Worked examples — Vector representation — magnitude, direction, components

Before we start, one piece of notation we lean on everywhere below. When we write $\vec{A}=A_x\hat{i}+A_y\hat{j}$, the little hats $\hat{i}$ and $\hat{j}$ are unit vectors: tiny arrows of length exactly $1$, one pointing along $+x$ ($\hat{i}$) and one along $+y$ ($\hat{j}$). The hat "​$\hat{}$​" is the universal symbol for "this arrow has length 1" — pure direction, no size. So "$6\hat{i}+8\hat{j}$" literally means "go $6$ steps in the $+x$ direction, then $8$ steps in the $+y$ direction." (More on this basis in Unit Vectors i, j, k.)

With that: a vector is an arrow carrying a size (its magnitude $A\ge 0$) and a heading (its direction angle $\theta$, measured anticlockwise from the $+x$ axis). Its components $A_x,A_y$ are the shadows it casts on the horizontal and vertical axes. The two recipes we drill:

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The scenario matrix

Every problem this topic can throw lives in one of these cells. We will hit all of them.

#	Cell (scenario class)	What makes it tricky	Covered by
1	Quadrant I ($A_x>0,A_y>0$)	the easy baseline	Ex A
2	Quadrant II ($A_x<0,A_y>0$)	calculator lies, add $180°$	Ex B
3	Quadrant III ($A_x<0,A_y<0$)	both negative, still add $180°$	Ex C
4	Quadrant IV ($A_x>0,A_y<0$)	negative or $+360°$	Ex D
5	Degenerate: purely axial ($A_x=0$ or $A_y=0$)	${\tan }^{-1}$ blows up / ratio is $0$	Ex E
6	Zero vector ($A_x=A_y=0$)	direction is undefined	Ex E
7	Limiting angle ($\theta \to 90°$)	component behaviour at the boundary	Ex F
8	Real-world word problem (add two vectors)	translate words $\to$ components	Ex G
9	Exam twist: angle from the wrong axis	measured from $+y$, not $+x$	Ex H
10	Exam twist: unit vector + reconstruction	strip size, then rebuild	Ex I

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Ex A — Quadrant I (the baseline) cell 1

In the figure the orange arrow is $\vec{A}$; its plum shadow on the x-axis is $A_x=6$ (adjacent to $\theta$) and its teal dashed shadow up to the tip is $A_y=8$ (opposite $\theta$). The little ink arc at the origin is the angle $\theta$ we're solving for.

1. Magnitude. $A=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10$. Why this step? The plum and teal shadows are the two legs of the right triangle in the figure; Pythagoras turns those legs into the hypotenuse, which is the orange arrow's length.
2. Raw angle. $\theta ={\tan }^{-1}(8/6)={\tan }^{-1}(1.333)=53.13°$. Why this step? $\tan \theta =\frac{\text{opposite}}{\text{adjacent}}=\frac{A_y}{A_x}$ measures how steeply the orange arrow climbs (teal over plum); ${\tan }^{-1}$ asks "which angle has this steepness?"
3. Quadrant check. $A_x>0,A_y>0\Rightarrow$ Quadrant I $\Rightarrow$ no fix. Final $\theta =53.13°$. Why this step? The calculator returns angles in $(-90°,90°)$, which is exactly right for QI, so the ink arc in the figure needs no correction.

Verify: rebuild the components — $10\cos 53.13°=6.00$, $10\sin 53.13°=8.00$. ✓ Back to the start. Units: components share the vector's units (here pure numbers).

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Ex B — Quadrant II (calculator lies) cell 2

In the figure the orange arrow is the true vector $\vec{A}$ pointing up-left; the faint plum arrow pointing down-right is the calculator's guess ($-67.4°$). Notice both arrows lie on one straight line through the origin — that's why they share the same $\tan$ ratio, and exactly why the calculator confuses them. The teal arc is the corrected angle.

1. Magnitude. $A=\sqrt{(-5{)}^2+12^2}=\sqrt{25+144}=\sqrt{169}=13$. Why this step? Pythagoras uses the squares of the legs, so the sign of $A_x$ vanishes — magnitude is always $\ge 0$.
2. Raw angle. ${\theta }_{\text{raw}}={\tan }^{-1}\left(\frac{12}{-5}\right)={\tan }^{-1}(-2.4)=-67.38°$. Why this step? ${\tan }^{-1}$ only knows the ratio $-2.4$, so it returns the plum down-right arrow in the figure — the wrong direction, but on the same line.
3. Quadrant fix. $A_x<0\Rightarrow$ add $180°$: $\theta =-67.38°+180°=112.62°$. Why this step? $\tan$ repeats every $180°$; rotating the plum guess by half a turn lands exactly on the orange arrow, without changing the ratio.

