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

WHAT we do: multiply matching slots, add. $\mathbf{a}\cdot \mathbf{b}=(3)(4)+(-1)(2)+(2)(-1)=12-2-2=\boxed{8}.$ WHY: the component form is "line the vectors up and sum the products." Nothing geometric needed yet.

WHY this tool: length comes from dotting a vector with itself — that is the Pythagorean theorem hiding inside the dot product. $\Vert \mathbf{v}\Vert =\sqrt{\mathbf{v}\cdot \mathbf{v}}=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=\boxed{10}.$

WHY: two vectors are perpendicular exactly when their dot product is $0$ (because $\cos 90^{\circ }=0$). $\mathbf{a}\cdot \mathbf{b}=(2)(3)+(-3)(2)=6-6=0.$ Dot product is zero → $\boxed{\text{yes, perpendicular}}$. See Orthogonality and orthonormal bases for the general idea.

WHY the cosine formula: we want an angle, and the only bridge from components to an angle is $\mathbf{a}\cdot \mathbf{b}=\Vert \mathbf{a}\Vert \Vert \mathbf{b}\Vert \cos \theta$. Solve it for $\cos \theta$.

• $\mathbf{a}\cdot \mathbf{b}=(1)(1)+(\sqrt{3})(0)=1.$
• $\Vert \mathbf{a}\Vert =\sqrt{1+3}=2,\Vert \mathbf{b}\Vert =1.$ $\cos \theta =\frac{\mathbf{a}\cdot \mathbf{b}}{\Vert \mathbf{a}\Vert \Vert \mathbf{b}\Vert }=\frac{1}{2\cdot 1}=\frac{1}{2}\Rightarrow \theta =\boxed{60^{\circ }}.$ WHAT it looks like: in the figure, the yellow arrow $\mathbf{b}$ lies flat on the $x$-axis and the blue arrow $\mathbf{a}$ rises steeply; the red wedge between them is the $60^{\circ }$ we solved for. Notice the arrow lengths ($\Vert \mathbf{b}\Vert =1$, $\Vert \mathbf{a}\Vert =2$) do not change the angle — only their directions do, which is why the length factors cancelled out of $\cos \theta$.

WHY the cosine formula again: we're asked for $\cos \theta$, so we run the same bridge $\cos \theta =\frac{\mathbf{a}\cdot \mathbf{b}}{\Vert \mathbf{a}\Vert \Vert \mathbf{b}\Vert }$ — now in 3D, where we cannot easily draw the angle, so the algebra is our only handle on it.

• Numerator (overlap): $\mathbf{a}\cdot \mathbf{b}=(2)(0)+(1)(3)+(-2)(4)=0+3-8=-5.$ Why: multiply matching slots and add — the same machine as L1.1.
• Denominator (lengths): $\Vert \mathbf{a}\Vert =\sqrt{4+1+4}=3,\Vert \mathbf{b}\Vert =\sqrt{0+9+16}=5.$ Why: each length is $\sqrt{\text{vector}\cdot \text{itself}}$, the Pythagorean idea in 3D. $\cos \theta =\frac{-5}{3\cdot 5}=\frac{-5}{15}=-\frac{1}{3}.$ WHY the sign matters: the numerator carries the sign, and a negative cosine means $\theta >90^{\circ }$ — the vectors lean away from each other (obtuse angle). $\boxed{\cos \theta =-\frac{1}{3}}$ (so $\theta \approx 109.47^{\circ }$). Sanity check: $\left|-\frac{1}{3}\right|\le 1$, as any genuine cosine must be — a reassurance that we divided by the lengths correctly.

