Worked examples — Vector addition — triangle law, parallelogram law

Before anything, let me re-earn three words so nobody is lost:

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The scenario matrix

Two arrows can meet in a surprisingly small number of genuinely different ways. Here is every cell we must hit, and which worked example covers it.

#	Case class	What is special about it	Covered by
C1	Acute angle ($0^{\circ }<\theta <90^{\circ }$)	cross term $2AB\cos \theta$ is positive	Ex 2
C2	Right angle ($\theta =90^{\circ }$)	cross term is zero → pure Pythagoras	Ex 1
C3	Obtuse angle ($90^{\circ }<\theta <180^{\circ }$)	cross term is negative → resultant shrinks	Ex 3
C4	Degenerate: same direction ($\theta =0^{\circ }$)	arrows collinear, resultant is maximum $A+B$	Ex 4
C5	Degenerate: opposite ($\theta =180^{\circ }$)	arrows anti-parallel, resultant is minimum $|A-B|$	Ex 4
C6	Zero input (one arrow has length $0$)	adding "nothing" — identity check	Ex 4
C7	Inverse problem (given $R$, find $\theta$)	solve the formula backwards	Ex 5
C8	Direction lands in a tricky quadrant	$R_x<0$, so naive $\arctan$ misleads	Ex 6
C9	Real-world word problem	boat crossing a river (velocity + velocity)	Ex 7
C10	Exam twist	"resultant is perpendicular to $\vec{A}$" condition	Ex 8

We now hit every cell.

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Example 1 — Right angle (cell C2)

Forecast: guess the resultant before reading on. Is it bigger or smaller than $6+8=14$? (It must be smaller — they don't fully help each other.)

1. Kill the cross term. Why this step? At $\theta =90^{\circ }$, $\cos 90^{\circ }=0$, so $2AB\cos \theta =0$. The formula collapses to $R=\sqrt{A^2+B^2}$ — plain Pythagoras, because the two arrows form a right angle exactly like the two short sides of a right triangle. $R=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10\text{N}.$
2. Direction. Why this step? $\tan \alpha =\frac{B\sin \theta }{A+B\cos \theta }$; with $\sin 90^{\circ }=1$, $\cos 90^{\circ }=0$ this is $\frac{8}{6}$. $\tan \alpha =\frac{8}{6}=1.333\Rightarrow \alpha =\arctan (1.333)\approx 53.13^{\circ }.$

Verify: $10$ lies inside the allowed band $[|6-8|,6+8]=[2,14]$. ✔ Units: N + N → N. ✔ And $53.13^{\circ }$ tilts toward north — sensible, because the northward pull ($8$) is the stronger of the two.

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Example 2 — Acute angle (cell C1)

Forecast: the arrows are equal and lean toward each other. Where should the answer point? (Right down the middle — symmetry demands it.)

1. Magnitude. Why this step? $\cos 60^{\circ }=0.5$, so the cross term $2AB\cos \theta =2(5)(5)(0.5)=25$ is positive and adds length. $R=\sqrt{25+25+25}=\sqrt{75}=5\sqrt{3}\approx 8.66\text{N}.$
2. Direction. Why this step? Same formula; $\sin 60^{\circ }\approx 0.866$. $\tan \alpha =\frac{5(0.866)}{5+5(0.5)}=\frac{4.330}{7.5}=0.5774\Rightarrow \alpha =30^{\circ }.$

Verify: $30^{\circ }$ is exactly half of $60^{\circ }$ — the resultant bisects the angle, which is what equal vectors must do. ✔ $8.66$ sits inside $[0,10]$. ✔

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Example 3 — Obtuse angle (cell C3)

Forecast: same magnitudes as Example 2 but wider apart. Bigger or smaller resultant? (Smaller — they now fight each other more.)

1. Magnitude. Why this step? $\cos 120^{\circ }=-0.5$, so the cross term is now negative: $2(5)(5)(-0.5)=-25$. This is the whole point of obtuse angles — the arrows partly cancel. $R=\sqrt{25+25-25}=\sqrt{25}=5\text{N}.$
2. Direction. $\sin 120^{\circ }\approx 0.866$, $\cos 120^{\circ }=-0.5$: $\tan \alpha =\frac{5(0.866)}{5+5(-0.5)}=\frac{4.330}{2.5}=1.732\Rightarrow \alpha =60^{\circ }.$

Verify: Compare with Ex 2: same lengths, wider angle → smaller resultant ($5<8.66$). ✔ The bisector logic still holds: $\alpha =60^{\circ }=\frac{1}{2}(120^{\circ })$. ✔

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Example 4 — The degenerate trio (cells C4, C5, C6)

Forecast: collinear arrows behave like ordinary numbers. Predict all three before the algebra.

