How does these two ways of finding scalar (dot) product relate?

Question

How does

a · b = a1b1 + a2b2 + a3b3 

relate to

 → a ⋅ → b = || → a || ||→ b || cos θ

I don't know what to google to find this demonstration of how the two methods relate to eachother or imply eachother.

Answer 1

We want to show $$\vec u\cdot\vec v=u_1v_1+u_2v_2+u_3v_3.$$

Define $\vec{u}=(u_1,u_2,u_3)=u_1\hat i+u_2\hat j+u_3\hat k$ and $\vec{v}=(v_1,v_2,v_3)=v_1\hat i+v_2\hat j+v_3\hat k$, where $\hat i, \hat j$, and $\hat k$ are the unit vectors of the $x$, $y$, and $z$-axes. Then we have \begin{align} \vec{u}\cdot\vec{v}&=(u_1\hat i+u_2\hat j+u_3\hat k)\cdot(v_1\hat i+v_2\hat j+v_3\hat k)\\ &=\color{blue}{u_1v_1\hat i\cdot\hat i}+\color{red}{u_1v_2\hat i\cdot\hat j}+\color{red}{u_1v_3\hat i\cdot\hat k}\\ &\qquad+\color{red}{u_2v_1\hat j\cdot\hat i}+\color{blue}{u_2v_2\hat j\cdot\hat j}+\color{red}{u_2v_3\hat j\cdot\hat k}\\ &\qquad\qquad+\color{red}{u_3v_1\hat k\cdot\hat i}+\color{red}{u_3v_2\hat k\cdot\hat j}+\color{blue}{u_3v_3\hat k\cdot\hat k}\\ &=\color{blue}{u_1v_1}+\color{blue}{u_2v_2}+\color{blue}{u_3v_3} \end{align} as desired (since we have $\hat a\cdot\hat a=\vert\hat a\vert^2=1$ and perpendicular vectors have a dot product of $0$).

The geometric definition is a result of the Law of Cosines $$c^2=a^2+b^2-2ab\cos\gamma.$$

Define $\vec w=\vec u-\vec v$ and let $\theta$ represent the angle between $\vec u$ and $\vec v$. As such, we have \begin{align} \vec w&=\vec u-\vec v\\ \vec w\cdot\vec w&=(\vec u-\vec v)\cdot(\vec u-\vec v)\\ \vert\vec w\vert^2&=\vec u\cdot\vec u-2\vec u\cdot\vec v+\vec v\cdot\vec v\\ \vert\vec w\vert^2&=\vert\vec u\vert^2+\vert\vec v\vert^2-2\vec u\cdot\vec v. \end{align}

Matching the two equations, it follows that we can equate $$\vec u\cdot\vec v=\vert\vec u\vert\vert\vec v\vert\cos\theta.$$

Sometimes the algebraic and geometric definitions are used in conjunction to solve problems that involve variable and given vector components and angles between vectors.