Verify: $13\cos 112.62°=-5.00$, $13\sin 112.62°=12.00$. ✓ Signs match the original $(-5,+12)$.

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Ex C — Quadrant III (both negative) cell 3

In the figure the orange arrow points down-left (the true $\vec{A}$); the faint plum arrow up-right is again the calculator's guess ($36.9°$). Same story as Ex B — two arrows, one line, one shared ratio. The teal arc sweeps round to the corrected angle.

1. Magnitude. $A=\sqrt{(-8{)}^2+(-6{)}^2}=\sqrt{64+36}=\sqrt{100}=10$. Why this step? Squaring each leg erases both minus signs, so Pythagoras gives the same length as the plum mirror arrow in QI — magnitude never carries the direction's sign.
2. Raw angle. ${\theta }_{\text{raw}}={\tan }^{-1}\left(\frac{-6}{-8}\right)={\tan }^{-1}(0.75)=36.87°$. Why this step? The two minus signs cancel inside the ratio, so the calculator returns the plum up-right arrow — it cannot see that both legs actually point backward.
3. Quadrant fix. $A_x<0\Rightarrow$ add $180°$: $\theta =36.87°+180°=216.87°$. Why this step? Same rule as Ex B: whenever $A_x<0$ (QII or QIII) the calculator is a half-turn off — rotating the plum arrow $180°$ lands on the orange one.

Verify: $10\cos 216.87°=-8.00$, $10\sin 216.87°=-6.00$. ✓ Both negative, as required.

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Ex D — Quadrant IV (negative or +360°) cell 4

In the figure the orange arrow dips below the x-axis into QIV; the teal arc curls clockwise from the $+x$ axis, showing the angle as a negative sweep of $-45°$. The plum dashed shadows mark $A_x=5$ (rightward) and $A_y=-5$ (downward).

1. Magnitude. $A=\sqrt{5^2+(-5{)}^2}=\sqrt{25+25}=\sqrt{50}=5\sqrt{2}\approx 7.07$. Why this step? Pythagoras squares each leg, so the negative $A_y$ (the downward plum shadow) contributes a positive $25$ just like the positive $A_x$ — the length is blind to the downward direction.
2. Raw angle. $\theta ={\tan }^{-1}\left(\frac{-5}{5}\right)={\tan }^{-1}(-1)=-45°$. Why this step? $A_x>0$ means the arrow really is on the right half, exactly where ${\tan }^{-1}$ lives — the teal clockwise arc needs no fix.
3. Two valid names. $\theta =-45°$ or equivalently $-45°+360°=315°$. Why this step? Angles are cyclic; $-45°$ and $315°$ point the identical way. Use whichever your problem's convention prefers.

Verify: $5\sqrt{2}\cos (-45°)=5\sqrt{2}\cdot \frac{1}{\sqrt{2}}=5.00$, $5\sqrt{2}\sin (-45°)=-5.00$. ✓

Recall The one-line quadrant rule

When does the calculator's ${\tan }^{-1}$ need $+180°$? Whenever $A_x<0$ (Quadrants II and III). If $A_x>0$ it's already correct.

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Ex E — Degenerate & zero vectors cells 5, 6

In the figure the teal arrow points straight up ($\vec{P}$), the orange arrow points straight left ($\vec{Q}$), and the plum dot at the origin is $\vec{Z}$ — an arrow with no length and therefore no tip to aim.

1. (i) purely $+y$. $P=\sqrt{0^2+7^2}=7$. The teal arrow points straight up, so $\theta =90°$. Why this step? $A_x=0$ makes $A_y/A_x$ a division by zero — the calculator errors. But the picture is unambiguous: straight up is $90°$.
2. (ii) purely $-x$. $Q=\sqrt{(-4{)}^2+0^2}=4$. The orange arrow points straight left, so $\theta =180°$. Why this step? $A_y=0$ gives ratio $0$, whose ${\tan }^{-1}$ is $0°$ — but $A_x<0$ means we add $180°$: $0°+180°=180°$. The quadrant rule still saves us.
3. (iii) zero vector. $Z=\sqrt{0^2+0^2}=0$. Direction is undefined. Why this step? The plum dot has length $0$, so it has no tip — it points nowhere. Never quote an angle for the zero vector.

