Dot product — formula, cosine formula, Cauchy-Schwarz inequality proof

1. The algebraic definition

WHAT is this really? Multiply matching components, add them up.

WHY this and not something else? Because it gives us the length for free. Set $\mathbf{b}=\mathbf{a}$: $\mathbf{a}\cdot\mathbf{a} = a_1^2 + a_2^2 + \cdots + a_n^2 = \|\mathbf{a}\|^2$ This is just the Pythagorean theorem! So the dot product contains the notion of distance. That is the design goal it was built to satisfy.

Basic properties (all provable from the sum)

• Commutative: $\mathbf{a}\cdot\mathbf{b}=\mathbf{b}\cdot\mathbf{a}$ (since $a_ib_i=b_ia_i$).
• Distributive: $\mathbf{a}\cdot(\mathbf{b}+\mathbf{c})=\mathbf{a}\cdot\mathbf{b}+\mathbf{a}\cdot\mathbf{c}$.
• Scalar pull-out: $(\lambda\mathbf{a})\cdot\mathbf{b}=\lambda(\mathbf{a}\cdot\mathbf{b})$.
• Positive: $\mathbf{a}\cdot\mathbf{a}\ge 0$, and $=0$ only if $\mathbf{a}=\mathbf{0}$.

2. The cosine (geometric) formula — derived, not memorized

HOW do we get this? Use the Law of Cosines on the triangle formed by $\mathbf{a}$, $\mathbf{b}$ and the side $\mathbf{a}-\mathbf{b}$.

Law of cosines on that triangle: $\|\mathbf{a}-\mathbf{b}\|^2 = \|\mathbf{a}\|^2 + \|\mathbf{b}\|^2 - 2\|\mathbf{a}\|\|\mathbf{b}\|\cos\theta$ Why this step? The vector $\mathbf{a}-\mathbf{b}$ is the side opposite to angle $\theta$.

Now expand the left side algebraically using the dot product: $\|\mathbf{a}-\mathbf{b}\|^2 = (\mathbf{a}-\mathbf{b})\cdot(\mathbf{a}-\mathbf{b}) = \mathbf{a}\cdot\mathbf{a} - 2\,\mathbf{a}\cdot\mathbf{b} + \mathbf{b}\cdot\mathbf{b}$ $= \|\mathbf{a}\|^2 - 2\,\mathbf{a}\cdot\mathbf{b} + \|\mathbf{b}\|^2$ Why this step? We use distributivity and $\mathbf{a}\cdot\mathbf{a}=\|\mathbf{a}\|^2$.

Set the two expressions for $\|\mathbf{a}-\mathbf{b}\|^2$ equal. The $\|\mathbf{a}\|^2$ and $\|\mathbf{b}\|^2$ cancel: $-2\,\mathbf{a}\cdot\mathbf{b} = -2\|\mathbf{a}\|\|\mathbf{b}\|\cos\theta \quad\Longrightarrow\quad \boxed{\mathbf{a}\cdot\mathbf{b}=\|\mathbf{a}\|\|\mathbf{b}\|\cos\theta}$

Angle extraction: $\cos\theta = \frac{\mathbf{a}\cdot\mathbf{b}}{\|\mathbf{a}\|\,\|\mathbf{b}\|}, \qquad \mathbf{a}\perp\mathbf{b}\iff \mathbf{a}\cdot\mathbf{b}=0$

3. Cauchy–Schwarz inequality

WHY is it "obviously" true geometrically? Because $\mathbf{a}\cdot\mathbf{b}=\|\mathbf{a}\|\|\mathbf{b}\|\cos\theta$ and $|\cos\theta|\le 1$. Done!

But that geometric "proof" secretly assumes we already have an angle $\theta$ — which is fine in $\mathbb{R}^2,\mathbb{R}^3$ but not obvious in abstract spaces. So we prove it purely algebraically, which then justifies defining the angle.

Derivation from scratch (the discriminant trick)

Consider the real function for any real $t$: $f(t) = \|\mathbf{a} - t\mathbf{b}\|^2 \ge 0$ Why start here? A squared length is never negative — that single fact is the whole engine.

Expand it: $f(t) = (\mathbf{a}-t\mathbf{b})\cdot(\mathbf{a}-t\mathbf{b}) = \|\mathbf{a}\|^2 - 2t\,(\mathbf{a}\cdot\mathbf{b}) + t^2\|\mathbf{b}\|^2$

This is a quadratic in $t$ of the form $At^2+Bt+C$ with $A=\|\mathbf{b}\|^2,\quad B=-2(\mathbf{a}\cdot\mathbf{b}),\quad C=\|\mathbf{a}\|^2.$

Since $f(t)\ge 0$ for all $t$, the parabola never crosses below the axis, so its discriminant must satisfy $B^2-4AC\le 0$: $4(\mathbf{a}\cdot\mathbf{b})^2 - 4\|\mathbf{b}\|^2\|\mathbf{a}\|^2 \le 0$ Why this step? A non-negative upward parabola has at most one real root → discriminant $\le 0$.

Divide by 4 and rearrange: $(\mathbf{a}\cdot\mathbf{b})^2 \le \|\mathbf{a}\|^2\|\mathbf{b}\|^2$ Take square roots (both sides $\ge0$): $|\mathbf{a}\cdot\mathbf{b}| \le \|\mathbf{a}\|\,\|\mathbf{b}\| \qquad\blacksquare$

Equality case: $B^2-4AC=0$ means $f(t)=0$ for some $t=t_0$, i.e. $\|\mathbf{a}-t_0\mathbf{b}\|=0\Rightarrow \mathbf{a}=t_0\mathbf{b}$. So equality $\iff$ vectors are parallel. ✔

4. Worked examples

5. Common mistakes (steel-manned)

6. Flashcards

Concept Map

Hinglish (regional understanding)