WHAT a projection is: the shadow $\mathbf{a}$ casts along the direction of $\mathbf{b}$. Its signed length is $\frac{\mathbf{a}\cdot \mathbf{b}}{\Vert \mathbf{b}\Vert }$ — see Vector projection. WHY that formula: $\mathbf{a}\cdot \mathbf{b}=\Vert \mathbf{a}\Vert \Vert \mathbf{b}\Vert \cos \theta$, and $\Vert \mathbf{a}\Vert \cos \theta$ is exactly the adjacent side (the shadow). Dividing by $\Vert \mathbf{b}\Vert$ strips out $\mathbf{b}$'s length so only the shadow survives. Edge case first — WHY $\Vert \mathbf{b}\Vert \ne 0$ is required: the formula divides by $\Vert \mathbf{b}\Vert$, so it is only defined when $\mathbf{b}\ne 0$. Projecting onto the zero vector is meaningless — the zero vector has no direction to cast a shadow along, and the formula would demand division by $0$. Always confirm $\mathbf{b}\ne 0$ before projecting. Here $\Vert \mathbf{b}\Vert =1\ne 0$, so we are safe. $\frac{\mathbf{a}\cdot \mathbf{b}}{\Vert \mathbf{b}\Vert }=\frac{(3)(1)+(4)(0)}{1}=\boxed{3}.$ WHAT to see in the figure: the blue arrow is $\mathbf{a}=(3,4)$; drop a white dashed vertical line straight down onto the direction of $\mathbf{b}$ (the $x$-axis). Where it lands (red dot at $x=3$) marks the tip of the green shadow of length $3$. The shadow is shorter than $\mathbf{a}$ itself ($\Vert \mathbf{a}\Vert =5$) because part of $\mathbf{a}$ points up, away from $\mathbf{b}$ — projection keeps only the part along $\mathbf{b}$.

WHY subtract: the equation says $\mathbf{a}$ and $\mathbf{b}$ have identical overlap with everything. Rewrite using distributivity: $\mathbf{a}\cdot \mathbf{c}-\mathbf{b}\cdot \mathbf{c}=0\Rightarrow (\mathbf{a}-\mathbf{b})\cdot \mathbf{c}=0\text{for all }\mathbf{c}.$ The key move: "for all $\mathbf{c}$" includes the choice $\mathbf{c}=\mathbf{a}-\mathbf{b}$ itself. Plug it in: $(\mathbf{a}-\mathbf{b})\cdot (\mathbf{a}-\mathbf{b})=\Vert \mathbf{a}-\mathbf{b}{\Vert }^2=0.$ A squared length is zero only for the zero vector, so $\mathbf{a}-\mathbf{b}=0$, i.e. $\boxed{\mathbf{a}=\mathbf{b}}$. $■$

WHY expand with the dot product: every length-squared is a dot product, and dot products distribute, so we can just multiply out. $\Vert \mathbf{a}+\mathbf{b}{\Vert }^2=(\mathbf{a}+\mathbf{b})\cdot (\mathbf{a}+\mathbf{b})=\Vert \mathbf{a}{\Vert }^2+2\mathbf{a}\cdot \mathbf{b}+\Vert \mathbf{b}{\Vert }^2.$ $\Vert \mathbf{a}-\mathbf{b}{\Vert }^2=(\mathbf{a}-\mathbf{b})\cdot (\mathbf{a}-\mathbf{b})=\Vert \mathbf{a}{\Vert }^2-2\mathbf{a}\cdot \mathbf{b}+\Vert \mathbf{b}{\Vert }^2.$ WHAT happens on adding: the $\pm 2\mathbf{a}\cdot \mathbf{b}$ terms cancel: $\Vert \mathbf{a}+\mathbf{b}{\Vert }^2+\Vert \mathbf{a}-\mathbf{b}{\Vert }^2=2\Vert \mathbf{a}{\Vert }^2+2\Vert \mathbf{b}{\Vert }^2.■$ Picture: the two diagonals of a parallelogram relate to its four sides — this is the algebraic version of that geometry.

WHY this identity: $\Vert \mathbf{a}-\mathbf{b}{\Vert }^2$ expands into things we already know. $\Vert \mathbf{a}-\mathbf{b}{\Vert }^2=\Vert \mathbf{a}{\Vert }^2-2\mathbf{a}\cdot \mathbf{b}+\Vert \mathbf{b}{\Vert }^2=25-12+9=22.$ $\Vert \mathbf{a}-\mathbf{b}\Vert =\sqrt{22}\approx \boxed{4.69}.$ (This is exactly the Law of cosines in disguise — the side opposite the angle.)

