The Scalar (Dot) Product
Two vectors walk into a room. How much do they "agree"? The dot product $\mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2$ gives a single number — a scalar — that captures how aligned two vectors are. It also connects to geometry through $\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}||\mathbf{v}|\cos\theta$, unlocking the angle between any two vectors.
Consider $\mathbf{u} = (1, 0)$ (pointing right) and $\mathbf{v} = (0, 1)$ (pointing up). Without any formula — do you think these two vectors are "aligned" or "unaligned"? What number would you use to measure their agreement? Write your reasoning below.
The scalar (dot) product takes two vectors and returns a scalar (a single number). Two equivalent definitions:
Algebraic (component) form — multiply matching components and add:
$\mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2$ (2D)
$\mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3$ (3D)
Geometric form — relates to the angle $\theta$ between them:
$\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}||\mathbf{v}|\cos\theta$
Key facts
- $\mathbf{u}\cdot\mathbf{v} = u_1v_1+u_2v_2$ (2D) and $u_1v_1+u_2v_2+u_3v_3$ (3D)
- $\mathbf{u}\cdot\mathbf{v} = |\mathbf{u}||\mathbf{v}|\cos\theta$ — geometric definition
- The dot product is commutative: $\mathbf{u}\cdot\mathbf{v}=\mathbf{v}\cdot\mathbf{u}$
Concepts
- Why $\mathbf{u}\cdot\mathbf{v}=0$ when vectors are perpendicular ($\theta=90°$)
- Why $\mathbf{u}\cdot\mathbf{u} = |\mathbf{u}|^2$
- The sign of the dot product and what it tells us about the angle
Skills
- Evaluate the dot product of two vectors in 2D or 3D
- Apply properties: commutativity, distributivity, scalar factoring
- Use $\mathbf{u}\cdot\mathbf{v}=0$ to test for perpendicularity
For vectors $\mathbf{u} = u_1\mathbf{i}+u_2\mathbf{j}$ and $\mathbf{v} = v_1\mathbf{i}+v_2\mathbf{j}$:
Example 1: $\mathbf{u}=(2,3)$ and $\mathbf{v}=(4,-1)$: $\mathbf{u}\cdot\mathbf{v} = (2)(4)+(3)(-1) = 8-3 = 5$.
Example 2 — perpendicular vectors: $\mathbf{i}=(1,0)$ and $\mathbf{j}=(0,1)$: $\mathbf{i}\cdot\mathbf{j}=(1)(0)+(0)(1)=0$. ✓ Perpendicular vectors always give dot product zero.
Example 3 — self dot product: $\mathbf{u}=(3,4)$: $\mathbf{u}\cdot\mathbf{u}=9+16=25=|\mathbf{u}|^2$. ✓
For vectors $\mathbf{u} = u_1\mathbf{i}+u_2\mathbf{j}$ and $\mathbf{v} = v_1\mathbf{i}+v_2\mathbf{j}$:
Pause — copy the dot product definition: $\vec{u}\cdot\vec{v}=u_1v_1+u_2v_2$ (2D) and $u_1v_1+u_2v_2+u_3v_3$ (3D) with a worked numerical example into your book.
Quick check: Evaluate $\mathbf{u}\cdot\mathbf{v}$ where $\mathbf{u} = 3\mathbf{i} + 2\mathbf{j}$ and $\mathbf{v} = -\mathbf{i} + 4\mathbf{j}$.
We just saw the component definition: $\vec{u}\cdot\vec{v}=u_1v_1+u_2v_2$ (2D) or $u_1v_1+u_2v_2+u_3v_3$ (3D), giving a scalar result. That raises a question: the dot product satisfies several algebraic laws — which ones are the same as for ordinary multiplication, and is there a law that fails (e.g. is there an "inverse" for the dot product)? This card answers it → commutativity $\vec{u}\cdot\vec{v}=\vec{v}\cdot\vec{u}$ and distributivity hold; but division is not defined because the output is a scalar.
The dot product satisfies several useful algebraic properties. For vectors $\mathbf{u}$, $\mathbf{v}$, $\mathbf{w}$ and scalar $\lambda$:
- Commutative: $\mathbf{u}\cdot\mathbf{v} = \mathbf{v}\cdot\mathbf{u}$
- Distributive: $\mathbf{u}\cdot(\mathbf{v}+\mathbf{w}) = \mathbf{u}\cdot\mathbf{v} + \mathbf{u}\cdot\mathbf{w}$
- Scalar factor: $(\lambda\mathbf{u})\cdot\mathbf{v} = \lambda(\mathbf{u}\cdot\mathbf{v})$
- Self dot: $\mathbf{u}\cdot\mathbf{u} = |\mathbf{u}|^2 \geq 0$
- Zero vector: $\mathbf{0}\cdot\mathbf{u} = 0$ for all $\mathbf{u}$
In 3D: $\mathbf{i}\cdot\mathbf{i} = \mathbf{j}\cdot\mathbf{j} = \mathbf{k}\cdot\mathbf{k} = 1$ and $\mathbf{i}\cdot\mathbf{j} = \mathbf{j}\cdot\mathbf{k} = \mathbf{i}\cdot\mathbf{k} = 0$.
