Im trying to follow this proof
https://www.youtube.com/watch?v=9aomoLESkeI
The intention is to prove that the geometric and algebraic definitions of the dot product are equivalent.
The author does most of the proof using mostly coordinate notation ($<a_x,a_y,a_z>$), I wanted to simplify things by using vector notation ($\vec{A}$) until the very end and I found I cant do it, not sure why.
First I declare the law of cosines using the vectors $\vec{A},\vec{B}$.
I declare vector $\vec{C}$ to be
$\vec{C}=\vec{A}-\vec{B}$
So, law of cosines:
$$
|A|^2+|B|^2-2|A||B|cos(\theta)=|A-B|^2
$$
I expand the right hand side
$$
\require{cancel}
\text{Expand right side: }
|A|^2+|B|^2-2|A||B|cos(\theta)=|A|^2+|B|^2-2|A||B| \\
\text{simplify: }\cancel{|A|^2}+\cancel{|B|^2}\cancel{-2}|A||B|cos(\theta)=\cancel{|A|^2}+\cancel{|B|^2}\cancel{-2}|A||B|\\
\text{then: }|A||B|cos(\theta)=|A||B|
$$
Ok so thats a funny result that could only be true if $\theta$ is $0$
whats the meaning of this?
can somehow $|A||B|$ be transformed into $a_xb_x+a_yb_y+c_xc_y$? the proof on the video would suggest so, but i dont see how I can do that from
$$
|A||B|= \sqrt{a_x^2+a_y^2+a_z^2}\sqrt{b_x^2+b_y^2+b_z^2}
$$
maybe im missing some simple algebraic trick?
Best Answer
Note that, in general,
$$|A-B|^2 \neq |A|^2 + |B|^2 - 2|A| |B| \tag{1}\label{eq1A}$$
Instead, as stated in Aretino's question comment, you have
$$|A - B|^2 = (A - B)\cdot (A - B) = |A|^2 + |B|^2 - 2A \cdot B \tag{2}\label{eq2A}$$
This shows it's only true when $A$ and $B$ are parallel, i.e., $\theta = 0$, as you found.
In a similar concept, consider the triangle inequality which says
$$|A + B| \leq |A| + |B| \tag{3}\label{eq3A}$$
Squaring both sides gives
$$|A + B|^2 \leq |A|^2 + |B|^2 + 2|A| |B| \tag{4}\label{eq4A}$$
In particular, the $2$ sides are not equal in general.