WHY expand carefully: the scalars $2$ and $-3$ pull out of the dot product, so track them. $\Vert 2\mathbf{a}-3\mathbf{b}{\Vert }^2=(2\mathbf{a}-3\mathbf{b})\cdot (2\mathbf{a}-3\mathbf{b})=4\Vert \mathbf{a}{\Vert }^2-12\mathbf{a}\cdot \mathbf{b}+9\Vert \mathbf{b}{\Vert }^2.$ Substitute $\Vert \mathbf{a}{\Vert }^2=1,\Vert \mathbf{b}{\Vert }^2=1,\mathbf{a}\cdot \mathbf{b}=\frac{1}{2}$: $=4(1)-12\left(\frac{1}{2}\right)+9(1)=4-6+9=7.$ $\Vert 2\mathbf{a}-3\mathbf{b}\Vert =\sqrt{7}\approx \boxed{2.65}.$

WHY set the dot product to zero: perpendicular $⟺$ $\mathbf{a}\cdot \mathbf{b}=0$. $\mathbf{a}\cdot \mathbf{b}=3k+2k+(-4)=5k-4=0\Rightarrow k=\frac{4}{5}=\boxed{0.8}.$ One linear equation → one value of $k$. (If the components had produced a quadratic, we would keep every root.)

Start from the square (lengths are non-negative, so squaring is reversible): $\Vert \mathbf{a}+\mathbf{b}{\Vert }^2=\Vert \mathbf{a}{\Vert }^2+2\mathbf{a}\cdot \mathbf{b}+\Vert \mathbf{b}{\Vert }^2.$ WHY Cauchy–Schwarz enters: we need an upper bound on the middle term. Cauchy–Schwarz gives $\mathbf{a}\cdot \mathbf{b}\le |\mathbf{a}\cdot \mathbf{b}|\le \Vert \mathbf{a}\Vert \Vert \mathbf{b}\Vert$. So: $\Vert \mathbf{a}+\mathbf{b}{\Vert }^2\le \Vert \mathbf{a}{\Vert }^2+2\Vert \mathbf{a}\Vert \Vert \mathbf{b}\Vert +\Vert \mathbf{b}{\Vert }^2=(\Vert \mathbf{a}\Vert +\Vert \mathbf{b}\Vert {)}^2.$ Both sides non-negative → take square roots: $\Vert \mathbf{a}+\mathbf{b}\Vert \le \Vert \mathbf{a}\Vert +\Vert \mathbf{b}\Vert .■$ This is the vector version of "the shortest path is a straight line" — see Triangle inequality.

WHY no angle: in an abstract inner product space there is no picture of an angle yet — so we must argue from algebra alone. The single engine is: a squared length is never negative.

Step 1 — build a non-negative machine. For any real number $t$, $f(t)=\Vert \mathbf{a}-t\mathbf{b}{\Vert }^2\ge 0.$ Step 2 — expand into a quadratic in $t$: $f(t)=\Vert \mathbf{a}{\Vert }^2-2t(\mathbf{a}\cdot \mathbf{b})+t^2\Vert \mathbf{b}{\Vert }^2=A{t}^2+Bt+C,$ $A=\Vert \mathbf{b}{\Vert }^2,B=-2(\mathbf{a}\cdot \mathbf{b}),C=\Vert \mathbf{a}{\Vert }^2.$ Step 3 — force the discriminant $\le 0$. An upward parabola ($A\ge 0$) that never dips below the axis can touch it at most once, so it has at most one real root: $B^2-4AC\le 0\Rightarrow 4(\mathbf{a}\cdot \mathbf{b}{)}^2-4\Vert \mathbf{b}{\Vert }^2\Vert \mathbf{a}{\Vert }^2\le 0.$ Step 4 — tidy up. Divide by $4$, move the second term over, take square roots (both sides $\ge 0$): $(\mathbf{a}\cdot \mathbf{b}{)}^2\le \Vert \mathbf{a}{\Vert }^2\Vert \mathbf{b}{\Vert }^2\Rightarrow |\mathbf{a}\cdot \mathbf{b}|\le \Vert \mathbf{a}\Vert \Vert \mathbf{b}\Vert .■$ Equality case: discriminant $=0$ means $f(t_0)=0$ for some $t_0$, i.e. $\Vert \mathbf{a}-t_0\mathbf{b}\Vert =0$, so $\mathbf{a}=t_0\mathbf{b}$ — the vectors are parallel. (Degenerate check: if $\mathbf{b}=0$ then $A=0$, $f$ is linear/constant $\ge 0$, and both sides are $0$ — the inequality holds as $0\le 0$.) WHAT to see in the figure: the blue curve is $f(t)$ for a non-parallel pair — it floats entirely above the $t$-axis (its minimum is positive), which is exactly "discriminant $<0$" and gives the strict inequality. The yellow curve is the parallel case: it just kisses the axis at one point (red dot, $t=t_0$), so the discriminant is $0$ and Cauchy–Schwarz becomes an equality. The picture is the proof: "never below the axis" forces the discriminant condition.

