultant is a zero-length arrow. But magnitudes are just lengths; adding two lengths can never cancel. This is the whole reason we add components (which carry sign) rather than magnitudes. Equality would only hold if the arrows were parallel and same-direction.

Recall Solution

$v_{x} = 20 cos 60° = 20 \times 0.5 = 10 m/s$ $v_{y} = 20 sin 60° = 20 \times 0.8660 = 17.32 m/s$ $\frac{v _{y}}{v _{x}} = \frac{17.32}{10} = 1.732 = tan 60° ✓$ WHY the ratio is $tan θ$ : dividing the two component equations, the $20$ cancels: $\frac{20 sin 60°}{20 cos 60°} = tan 60°$ . The launch magnitude drops out — only the slope of the velocity survives, which is exactly what $tan$ measures. This split is the foundation of Projectile Motion.

Level 5 — Mastery

Recall Solution

WHAT goes wrong: $A_{y} / A_{x} = - 7/0$ is undefined — the calculator refuses. But the vector is perfectly real: it points straight down. WHY the arrow is straight down: $A_{x} = 0$ means no horizontal shadow, and $A_{y} = - 7 < 0$ means the whole arrow points in $- y$ . Straight down is $θ = 270°$ (or equivalently $- 90°$ ). WHY the formula breaks: as the arrow becomes vertical, its slope $A_{y} / A_{x} \to \pm \infty$ ; $tan$ shoots to infinity at $90°$ and $270°$ , so the inverse has no finite input there. This is the limiting behaviour you must handle by inspection, not by plugging in. The safe rule: if $A_{x} = 0$ : $θ = 90°$ when $A_{y} > 0$ , $θ = 270°$ (or $- 90°$ ) when $A_{y} < 0$ , and $θ$ is undefined if $A_{y} = 0$ too (the zero vector has no direction).

Recall Solution

First, the notation: the hat " $^$ " on top of any letter is the standard flag for "unit vector in that direction". So $\hat{A}$ is to $A$ what $\hat{i}$ is to the $+ x$ axis: same aim, length trimmed to 1. By the definition $\hat{A} = A / A$ , its components are $(A_{x} / A, A_{y} / A)$ . Compute its magnitude directly: $∣ \hat{A} ∣ = (\frac{A _{x}}{A})^{2} + (\frac{A _{y}}{A})^{2} = \frac{A _{x}^{2} + A _{y}^{2}}{A ^{2}} = \frac{A _{x}^{2} + A _{y}^{2}}{A} .$ But $A_{x}^{2} + A_{y}^{2}$ is by definition the magnitude $A$ . So $∣ \hat{A} ∣ = \frac{A}{A} = 1.$ WHY it's always 1: we divided every component by the very number that is the length. Squaring and re-summing gives $A^{2}$ upstairs, cancelling $A^{2}$ downstairs — the direction (the ratio $A_{x} : A_{y}$ ) is preserved, but the size is normalised to 1. Works for every quadrant because only squares appear.

Recall Solution

Original: $A_{x} = A cos θ, A_{y} = A sin θ$ . After rotating by $+ 90°$ , the new angle is $θ + 90°$ : $A_{x}^{'} = A cos (θ + 90°), A_{y}^{'} = A sin (θ + 90°) .$ Use the shift identities $cos (θ + 90°) = - sin θ$ and $sin (θ + 90°) = cos θ$ : $A_{x}^{'} = - A sin θ = - A_{y}, A_{y}^{'} = A cos θ = A_{x} .$ So $(A_{x}, A_{y}) \to (- A_{y}, A_{x})$ . WHY this makes sense geometrically: turning the arrow a quarter-turn anticlockwise sends "right" to "up" and "up" to "left". The old x-shadow becomes the new y-shadow ( $A_{y}^{'} = A_{x}$ ), and the old y-shadow flips to the new negative x ( $A_{x}^{'} = - A_{y}$ ). This swap-and-negate is the seed of the cross product's perpendicularity.

Recall check

Recall Quick self-quiz (open after attempting)

Why do we add components, not magnitudes? ::: Because arrows carry direction and can cancel; components carry sign, magnitudes don't. What correction does $tan^{- 1}$ need when $A_{x} < 0$ ? ::: Add $180°$ (the vector is in Quadrant II or III). Direction of $A = (0, - 7)$ ? ::: Straight down, $270°$ (formula fails — do it by inspection). Magnitude of any unit vector? ::: Exactly $1$ , always. What does the hat in $\hat{A}$ mean? ::: "Unit vector in the direction of $A$ ", i.e. $A /∣ A ∣$ . Rotating $(A_{x}, A_{y})$ by $+ 90°$ gives? ::: $(- A_{y}, A_{x})$ .

Connections

Vector representation — magnitude, direction, components — the parent theory this drills.
Scalars vs Vectors — why a lone number can't do this job.
Vector Addition — Triangle & Parallelogram Law — used in L4·Q1.
Resolution of Forces — every component split above is resolution.
Projectile Motion — the L4·Q3 velocity split.
Dot Product and Cross Product — the L5·Q3 rotation foreshadows it.
Unit Vectors i, j, k — the $\hat{i}, \hat{j}$ basis underlying every answer.