Verify: $7\cos 90°=0,7\sin 90°=7$ ✓; $4\cos 180°=-4,4\sin 180°=0$ ✓; the zero vector has $0=0$ and no angle. ✓

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Ex F — Limiting behaviour as $\theta \to 90°$ cell 7

In the figure three arrows of equal length $10$ fan upward — plum at $0°$ (flat along $+x$), teal at $45°$ (the diagonal), orange at $89°$ (almost vertical). The dotted vertical drop-lines are each arrow's horizontal shadow $A_x$: watch how they get shorter and shorter as the fan sweeps up, exactly the trend the forecast predicts.

1. $\theta =0°$: $A_x=10\cos 0°=10.00$, $A_y=10\sin 0°=0.00$. Why this step? $\cos 0°=1$ and $\sin 0°=0$: the plum arrow lies flat along $+x$, so its shadow is the whole length horizontally and nothing vertically.
2. $\theta =45°$: $A_x=10\cos 45°=7.07$, $A_y=10\sin 45°=7.07$. Why this step? $\cos 45°=\sin 45°=\frac{1}{\sqrt{2}}\approx 0.707$: at the teal diagonal both shadows are equal, splitting the length evenly.
3. $\theta =89°$: $A_x=10\cos 89°=0.175$, $A_y=10\sin 89°=9.998$. Why this step? $\cos 89°\approx 0.017$ has almost died while $\sin 89°\approx 0.9998$ is nearly $1$: the orange arrow's horizontal shadow is the tiny stub near the y-axis in the figure, and the vertical part is nearly full.
4. $\theta =90°$: $A_x=0.00$, $A_y=10.00$. Why this step? $\cos 90°=0$ and $\sin 90°=1$ — the limit the fan is heading toward. The horizontal shadow has been squeezed to nothing; the whole length now lives in $y$.

Verify: at every angle $A_x^2+A_y^2=100$ (magnitude conserved under rotation): check $\theta =45°$: $7.07^2+7.07^2\approx 100$. ✓

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Ex G — Word problem: two displacements cell 8

In the figure the plum arrow is leg 1 ($30$ m at $40°$), the teal arrow is leg 2 ($20$ m straight up) drawn tip-to-tail from the plum tip, and the orange arrow is the resultant $\vec{R}$ from the start to the final tip. The orange arrow is visibly steeper than the plum one — the northward leg tilted the total upward, just as forecast.

1. Leg 1 components. $x_1=30\cos 40°=22.98$, $y_1=30\sin 40°=19.28$ (east, north). Why this step? East is our $+x$, north is $+y$; the angle is measured from east ($+x$), so east gets cosine.
2. Leg 2 components. Due north $\Rightarrow x_2=0,y_2=20$. Why this step? "Due north" is straight up (the teal arrow): no eastward part.
3. Add componentwise. $R_x=22.98+0=22.98$, $R_y=19.28+20=39.28$. Why this step? Components along the same axis add like plain numbers — that's the whole reason we resolved (see Vector Addition — Triangle & Parallelogram Law).
4. Rebuild magnitude. $R=\sqrt{22.98^2+39.28^2}=\sqrt{528.3+1543.0}=\sqrt{2071.3}=45.51\text{ m}$. Why this step? The summed components $R_x,R_y$ are the legs of the resultant's right triangle (orange in the figure); Pythagoras turns those legs into the single straight-line distance the hiker ends up from the start.
5. Rebuild angle. $\theta ={\tan }^{-1}(39.28/22.98)={\tan }^{-1}(1.709)=59.66°$. Both components positive $\Rightarrow$ QI, no fix. Why this step? ${\tan }^{-1}$ of the resultant's opposite-over-adjacent ratio recovers its heading; both components are positive so the orange arrow is in QI and no $180°$ correction is needed. (A note on rounding: if you keep more decimals in $x_1,y_1$ the ratio is $1.709$ and the angle is $59.66°$; rounding the components hard to $22.98$ and $39.28$ before dividing can nudge the answer to about $59.8°$ — always carry full precision until the last line.)

Verify: rebuild — $45.51\cos 59.66°=22.98$, $45.51\sin 59.66°=39.28$ ✓. Sanity: $45.51\text{ m}<30+20=50\text{ m}$ (arrows didn't fully align, so total is less than the naive sum) ✓, and $59.66°>40°$ as forecast ✓. Units: metres throughout.

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Ex H — Exam twist: angle from the wrong axis cell 9

In the figure the orange arrow is the force; the teal arc hugs the $+y$ axis, showing the $25°$ is measured from vertical, not from the x-axis. The plum dashed shadows are $F_x$ (short, horizontal — opposite the $25°$) and $F_y$ (tall, vertical — adjacent to the $25°$). The picture makes it obvious which leg is adjacent.