WHY verify numerically: a general theorem earns trust by surviving concrete tests. We recompute each piece and watch Cauchy–Schwarz hold with room to spare.

• Overlap: $\mathbf{a}\cdot \mathbf{b}=(1)(-2)+(2)(1)+(3)(1)=-2+2+3=3.$ Why: the component machine, as always.
• Lengths: $\Vert \mathbf{a}\Vert =\sqrt{1+4+9}=\sqrt{14},\Vert \mathbf{b}\Vert =\sqrt{4+1+1}=\sqrt{6}.$ Why: $\Vert \mathbf{v}\Vert =\sqrt{\mathbf{v}\cdot \mathbf{v}}$.
• The two sides of Cauchy–Schwarz: $|\mathbf{a}\cdot \mathbf{b}|=3,\Vert \mathbf{a}\Vert \Vert \mathbf{b}\Vert =\sqrt{14}\sqrt{6}=\sqrt{84}\approx \mathrm{9.17.}$ Why check both? the inequality compares exactly these two numbers; we want to see the left one sitting below the right.
• Strict or equal? $3\le 9.17$ holds, and it is strict because $\mathbf{a}$ is not a scalar multiple of $\mathbf{b}$ (no single $t$ makes $\mathbf{a}=t\mathbf{b}$) — recall equality needs parallel vectors, matching the L5.1 equality case. $\cos \theta =\frac{\mathbf{a}\cdot \mathbf{b}}{\Vert \mathbf{a}\Vert \Vert \mathbf{b}\Vert }=\frac{3}{\sqrt{84}}\approx 0.327\Rightarrow \theta \approx \boxed{70.9^{\circ }}.$ Sanity: $0.327\le 1$, a valid cosine — the same $|\cos \theta |\le 1$ fact that Cauchy–Schwarz guarantees in general.

WHY square the sum: the condition involves a sum of vectors; dotting that sum with itself turns it into exactly the pairwise products we want. $0=\Vert \mathbf{u}+\mathbf{v}+\mathbf{w}{\Vert }^2=(\mathbf{u}+\mathbf{v}+\mathbf{w})\cdot (\mathbf{u}+\mathbf{v}+\mathbf{w}).$ Expand — the three self-terms are lengths ($=1$ each since unit vectors) and each cross pair appears twice: $0=\underset{1+1+1=3}{\underbrace{\Vert \mathbf{u}{\Vert }^2+\Vert \mathbf{v}{\Vert }^2+\Vert \mathbf{w}{\Vert }^2}}+2(\mathbf{u}\cdot \mathbf{v}+\mathbf{v}\cdot \mathbf{w}+\mathbf{w}\cdot \mathbf{u}).$ Solve: $\mathbf{u}\cdot \mathbf{v}+\mathbf{v}\cdot \mathbf{w}+\mathbf{w}\cdot \mathbf{u}=-\frac{3}{2}=\boxed{-1.5}.$ Sanity: the vectors point $120^{\circ }$ apart (like the Mercedes logo), and $3\cos 120^{\circ }=3(-\frac{1}{2})=-\frac{3}{2}$. ✔

Component dot ::: multiply matching slots, then add → a scalar Length ::: $\Vert \mathbf{a}\Vert =\sqrt{\mathbf{a}\cdot \mathbf{a}}$ Angle ::: $\cos \theta =\frac{\mathbf{a}\cdot \mathbf{b}}{\Vert \mathbf{a}\Vert \Vert \mathbf{b}\Vert }$ Perpendicular test ::: $\mathbf{a}\cdot \mathbf{b}=0$ Projection length onto $\mathbf{b}$ ::: $\frac{\mathbf{a}\cdot \mathbf{b}}{\Vert \mathbf{b}\Vert }$, valid only when $\mathbf{b}\ne 0$ Cauchy–Schwarz ::: $|\mathbf{a}\cdot \mathbf{b}|\le \Vert \mathbf{a}\Vert \Vert \mathbf{b}\Vert$, equality iff parallel C–S proof engine ::: $f(t)=\Vert \mathbf{a}-t\mathbf{b}{\Vert }^2\ge 0$ → discriminant $\le 0$ Squaring a vector sum ::: each cross pair counts twice