Inner product spaces — dot product generalization: Worked examples

This page is a drill through every kind of situation an inner-product question can throw at you. We start by listing all the case classes in one table, then work an example for each cell so you never meet a scenario cold. Everything here rests on the parent note: an inner product $⟨ \cdot, \cdot ⟩$ is a rule that eats two vectors and returns one number, obeying symmetry, linearity in the first slot, and positive-definiteness ( $⟨ v, v ⟩ \geq 0$ , and $= 0$ only for the zero vector).

Reminder of the derived geometry we will keep using: $∥ v ∥ = ⟨ v, v ⟩, cos θ = \frac{⟨ u , v ⟩}{∥ u ∥ ∥ v ∥}, u ⊥ v ⟺ ⟨ u, v ⟩ = 0.$

The scenario matrix

Here is every case class this topic can put in front of you. Each later example is tagged with the cell it lands in.

#	Case class	What makes it tricky	Example
A	Standard positive angle ( $⟨ u, v ⟩ > 0$ )	cosine in $(0, 1)$ , acute angle	Ex 1
B	Negative inner product ( $⟨ u, v ⟩ < 0$ )	obtuse angle, sign of cosine	Ex 2
C	Orthogonal case ( $⟨ u, v ⟩ = 0$ )	right angle in that geometry	Ex 3
D	Degenerate / zero input ( $v = 0$ )	cosine formula divides by zero	Ex 4
E	Non-standard weights stretch axes	usual right angle stops being right	Ex 5
F	Function space (integral)	orthogonality by integrating a period	Ex 6
G	Limiting / boundary of Cauchy–Schwarz (equality)	$cos θ = \pm 1$ , parallel vectors	Ex 7
H	Real-world word problem	translate words into an inner product	Ex 8
I	Exam twist: "is this an inner product?"	one axiom quietly fails	Ex 9

We now hit every cell.

Cell A — acute angle, everything positive

Worked example Ex 1 (Cell A): angle between

$(3, 1)$ and $(1, 2)$ under the standard dot product Use $⟨ u, v ⟩ = u_{1} v_{1} + u_{2} v_{2}$ . Find the angle $θ$ .

Forecast: the arrows both point up-and-right, so guess: acute, cosine positive, somewhere near $4 5^{\circ}$ ?

Inner product. $⟨ u, v ⟩ = (3) (1) + (1) (2) = 5$ . Why this step? This one number carries all the geometry; positive already tells us the angle is acute (Cell A).
Norms. $∥ u ∥ = 3^{2} + 1^{2} = 10$ , $∥ v ∥ = 1^{2} + 2^{2} = 5$ . Why this step? $cos θ$ needs the lengths to normalise away "how long" so only "which direction" remains.
Cosine. $cos θ = \frac{5}{10 5} = \frac{5}{50} = \frac{5}{5 2} = \frac{1}{2}$ . Why this step? Divide the inner product by the product of lengths — the definition of angle.
Angle. $θ = arccos \frac{1}{2} = 4 5^{\circ}$ . Why this step? $arccos$ answers "which angle has this cosine?" — it undoes the cosine.

Verify: $\frac{1}{2} \approx 0.707 \in (- 1, 1)$ ✓ (Cauchy–Schwarz respected), and $0 < 0.707 < 1$ ✓ so acute, matching the forecast. See Cauchy-Schwarz Inequality.

Cell B — obtuse angle, negative inner product

Cell C — orthogonality (right angle in this geometry)

Cell D — degenerate / zero vector

Worked example Ex 4 (Cell D): what is the angle between

$v = (2, 3)$ and the zero vector $0$ ? Standard dot product.

Forecast: the zero vector has no direction — does the angle even exist? Guess: undefined.

Inner product. $⟨ v, 0 ⟩ = (2) (0) + (3) (0) = 0$ . Why this step? Every inner product gives $0$ against the zero vector (linearity: $⟨ v, 0 \cdot 0 ⟩ = 0$ ).
Norm of $0$ . $∥ 0 ∥ = ⟨ 0, 0 ⟩ = 0 = 0$ . Why this step? Positive-definiteness says only the zero vector has zero length — here it does.
Try the cosine. $cos θ = \frac{0}{∥ v ∥ \cdot 0} = \frac{0}{0}$ . Why this step? The denominator is $0$ , so the formula divides by zero — undefined.

