Given vectors via coordinates, a=\a_x,a_y,a_z\…
Given vectors via coordinates, $\vec{a}=\{a_x,a_y,a_z\}$, $\vec{b}=\{b_x,b_y,b_z\}$, prove that $1^\circ\quad \vec{a}\cdot\vec{b} = a_x b_x + a_y b_y + a_z b_z$ $2^\circ\quad \cos\angle(\vec{a},\vec{b}) = \frac{a_x b_x+a_y b_y+a_z b_z}{\sqrt{a_x^2+a_y^2+a_z^2}\cdot\sqrt{b_x^2+b_y^2+b_z^2}}$.
Explanation of Method
Coordinate formulas for the scalar product and the angle between vectors.
Step 1: Proof of $1^\circ$
Write $\vec{a} = a_x\vec{i}+a_y\vec{j}+a_z\vec{k}$, $\vec{b} = b_x\vec{i}+b_y\vec{j}+b_z\vec{k}$ and expand with the distributive law (Problem 41). Each of the nine terms has the form $a_\bullet b_\bullet(\vec e\cdot\vec e\,')$; by Problem 44 the six mixed products are $0$ and the three squares are $1$:
| $\cdot$ | $b_x\vec{i}$ | $b_y\vec{j}$ | $b_z\vec{k}$ |
|---|---|---|---|
| $a_x\vec{i}$ | $a_x b_x$ | $0$ | $0$ |
| $a_y\vec{j}$ | $0$ | $a_y b_y$ | $0$ |
| $a_z\vec{k}$ | $0$ | $0$ | $a_z b_z$ |
\[\vec{a}\cdot\vec{b} = a_x b_x + a_y b_y + a_z b_z\]
Step 2: Proof of $2^\circ$
By Problem 42, $\cos\theta = \dfrac{\vec a\cdot\vec b}{\sqrt{\vec a\cdot\vec a}\sqrt{\vec b\cdot\vec b}}$. By $1^\circ$, $\vec a\cdot\vec a=a_x^2+a_y^2+a_z^2$ and $\vec b\cdot\vec b=b_x^2+b_y^2+b_z^2$, so
\[\cos\angle(\vec{a},\vec{b}) = \frac{a_x b_x+a_y b_y+a_z b_z}{\sqrt{a_x^2+a_y^2+a_z^2}\cdot\sqrt{b_x^2+b_y^2+b_z^2}}\]
- $\vec a=a_x\vec i+a_y\vec j+a_z\vec k$
- $\vec b=b_x\vec i+b_y\vec j+b_z\vec k$
Step 3: Consequence: Cauchy–Schwarz inequality
Since $|\cos\theta|\le 1$: $|a_x b_x+a_y b_y+a_z b_z| \le \sqrt{a_x^2+a_y^2+a_z^2}\cdot\sqrt{b_x^2+b_y^2+b_z^2}$.
Result
\[\boxed{\,\vec{a}\cdot\vec{b} = a_x b_x + a_y b_y + a_z b_z\,}\]
\[\boxed{\,\cos\angle(\vec{a},\vec{b}) = \frac{a_x b_x+a_y b_y+a_z b_z}{\sqrt{a_x^2+a_y^2+a_z^2}\cdot\sqrt{b_x^2+b_y^2+b_z^2}}\,}\]