I am about to enter the class Engineering Physics II. Alas, much of my mastery of vector manipulation is predicated on something I don't understand that must be taken as an assumption for me to proceed; that's killing me. With that prelude, I can't seem to associate the sensible geometric definition of the dot product: $$AB\cos{\theta}$$ with the (seemingly) painfully arbitrary algebraic definition of the dot product, derivable from: $$(A_x\hat{i} + A_y\hat{j}+A_z\hat{z}) \bullet (B_x\hat{i} + B_y\hat{j}+B_z\hat{z})$$
I simply can't wrap my mind around how the geometry translates to this is any terms. (I can come to the conclusion that this statement simplifies to $A_xB_x+A_yB_y+A_zB_z=\vec{A}\bullet\vec{B}$, as $\hat{i}\hat{i}=0$, among others. But I CANNOT come to terms with this or any other more"top-level" statement.) I can neither prove this algebraically nor think, again, of a geometric case which can be constructed to demonstrate this. To prove that I tried, let me post some of my mathematical convulsions:
First, try to put the equations in more similar terms:
$$\sqrt{A_x^2+A_y^2+A_z^2}\sqrt{B_x^2+B_y^2+B_z^2}\cos{\theta} = (A_x\hat{i} + A_y\hat{j}+A_z\hat{z}) \bullet (B_x\hat{i} + B_y\hat{j}+B_z\hat{z}) \space \mathrm{where} \space \theta =|\angle{A}-\angle{B}|$$
But I also states the right side of the equation in more explicit terms
$$ (\sqrt{A_x^2+A_y^2+A_z^2}\hat{i}+ \sqrt{B_x^2+B_y^2+B_z^2}\hat{j} +\small{\small{\bullet\bullet\bullet}} \space \mathrm{ect}$$
For the sake of it being thus far fruitless, I have excluded the diagram.
Thank you for any help!