1. Identify the adjacent side. The teal arc's $25°$ sits between the force and the $+y$ axis, so the $y$-side is adjacent to $25°$. Why this step? Cosine = adjacent/hypotenuse by definition; the adjacent leg is the one the angle opens from — here the tall vertical plum shadow.
2. $y$ gets cosine. $F_y=50\cos 25°=45.32\text{ N}$. Why this step? $F_y$ is the leg adjacent to the $25°$ angle, and adjacent/hypotenuse is exactly cosine — so this projection uses $\cos$, not $\sin$.
3. $x$ gets sine. $F_x=50\sin 25°=21.13\text{ N}$. Why this step? The short horizontal plum shadow $F_x$ is opposite the $25°$, and opposite/hypotenuse is sine.

Alternative (convert the angle): from $+x$ the angle is $90°-25°=65°$, giving $F_x=50\cos 65°=21.13$, $F_y=50\sin 65°=45.32$ — identical. ✓

Verify: $\sqrt{21.13^2+45.32^2}=\sqrt{446.5+2054}=\sqrt{2500}=50.00\text{ N}$. ✓ Recovers the magnitude. See Resolution of Forces for more of these.

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Ex I — Exam twist: unit vector then reconstruct cell 10

In the figure the long orange arrow is $\vec{p}$ (length $15$) pointing down-right into QIV; the short teal arrow riding along the very same line is $\hat{p}$ — identical direction, shrunk to length $1$. Same heading, different size: that is the whole idea of a unit vector.

1. Magnitude. $p=\sqrt{9^2+(-12{)}^2}=\sqrt{81+144}=\sqrt{225}=15$. Why this step? We need the orange arrow's length before we can shrink it to $1$; Pythagoras on the two legs $9$ and $-12$ gives that length (the negative squares away).
2. Unit vector. $\hat{p}=\frac{\vec{p}}{p}=\frac{9}{15}\hat{i}+\frac{-12}{15}\hat{j}=0.6\hat{i}-0.8\hat{j}$. Why this step? Dividing every component by the same $p$ shrinks the length to $1$ (orange → teal in the figure) but leaves the ratio $A_y/A_x$ — the direction — untouched (see Unit Vectors i, j, k).
3. Check it's length 1. $\sqrt{0.6^2+(-0.8{)}^2}=\sqrt{0.36+0.64}=\sqrt{1}=1$. ✓ Why this step? A "unit" vector must have magnitude exactly $1$ by definition; running Pythagoras on its components confirms the teal arrow really is length $1$.
4. Reconstruct. $p\hat{p}=15(0.6\hat{i}-0.8\hat{j})=9\hat{i}-12\hat{j}=\vec{p}$. Why this step? magnitude $\times$ direction $=$ the whole vector — stretching the teal unit arrow back up by $15$ lands exactly on the orange $\vec{p}$. The two operations are inverses.

Verify: direction check — angle of $\vec{p}$ is ${\tan }^{-1}(-12/9)=-53.13°$ (QIV, $A_x>0$ so no fix); angle of $\hat{p}$ is ${\tan }^{-1}(-0.8/0.6)=-53.13°$ — same heading. ✓ Units of $\hat{p}$: none (pure direction); reconstructed $\vec{p}$ back in kg·m/s.

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Active recall

Recall Reveal

In Ex B, why add $180°$ but in Ex D leave the angle negative? ::: Ex B has $A_x<0$ (QII) so the calculator is a half-turn off; Ex D has $A_x>0$ (QIV) so ${\tan }^{-1}$ is already correct. In Ex E(iii), why no angle? ::: The zero vector has length $0$ and thus no direction — undefined. In Ex G, why is the resultant $45.5\text{ m}$ less than $50\text{ m}$? ::: The two legs weren't parallel, so they partly point in different directions; only parallel arrows add to the full sum. In Ex H, which trig function does the axis you measure FROM get? ::: Cosine (that axis is the adjacent side).

Connections

• Vector representation — magnitude, direction, components — the parent recipes drilled here.
• Scalars vs Vectors — why a single number can't do this job.
• Vector Addition — Triangle & Parallelogram Law — the engine behind Ex G.
• Resolution of Forces — Ex H in its natural mechanics habitat.
• Dot Product and Cross Product — the next tools built on components.
• Projectile Motion — splitting velocity like Ex F's limiting case.
• Unit Vectors i, j, k — the basis used in Ex I.