Dot product — formula, cosine formula, Cauchy-Schwarz inequality proof: Worked examples

The scenario matrix

Every worked example below is tagged with the cell it fills. The goal: cover the whole table.

Cell	Case class	Sign of $a \cdot b$	Angle $θ$	Example
C1	Acute — arrows lean the same way	positive	$0^{\circ} < θ < 9 0^{\circ}$	Ex 1
C2	Right angle — perpendicular	zero	$θ = 9 0^{\circ}$	Ex 2
C3	Obtuse — arrows lean apart	negative	$9 0^{\circ} < θ < 18 0^{\circ}$	Ex 3
C4	Parallel same-way (equality in C–S)	$= + ∥ a ∥∥ b ∥$	$θ = 0^{\circ}$	Ex 4
C5	Anti-parallel (equality, other sign)	$= - ∥ a ∥∥ b ∥$	$θ = 18 0^{\circ}$	Ex 4
C6	Degenerate — one vector is $0$	$= 0$ but not perpendicular	undefined	Ex 5
C7	Higher dimension ( $R^{3} / R^{4}$ )	any	any	Ex 6
C8	Real-world word problem (work / teamwork)	any	any	Ex 7
C9	Exam twist — solve for an unknown	forced to a value	forced	Ex 8
C10	Cauchy–Schwarz sanity, strict vs equality	—	—	Ex 9

Example 1 — Acute angle (Cell C1)

Worked example Positive overlap

$a = (3, 1), b = (1, 2)$ . Find $a \cdot b$ and the angle $θ$ .

Forecast: the arrows both point up-and-right, so guess a positive dot product and an angle less than $9 0^{\circ}$ .

$a \cdot b = 3 \cdot 1 + 1 \cdot 2 = 3 + 2 = 5.$ Why this step? The component formula: multiply matching slots, add. Positive, as forecast.
$∥ a ∥ = 3^{2} + 1^{2} = 10$ , $∥ b ∥ = 1^{2} + 2^{2} = 5$ . Why this step? Length $= a \cdot a$ ; we need lengths to strip them off and expose $cos θ$ .
$cos θ = \frac{5}{10 5} = \frac{5}{50} = \frac{5}{5 2} = \frac{1}{2} .$ Why this step? Rearranging the geometric formula for $cos θ$ isolates the angle.
$θ = arccos (\frac{1}{2}) = 4 5^{\circ} .$ Why this step? $arccos$ answers "which angle has this cosine?" — it undoes the cosine.

Verify: $0^{\circ} < 4 5^{\circ} < 9 0^{\circ}$ ✔ acute, matching the positive dot product. See the figure — the arrows visibly lean the same way.

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

Example 2 — Right angle (Cell C2)

Example 3 — Obtuse angle (Cell C3)

Example 4 — Parallel & anti-parallel (Cells C4, C5)

Worked example The two equality cases of Cauchy–Schwarz

Take $a = (2, 4)$ . Case C4: $b = (1, 2)$ (same direction, $a = 2 b$ ). Case C5: $c = (- 1, - 2)$ (opposite direction, $a = - 2 c$ ).

Forecast: parallel vectors make $∣ cos θ ∣ = 1$ , so Cauchy–Schwarz should be an equality, not a strict inequality. Same-way → $+$ ; opposite → $-$ .

$a \cdot b = 2 \cdot 1 + 4 \cdot 2 = 2 + 8 = 10.$ And $∥ a ∥∥ b ∥ = 20 \cdot 5 = 100 = 10.$ Why this step? We compare the dot product to the length-product to test equality.
So $cos θ = \frac{10}{10} = 1 \Rightarrow θ = 0^{\circ}$ — perfectly aligned (C4). Why this step? $cos θ = 1$ is the only value giving $θ = 0$ ; the arrows overlap in direction.
$a \cdot c = 2 (- 1) + 4 (- 2) = - 2 - 8 = - 10.$ And $∥ a ∥∥ c ∥ = 20 \cdot 5 = 10.$ Why this step? Same magnitudes, but the sign flips because the arrow is reversed.
$cos θ = \frac{- 10}{10} = - 1 \Rightarrow θ = 18 0^{\circ}$ — anti-parallel (C5). Why this step? $cos θ = - 1$ is the unique value at $θ = 18 0^{\circ}$ .