1. (a) Same direction, $\theta =0^{\circ }$ (C4). Why? $\cos 0^{\circ }=1$, so $R^2=A^2+B^2+2AB=(A+B{)}^2$. $R=\sqrt{49+9+2(7)(3)}=\sqrt{100}=10=A+B.$ This is the maximum any resultant can reach.
2. (b) Opposite, $\theta =180^{\circ }$ (C5). Why? $\cos 180^{\circ }=-1$, so $R^2=A^2+B^2-2AB=(A-B{)}^2$. $R=\sqrt{49+9-42}=\sqrt{16}=4=|A-B|=|7-3|.$ This is the minimum. (We write $|A-B|$ so the length stays positive even if $B>A$.)
3. (c) Adding zero, $B=0$ (C6). Why? An arrow of length $0$ is "stay put". The formula gives $R=\sqrt{A^2+0+0}=A$, and $\tan \alpha =\frac{0}{A}=0\Rightarrow \alpha =0^{\circ }$. $R=7,\alpha =0^{\circ }.$ The zero vector is the identity: $\vec{A}+\vec{0}=\vec{A}$.

Verify: every resultant of these two arrows must live in $[|7-3|,7+3]=[4,10]$. Cases (a) and (b) are the two ends of that band, and (c) gives $7\in [4,10]$. ✔

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Example 5 — Inverse problem: given the resultant, find the angle (cell C7)

Forecast: $6$, $8$, $10$... does that trio ring a Pythagorean bell? If so, what angle does Pythagoras demand?

1. Write the formula, treat $\theta$ as the unknown. Why this step? We know $R$ and both magnitudes; only $\cos \theta$ is missing, so isolate it. $R^2=A^2+B^2+2AB\cos \theta .$
2. Substitute. $100=36+64+2(6)(8)\cos \theta =100+96\cos \theta .$
3. Solve. Why this step? Subtract $100$ from both sides: $0=96\cos \theta \Rightarrow \cos \theta =0\Rightarrow \theta =90^{\circ }.$

Verify: Plug back: $\sqrt{36+64+0}=\sqrt{100}=10$. ✔ It matches the given resultant, and $6,8,10$ is the classic right triangle — consistent with $\theta =90^{\circ }$. ✔

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Example 6 — Direction in a tricky quadrant (cell C8)

Forecast: $B$ is much longer and points mostly backwards (up and to the left). So the resultant should point into the upper-left — past $90^{\circ }$. Watch how the naive arctan tries to hide this.

1. Break into components. Why this step? Because the resultant clearly won't sit near $\vec{A}$; going through $R_x,R_y$ lets us read the quadrant honestly. With $\cos 150^{\circ }=-0.8660$, $\sin 150^{\circ }=0.5$: $R_x=A+B\cos \theta =2+6(-0.8660)=2-5.196=-3.196,$ $R_y=B\sin \theta =6(0.5)=3.$
2. Magnitude. Why this step? Pythagoras on those components. $R=\sqrt{(-3.196{)}^2+3^2}=\sqrt{10.215+9}=\sqrt{19.215}\approx \mathrm{4.384.}$ (Cross-check with the direct formula: $R=\sqrt{4+36+2(2)(6)(-0.8660)}=\sqrt{40-20.78}=\sqrt{19.22}$ ✔.)
3. The direction trap. Why this step? $\tan \alpha =\frac{R_y}{R_x}=\frac{3}{-3.196}=-0.9387$. A calculator's $\arctan (-0.9387)$ returns $-43.2^{\circ }$ — which would point down-and-right, the wrong quadrant! Because $R_x<0$ and $R_y>0$, the arrow is really in the second quadrant (up-left). The fix: add $180^{\circ }$. $\alpha =-43.2^{\circ }+180^{\circ }\approx 136.8^{\circ }.$

Verify: $136.8^{\circ }$ is past $90^{\circ }$, matching our forecast of "upper-left". And $R\approx 4.38$ sits inside $[|2-6|,2+6]=[4,8]$. ✔ The lesson: whenever $A+B\cos \theta <0$, the resultant leans backward and you must add $180^{\circ }$ to the bare arctangent.