The dot product satisfies several useful algebraic properties. For vectors $\mathbf{u}$, $\mathbf{v}$, $\mathbf{w}$ and scalar $\lambda$:
Pause — copy the three dot product properties: (1) commutative $\vec{u}\cdot\vec{v}=\vec{v}\cdot\vec{u}$; (2) distributive over addition; (3) $\vec{a}\cdot\vec{a}=|\vec{a}|^2$ into your book.
Did you get this? True or false: if $\mathbf{u}\cdot\mathbf{v} = 0$ and both vectors are non-zero, the vectors must be perpendicular.
Worked examples · 3 in a row, reveal as you go
Find $\mathbf{a}\cdot\mathbf{b}$ where $\mathbf{a} = 4\mathbf{i} - 3\mathbf{j}$ and $\mathbf{b} = 2\mathbf{i} + 5\mathbf{j}$.
Find $\mathbf{p}\cdot\mathbf{q}$ where $\mathbf{p} = \mathbf{i} - 2\mathbf{j} + 3\mathbf{k}$ and $\mathbf{q} = 4\mathbf{i} + \mathbf{j} - \mathbf{k}$.
Show that $\mathbf{r} = 3\mathbf{i} + 4\mathbf{j}$ and $\mathbf{s} = -4\mathbf{i} + 3\mathbf{j}$ are perpendicular.
Fill the gap: For vectors $(3, -4)$ and $(4, 3)$, the dot product equals $(3)(4)+(-4)(3) =$ , so the vectors are perpendicular.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: the dot product of two non-zero vectors can equal zero.
Activities · practice with the ideas
Evaluate $\mathbf{a}\cdot\mathbf{b}$ where $\mathbf{a} = 5\mathbf{i} + 3\mathbf{j}$ and $\mathbf{b} = 2\mathbf{i} - 4\mathbf{j}$.
Evaluate $\mathbf{p}\cdot\mathbf{q}$ where $\mathbf{p} = -\mathbf{i}+2\mathbf{j}-2\mathbf{k}$ and $\mathbf{q} = 3\mathbf{i}+\mathbf{j}+2\mathbf{k}$.
Show that $\mathbf{u} = 2\mathbf{i} + 3\mathbf{j}$ and $\mathbf{v} = 6\mathbf{i} - 4\mathbf{j}$ are perpendicular.
Given $\mathbf{r} = (2, k)$ and $\mathbf{s} = (3, -6)$ are perpendicular, find $k$.
Verify that $\mathbf{u}\cdot\mathbf{u} = |\mathbf{u}|^2$ for $\mathbf{u} = 2\mathbf{i} - 3\mathbf{j} + \mathbf{k}$.
Odd one out: Three of these dot product evaluations are correct. Which one is NOT?
Earlier you guessed the dot product of $\mathbf{u}=(1,0)$ and $\mathbf{v}=(0,1)$.
The answer is $(1)(0)+(0)(1) = 0$. Perpendicular vectors always give a dot product of zero, because $\cos 90° = 0$. And for $\mathbf{u}=(1,0)$ dotted with itself: $(1)(1)+(0)(0) = 1 = |\mathbf{u}|^2$. The dot product measures alignment — maximum when parallel, zero when perpendicular.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Find $\mathbf{a}\cdot\mathbf{b}$ where $\mathbf{a} = -3\mathbf{i}+2\mathbf{j}$ and $\mathbf{b} = \mathbf{i}+5\mathbf{j}$. (2 marks)
Q2. Find all values of $k$ such that $\mathbf{u} = k\mathbf{i} + 3\mathbf{j}$ and $\mathbf{v} = 4\mathbf{i} - 2\mathbf{j}$ are perpendicular. (3 marks)
Q3. Using properties of the dot product, expand and simplify $|\mathbf{a}+\mathbf{b}|^2$. (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. $(5)(2)+(3)(-4)=10-12=-2$ · 2. $(-1)(3)+(2)(1)+(-2)(2)=-3+2-4=-5$ · 3. $(2)(6)+(3)(-4)=12-12=0$ ✓ · 4. $6k-6=0 \Rightarrow k=1$ (wait: $(2)(3)+(k)(-6)=0 \Rightarrow 6-6k=0 \Rightarrow k=1$) · 5. $\mathbf{u}\cdot\mathbf{u}=4+9+1=14$; $|\mathbf{u}|^2=\sqrt{14}^2=14$ ✓
Q1 (2 marks): $\mathbf{a}\cdot\mathbf{b} = (-3)(1)+(2)(5) = -3+10 = \mathbf{7}$ [2].
Q2 (3 marks): Set $\mathbf{u}\cdot\mathbf{v}=0$ [1]: $4k+3(-2)=0 \Rightarrow 4k-6=0 \Rightarrow k=\dfrac{3}{2}$ [2].
Q3 (3 marks): $|\mathbf{a}+\mathbf{b}|^2 = (\mathbf{a}+\mathbf{b})\cdot(\mathbf{a}+\mathbf{b})$ [1] $= \mathbf{a}\cdot\mathbf{a}+2\,\mathbf{a}\cdot\mathbf{b}+\mathbf{b}\cdot\mathbf{b}$ [1] $= |\mathbf{a}|^2+2\,\mathbf{a}\cdot\mathbf{b}+|\mathbf{b}|^2$ [1].
Five timed questions on the scalar product. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering dot product questions. Lighter alternative to the boss.
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