Worked examples — Dot product — formula, geometric meaning, work calculation

This page is a tour of every situation the dot product can throw at you. We start by listing all the case-classes in one table, then work an example for each — with a Forecast so you guess before you compute, and a Verify so you never trust a number blindly.

Everything here builds on the parent: Dot product — formula, geometric meaning, work calculation. If any symbol feels unfamiliar, peek at Vectors — components and unit vectors and Trigonometry — cosine and components first.

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The scenario matrix

Here is every distinct case-class a dot-product question can land in. Each worked example below is tagged with its cell letter.

We now hit every cell.

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Examples

C1 — Partly aligned (positive, acute angle)

1. Component form: $\vec{A}\cdot \vec{B}=(3)(2)+(1)(2)=6+2=8$. Why this step? Multiply matching components and add — fastest route to the number.
2. Magnitudes: $|\vec{A}|=\sqrt{3^2+1^2}=\sqrt{10}$, $|\vec{B}|=\sqrt{2^2+2^2}=\sqrt{8}$. Why this step? The geometric form needs both lengths to isolate $\cos \theta$.
3. $\cos \theta =\frac{8}{\sqrt{10}\sqrt{8}}=\frac{8}{\sqrt{80}}\approx 0.894$, so $\theta ={\cos }^{-1}(0.894)\approx 26.6^{\circ }$. Why this step? Rearranged geometric form turns the number into an angle.

Verify: $8>0$ ✓ positive as forecast. $26.6^{\circ }<90^{\circ }$ ✓ acute. Look at the figure — the blue shadow of $\vec{A}$ onto $\vec{B}$ lands in the same direction as $\vec{B}$, confirming the positive sign.

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C2 — Perpendicular (dot product is exactly zero)

1. $\vec{A}\cdot \vec{B}=(4)(-1)+(2)(2)=-4+4=0$. Why this step? Component form directly; a zero here is the perpendicularity test.
2. Since both vectors are non-zero and the dot product is $0$, $\cos \theta =0\Rightarrow \theta =90^{\circ }$. Why this step? Only $\cos 90^{\circ }=0$, so the geometric form forces a right angle.

Verify: Slopes: $\vec{A}$ rises $2/4=0.5$; $\vec{B}$ rises $2/(-1)=-2$. Product of slopes $=0.5\times (-2)=-1$, the classic perpendicular test ✓. In the figure the arrows meet at a clean right angle.

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C3 — Opposing (negative, obtuse angle)

1. $\vec{A}\cdot \vec{B}=(2)(-3)+(1)(1)=-6+1=-5$. Why this step? Component form; the negative sign is the whole story.
2. $|\vec{A}|=\sqrt{5}$, $|\vec{B}|=\sqrt{10}$, so $\cos \theta =\frac{-5}{\sqrt{5}\sqrt{10}}=\frac{-5}{\sqrt{50}}\approx -0.707$. Why this step? Divide by the magnitudes to isolate $\cos \theta$ (never skip the denominator!).
3. $\theta ={\cos }^{-1}(-0.707)\approx 135^{\circ }$. Why this step? A negative cosine lands in the second quadrant — obtuse, as predicted.

Verify: $-5<0$ ✓ negative. $135^{\circ }>90^{\circ }$ ✓ obtuse. In the figure the shadow of $\vec{A}$ falls in the opposite direction to $\vec{B}$ — that back-pointing shadow is what "negative" looks like.

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C4 — Parallel and anti-parallel (limiting cases)

1. $|\vec{A}|=\sqrt{3^2+4^2}=5$; $|\vec{B}|=\sqrt{6^2+8^2}=10$; $|\vec{C}|=10$. Why this step? We want to compare against the theoretical max $|\vec{A}||\vec{B}|$.
2. (i) $\vec{A}\cdot \vec{B}=(3)(6)+(4)(8)=18+32=50$. Why this step? Component form. Compare: $|\vec{A}||\vec{B}|=5\times 10=50$ — it hit the ceiling, so $\cos \theta =1$, $\theta =0^{\circ }$.
3. (ii) $\vec{A}\cdot \vec{C}=(3)(-6)+(4)(-8)=-18-32=-50$. Why this step? Same magnitude but $=-|\vec{A}||\vec{C}|$, so $\cos \theta =-1$, $\theta =180^{\circ }$.

Verify: $50=+|\vec{A}||\vec{B}|$ ✓ (max), $-50=-|\vec{A}||\vec{C}|$ ✓ (min). The dot product can never exceed $\pm |\vec{A}||\vec{B}|$ because $\cos \theta$ lives in $[-1,1]$ — these two are the boundary. See also Projection of a vector.