Verify: In both cases $∣ a \cdot b ∣ = ∥ a ∥∥ b ∥ = 10$ — equality in Cauchy–Schwarz exactly when parallel ✔. Contrast Ex 9, where it's strict.

Example 5 — Degenerate: the zero vector (Cell C6)

Worked example Why

$a \cdot b = 0$ is not always "perpendicular" $a = (5, - 7), b = 0 = (0, 0)$ . What is $a \cdot b$ , and what is the angle?

Forecast: everyone reflexively says " $0$ dot product ⇒ perpendicular." But a zero-length arrow has no direction — so there is no angle to speak of. Guess: dot product $0$ , but "perpendicular" is meaningless here.

$a \cdot b = 5 \cdot 0 + (- 7) \cdot 0 = 0.$ Why this step? Every term contains a factor $0$ , so the sum collapses to $0$ .
Try the angle formula: $cos θ = \frac{0}{∥ a ∥ \cdot 0} = \frac{0}{0}$ — undefined. Why this step? $∥ b ∥ = 0$ sits in the denominator. Division by zero means no angle exists.
Conclusion: $a \cdot 0 = 0$ for any $a$ , but this is a degenerate zero, not perpendicularity. Why this step? True perpendicularity (Ex 2) needs both vectors nonzero.

Verify: Pick any other $a$ , say $(100, 3)$ : $a \cdot 0 = 0$ still. The zero output is forced by $b = 0$ , independent of direction ✔.

Example 6 — Higher dimensions (Cell C7)

Worked example Same machine in

$R^{4}$ $a = (1, - 1, 2, 0), b = (3, 1, 1, 4)$ in $R^{4}$ . Find $a \cdot b$ and $cos θ$ .

Forecast: We can't draw 4D, but the formula doesn't care — it just sums four products. This is where the algebraic definition earns its keep (the parent stressed this).

$a \cdot b = 1 \cdot 3 + (- 1) \cdot 1 + 2 \cdot 1 + 0 \cdot 4 = 3 - 1 + 2 + 0 = 4.$ Why this step? Component formula extends to any dimension — one slot per coordinate.
$∥ a ∥ = 1 + 1 + 4 + 0 = 6, ∥ b ∥ = 9 + 1 + 1 + 16 = 27 = 33 .$ Why this step? $∥ \cdot ∥ = x \cdot x$ holds in every dimension.
$cos θ = \frac{4}{6 \cdot 3 3} = \frac{4}{3 18} = \frac{4}{9 2} \approx 0.3143.$ Why this step? The geometric formula still defines an angle in $R^{4}$ — and Cauchy–Schwarz (parent §3) is exactly what guarantees this ratio stays in $[- 1, 1]$ .
$θ = arccos (0.3143) \approx 71.6 8^{\circ} .$

Verify: Cauchy–Schwarz check: $∣ a \cdot b ∣ = 4$ and $∥ a ∥∥ b ∥ = 318 \approx 12.73$ , so $4 \leq 12.73$ ✔ — the cosine is safely inside $[- 1, 1]$ . See also Inner product spaces for abstract versions.

Example 7 — Real-world word problem (Cell C8)

Worked example Work done by a force

A box is pushed by a force $F = (6, 2)$ newtons while it slides along displacement $d = (4, 0)$ metres. Work is defined as $W = F \cdot d$ (joules). Compute $W$ , and find how much of the force actually helped.

Forecast: only the part of the force along the motion does work. The force leans slightly up, so some is "wasted" pushing into nothing. Guess $W$ a bit less than $∣ F ∣ \cdot ∣ d ∣$ .

$W = F \cdot d = 6 \cdot 4 + 2 \cdot 0 = 24$ joules. Why this step? Work is a dot product — it measures force-in-the-direction-of-motion, exactly what the dot product does.
$∥ F ∥ = 36 + 4 = 40 = 210, ∥ d ∥ = 4.$ Why this step? We need these to find the angle and the "useful fraction."
$cos θ = \frac{24}{2 10 \cdot 4} = \frac{24}{8 10} = \frac{3}{10} \approx 0.9487, θ \approx 18.4 3^{\circ} .$ Why this step? The angle tells us how well aligned the push is with the slide.
Force component along $d$ : $∥ F ∥ cos θ = 210 \cdot \frac{3}{10} = 6$ N. That's the projection of $F$ onto $d$ . Why this step? $W = (useful force) \times (distance) = 6 \times 4 = 24$ J — matches step 1, a nice consistency check.