Verify: the angle is genuinely undefined, not $9 0^{\circ}$ . People wrongly call $⟨ v, 0 ⟩ = 0$ "orthogonal", but the zero vector is orthogonal to everything and has no direction, so no angle exists. This is the degenerate boundary case.

Cell E — weighted inner product bends the geometry

Worked example Ex 5 (Cell E): are

$(1, 0)$ and $(0, 1)$ orthogonal under $⟨ u, v ⟩ = 3 u_{1} v_{1} + 2 u_{2} v_{2}$ ? And what is $∥ (1, 1) ∥$ ?

Forecast: in the usual geometry the axes are perpendicular. Does re-weighting them break that? Guess.

Inner product of the axis vectors. $⟨(1, 0), (0, 1)⟩ = 3 (1) (0) + 2 (0) (1) = 0$ . Why this step? Even with weights, the cross terms vanish because each vector lives on one axis — so the axes stay orthogonal here.
A vector that is NOT on an axis. Take $(1, 1)$ : $⟨(1, 1), (1, 1)⟩ = 3 (1) + 2 (1) = 5$ , so $∥ (1, 1) ∥ = 5$ . Why this step? The weights stretch the plane unevenly, so this length differs from the ordinary $2$ .
Show a bent angle. Under this inner product $(1, 1)$ and $(1, - 1)$ : $⟨(1, 1), (1, - 1)⟩ = 3 (1) (1) + 2 (1) (- 1) = 3 - 2 = 1 \neq = 0$ . Why this step? On plain paper $(1, 1) ⊥ (1, - 1)$ at $9 0^{\circ}$ ; here the value is $1 \neq = 0$ , so they are not orthogonal — the weights bent the geometry.

Verify: $∥ (1, 1) ∥ = 5 \approx 2.236 \neq = 2 \approx 1.414$ ✓, confirming distances changed. See Norms and Distance.

Cell F — function space, orthogonality by integration

Worked example Ex 6 (Cell F): show

$f (x) = 1$ and $g (x) = cos (2 π x)$ are orthogonal on $[0, 1]$ Use the integral inner product $⟨ f, g ⟩ = \int_{0}^{1} f (x) g (x) d x$ .

Forecast: a flat line and a full cosine wave — the wave spends equal time above and below zero, so guess: their product integrates to $0$ .

Set up. $⟨ f, g ⟩ = \int_{0}^{1} 1 \cdot cos (2 π x) d x$ . Why this step? Multiply the two functions pointwise, then sum (integrate) — that is the inner product for functions.
Integrate. $\int_{0}^{1} cos (2 π x) d x = [\frac{sin ( 2 π x )}{2 π}]_{0}^{1} = \frac{sin 2 π - sin 0}{2 π} = \frac{0 - 0}{2 π} = 0$ . Why this step? We integrate over exactly one full period, so the positive and negative humps cancel.

Verify: the integral is exactly $0$ ✓ — so the constant function is orthogonal to $cos (2 π x)$ . This equal-area cancellation is the engine behind Fourier Series.

Cell G — the equality boundary of Cauchy–Schwarz

Worked example Ex 7 (Cell G): angle between

$(2, 4)$ and $(1, 2)$ (parallel vectors) Standard dot product.

Forecast: $(2, 4)$ is exactly $2 \times (1, 2)$ — same direction. Guess: cosine hits its extreme value $+ 1$ , angle $0^{\circ}$ , and Cauchy–Schwarz becomes an equality.