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C5 — Degenerate: a zero vector

1. $\vec{A}\cdot \vec{B}=(0)(5)+(0)(-2)+(0)(7)=0$. Why this step? Every product contains a zero factor, so the sum is $0$.

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C6 — 3-D vectors: find the angle

1. $\vec{A}\cdot \vec{B}=(1)(2)+(2)(0)+(2)(-1)=2+0-2=0$. Why this step? Component form works identically in 3-D — just add the extra term.
2. Since the dot product is $0$ and neither vector is zero: $\cos \theta =0\Rightarrow \theta =90^{\circ }$. Why this step? Same perpendicularity test as 2-D; dimension doesn't change the logic.

Verify: $|\vec{A}|=\sqrt{1+4+4}=3$, $|\vec{B}|=\sqrt{4+0+1}=\sqrt{5}$, both non-zero ✓, so $90^{\circ }$ is genuine. The dot product doesn't care how many dimensions — the shadow idea is the same.

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C7 — Real-world work: lift, drag, and brake in one problem

1. Rope: $W_1=|{\vec{F}}_1||\vec{d}|\cos 30^{\circ }=(20)(8)(0.8660)\approx 138.6\text{ J}$. Why this step? Only the horizontal component $20\cos 30^{\circ }$ lies along $\vec{d}$; the dot product extracts it.
2. Friction: it points opposite to motion, so $\theta =180^{\circ }$: $W_2=(6)(8)\cos 180^{\circ }=(48)(-1)=-48\text{ J}$. Why this step? Anti-parallel ⇒ $\cos 180^{\circ }=-1$; friction removes energy (negative work is physical, not an error).
3. Gravity: $\theta =90^{\circ }$ to the horizontal path, so $W_g=(50)(8)\cos 90^{\circ }=0\text{ J}$. Why this step? Perpendicular force does no work — same reason a carried bag needs no work along level ground.

Verify: Signs match the forecast: $+138.6$, $-48$, $0$ ✓. Units: N·m = J ✓. Net work $=138.6-48+0=90.6\text{ J}$ goes into kinetic energy — see Work-Energy Theorem. The gravity-does-zero-work idea also drives Circular motion (centripetal force ⟂ velocity).

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C8 — Exam twist: solve for the unknown that makes them perpendicular

1. Perpendicular condition: $\vec{A}\cdot \vec{B}=0$. Why this step? Translate "perpendicular" into algebra via the zero-dot-product test.
2. $\vec{A}\cdot \vec{B}=(3)(2)+(k)(-6)=6-6k=0$. Why this step? Component form gives the equation to solve.
3. $6-6k=0\Rightarrow k=1$. Why this step? Isolate $k$.

Verify: Plug back: $(3)(2)+(1)(-6)=6-6=0$ ✓. So $\vec{A}=(3,1)$ is indeed perpendicular to $(2,-6)$.

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C9 — Self dot product = magnitude squared

1. $\vec{A}\cdot \vec{A}=(6)(6)+(-8)(-8)=36+64=100$. Why this step? Component form with $\vec{B}=\vec{A}$.
2. $\vec{A}\cdot \vec{A}=|\vec{A}{|}^2$, so $|\vec{A}|=\sqrt{100}=10$. Why this step? Geometric form with $\theta =0$: $|\vec{A}||\vec{A}|\cos 0=|\vec{A}{|}^2$.

Verify: Direct magnitude: $\sqrt{6^2+(-8{)}^2}=\sqrt{100}=10$ ✓. The self-dot is a shortcut to length-squared — no square root until the last step.

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Recall Which cell was which?

C1 acute-positive ::: $(3,1)\cdot (2,2)=8$, $\theta \approx 26.6^{\circ }$ C2 perpendicular-zero ::: $(4,2)\cdot (-1,2)=0$, $\theta =90^{\circ }$ C3 obtuse-negative ::: $(2,1)\cdot (-3,1)=-5$, $\theta \approx 135^{\circ }$ C4 parallel / anti-parallel limits ::: $+50$ (max) and $-50$ (min) C5 zero vector ::: $0$ by degeneracy, angle undefined C7 work signs ::: rope $+138.6$ J, friction $-48$ J, gravity $0$ J

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Connections

• Dot product — formula, geometric meaning, work calculation (parent)
• Vectors — components and unit vectors
• Trigonometry — cosine and components
• Projection of a vector
• Work-Energy Theorem
• Circular motion
• Cross product — area and torque