Verify: Units: N·m $=$ J ✔. If the force were straight up ( $F = (0, 2)$ ), $W = (0) (4) + (2) (0) = 0$ — no work, correct, since pushing sideways to the motion does nothing ✔.

Example 8 — Exam twist: solve for an unknown (Cell C9)

Worked example Find

$k$ making the vectors perpendicular For what value(s) of $k$ is $a = (k, 3)$ perpendicular to $b = (k, - 12)$ ?

Forecast: perpendicular ⇔ dot product $= 0$ . That's a single equation in $k$ — likely a quadratic, so expect up to two answers.

Perpendicular condition: $a \cdot b = 0$ . Why this step? This is the defining test for a right angle (both vectors will be nonzero for our solutions).
$a \cdot b = k \cdot k + 3 \cdot (- 12) = k^{2} - 36.$ Why this step? Apply the component formula symbolically, keeping $k$ as an unknown.
Set $k^{2} - 36 = 0 \Rightarrow k^{2} = 36 \Rightarrow k = \pm 6.$ Why this step? A quadratic gives both signs — a common exam trap is dropping $k = - 6$ .

Verify:

$k = 6$ : $a = (6, 3), b = (6, - 12)$ , dot $= 36 - 36 = 0$ ✔.
$k = - 6$ : $a = (- 6, 3), b = (- 6, - 12)$ , dot $= 36 - 36 = 0$ ✔. Both give nonzero vectors, so both are genuine right angles (orthogonal) ✔.

Example 9 — Cauchy–Schwarz: strict vs equality (Cell C10)

Worked example Watching the inequality tighten

Compare two pairs against $∣ a \cdot b ∣ \leq ∥ a ∥∥ b ∥$ : Pair A (not parallel): $a = (3, 4), b = (1, 2)$ . Pair B (parallel): $a = (3, 4), b = (6, 8)$ .

Forecast: Pair A should be strict ( $<$ ); Pair B, being parallel, should be an equality ( $=$ ).

Pair A: $∣ a \cdot b ∣ = ∣3 + 8∣ = 11.$ RHS $= ∥ a ∥∥ b ∥ = 5 \cdot 5 \approx 11.180.$ Why this step? Directly plug into both sides of the inequality.
$11 < 11.180$ — strict, because no scalar $t$ gives $a = t b$ (check: $3/1 \neq = 4/2$ ). Why this step? Strictness ⇔ non-parallel, per the parent's equality condition.
Pair B: $∣ a \cdot b ∣ = ∣3 \cdot 6 + 4 \cdot 8∣ = ∣18 + 32∣ = 50.$ RHS $= 5 \cdot 100 = 5 \cdot 10 = 50.$ Why this step? Here $b = 2 a$ exactly, so the "leftover perpendicular part" is zero and the bound is hit.
$50 = 50$ — equality, confirming Pair B is parallel.

Verify: $11 \leq 11.180$ ✔ (strict) and $50 \leq 50$ ✔ (equality). The discriminant proof (parent §3) predicts exactly this: equality iff $∥ a - t b ∥ = 0$ for some $t$ ✔.

Recap

Recall Which cell is which sign?

Positive dot product means angle is... ::: acute ( $0^{\circ} < θ < 9 0^{\circ}$ ) Zero dot product with both vectors nonzero means... ::: perpendicular ( $θ = 9 0^{\circ}$ ) Negative dot product means angle is... ::: obtuse ( $9 0^{\circ} < θ < 18 0^{\circ}$ ) $a \cdot 0 = 0$ tells us the angle is... ::: undefined (no direction for $0$ ), NOT perpendicular Cauchy–Schwarz is an equality exactly when... ::: the vectors are parallel

Related: Vector projection, Triangle inequality, Law of cosines, Cross product (the vector-valued cousin).