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Example 7 — Real-world word problem (cell C9)

Forecast: the boat aims straight across but the water carries it sideways. It cannot go faster than $3+4=7$. Guess where the true speed lands.

1. Identify the two vectors. Why this step? Velocities add exactly like displacements (they are "like quantities"). Here $A=3$ (across) and $B=4$ (downstream), meeting at $90^{\circ }$.
2. Magnitude. Why this step? Right angle → Pythagoras. $R=\sqrt{3^2+4^2+2(3)(4)\cos 90^{\circ }}=\sqrt{9+16}=5\text{m/s}.$
3. Direction. Why this step? Angle away from the "straight across" ($3\text{m/s}$) vector: $\tan \alpha =\frac{4\sin 90^{\circ }}{3+4\cos 90^{\circ }}=\frac{4}{3}=1.333\Rightarrow \alpha \approx 53.13^{\circ }\text{downstream of straight-across}.$

Verify: $5\text{m/s}$ is between $|3-4|=1$ and $3+4=7$. ✔ Units: m/s throughout. ✔ This is the same $3$-$4$-$5$ geometry as Example 1 — a reassuring echo (see Relative velocity for the full boat-and-river treatment).

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Example 8 — Exam twist: resultant perpendicular to $\vec{A}$ (cell C10)

Forecast: "$\vec{R}\perp \vec{A}$" is a disguised equation. What must the horizontal part of $\vec{R}$ be if $\vec{R}$ points straight up from $\vec{A}$? (It must vanish.)

1. Translate the geometry into algebra. Why this step? If $\vec{R}$ is perpendicular to $\vec{A}$ (our x-axis), then $\vec{R}$ has no x-component: $R_x=0$. $R_x=A+B\cos \theta =4+B\cos 120^{\circ }=4-0.5B=0.$
2. Solve for $B$. Why this step? One equation, one unknown. $4=0.5B\Rightarrow B=8.$
3. Find $R$ from the y-component. Why this step? Since $R_x=0$, the whole resultant is its y-component. $R=R_y=B\sin \theta =8\sin 120^{\circ }=8(0.8660)=6.928\approx 4\sqrt{3}.$

Verify: Check with the full magnitude formula: $R=\sqrt{4^2+8^2+2(4)(8)\cos 120^{\circ }}=\sqrt{16+64-32}=\sqrt{48}=4\sqrt{3}\approx 6.928$. ✔ And $\tan \alpha =\frac{B\sin \theta }{A+B\cos \theta }=\frac{6.928}{0}$ → undefined → $\alpha =90^{\circ }$, confirming $\vec{R}\perp \vec{A}$. ✔

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Recall Quick self-test across the whole matrix

Right-angle resultant of $6$ and $8$ ::: $10$, at $\arctan (8/6)\approx 53.13^{\circ }$ Equal vectors ($5,5$) at $60^{\circ }$: magnitude and direction ::: $5\sqrt{3}\approx 8.66$, bisecting at $30^{\circ }$ Equal vectors ($5,5$) at $120^{\circ }$: magnitude ::: $5$ Maximum and minimum resultant of $7$ and $3$ ::: $10$ (at $0^{\circ }$) and $4$ (at $180^{\circ }$) $6$ and $8$ giving resultant $10$: what is $\theta$? ::: $90^{\circ }$ (since $\cos \theta =0$) When must you add $180^{\circ }$ to the arctan direction? ::: When $R_x=A+B\cos \theta <0$ "Resultant perpendicular to $\vec{A}$" as an equation ::: $A+B\cos \theta =0$

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Connections

• Vectors — components and unit vectors — Examples 6 and 8 lean entirely on the component picture.
• Resolution of a vector into components — the $R_x,R_y$ split used throughout.
• Subtraction of vectors and the difference vector — Example 4(b) ($\theta =180^{\circ }$) is the seed of subtraction.
• Relative velocity — Example 7 is a first taste; the river problem lives here in full.
• Law of cosines — the magnitude formula is this rule in disguise; the inverse problem (Ex 5) is "solve the cosine rule for the angle".
• Scalar (dot) product — supplies the $\cos \theta$ hiding in the cross term.
• Vector addition — triangle law, parallelogram law — the parent note these examples serve.