Inner product. $⟨ u, v ⟩ = (2) (1) + (4) (2) = 10$ .
Norms. $∥ u ∥ = 4 + 16 = 20$ , $∥ v ∥ = 1 + 4 = 5$ .
Cosine. $cos θ = \frac{10}{20 5} = \frac{10}{100} = \frac{10}{10} = 1$ . Why this step? When one vector is a positive scalar multiple of the other, the discriminant trick in Cauchy–Schwarz gives equality — the parabola just touches zero.
Angle. $θ = arccos (1) = 0^{\circ}$ .

Verify: $∣ ⟨ u, v ⟩ ∣ = 10$ and $∥ u ∥∥ v ∥ = 10$ , so $∣ ⟨ u, v ⟩ ∣ = ∥ u ∥∥ v ∥$ — the equality case of Cauchy–Schwarz ✓, exactly as forecast. (Anti-parallel vectors like $(2, 4)$ and $(- 1, - 2)$ would give $cos θ = - 1$ , the other boundary.)

Cell H — real-world word problem

Worked example Ex 8 (Cell H): "similarity" of two products by feature scores

A phone $A$ scores $(camera, battery, price) = (8, 6, 4)$ and phone $B$ scores $(2, 9, 7)$ . Using cosine similarity (the inner-product angle), how alike are they?

Forecast: they disagree on camera but overlap on battery/price — guess a middling positive cosine, an acute angle far from $0^{\circ}$ .

Inner product. $⟨ A, B ⟩ = 8 (2) + 6 (9) + 4 (7) = 16 + 54 + 28 = 98$ . Why this step? The dot product adds up how much the two "agree" feature by feature — larger = more similar.
Norms. $∥ A ∥ = 64 + 36 + 16 = 116$ , $∥ B ∥ = 4 + 81 + 49 = 134$ . Why this step? Normalising removes "one phone just has bigger numbers", leaving only the pattern of scores.
Cosine similarity. $cos θ = \frac{98}{116 134} = \frac{98}{15544} \approx \frac{98}{124.68} \approx 0.786$ .
Angle. $θ = arccos (0.786) \approx 38. 2^{\circ}$ .

Verify: $0.786 \in (- 1, 1)$ ✓; acute and fairly similar, matching the forecast. This is literally how recommender systems compare items — geometry from an inner product. See Dot Product.

Cell I — exam twist: "is this an inner product?"

Worked example Ex 9 (Cell I): decide whether

$⟨ u, v ⟩ = u_{1} v_{1} - u_{2} v_{2}$ on $R^{2}$ is an inner product

Forecast: it looks symmetric and linear — tempting to say yes. But check every axiom before committing.

Symmetry. $u_{1} v_{1} - u_{2} v_{2} = v_{1} u_{1} - v_{2} u_{2}$ ✓. Why this step? Multiplication commutes, so swapping $u, v$ changes nothing.
Linearity in slot 1. $⟨ c u + w, v ⟩ = c (u_{1} v_{1} - u_{2} v_{2}) + (w_{1} v_{1} - w_{2} v_{2})$ ✓. Why this step? Each term is linear in the first vector's components.
Positive-definiteness — the trap. Test $v = (0, 1)$ : $⟨(0, 1), (0, 1)⟩ = 0 - 1 = - 1 < 0$ . Why this step? Axiom 3 demands $⟨ v, v ⟩ \geq 0$ ; here it is negative, so length would be imaginary.

Verify: $⟨(0, 1), (0, 1)⟩ = - 1 < 0$ ✓ (the failure is real). Not an inner product — positive-definiteness fails. This is the single axiom that exam twists most often break.

Recall Quick self-test across all cells

Sign of $cos θ$ when $⟨ u, v ⟩ < 0$ ::: negative → obtuse angle (Cell B). Angle between a vector and $0$ ::: undefined — division by zero (Cell D). Do axes stay orthogonal under $⟨ u, v ⟩ = 3 u_{1} v_{1} + 2 u_{2} v_{2}$ ::: yes, but off-axis right angles get bent (Cell E). What makes Cauchy–Schwarz an equality ::: the two vectors are scalar multiples (parallel/anti-parallel) (Cell G). Which axiom kills $u_{1} v_{1} - u_{2} v_{2}$ ::: positive-definiteness (Cell I).