Vector dot product definitions

Question

For 2 vectors a and b, the dot product is defined as either:

a · b = ax × bx + ay × by

Or:

a · b = |a| × |b| × cos(θ) where θ is the angle between two vectors.

How can I prove these two formulas are the same?

Answer 1

I hate mathematical proofs, because I'm never sure which axioms are assumed to be known, and which need to be (re)stated, and especially because I'm so verbose my "proofs" are really slow plodding and never elegant, but perhaps this helps you understand the path I would take.

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The two statements are $$\vec{a} \cdot \vec{b} = a_x b_x + a_y b_y \tag{1}\label{na1}$$ $$\vec{a} \cdot \vec{b} = \lvert\vec{a}\rvert \, \lvert\vec{b}\rvert \cos \theta \tag{2}\label{na2}$$

Let $$\vec{a} = ( a_x , a_y ) = ( r_a \cos\theta_a , r_a \sin\theta_a ) \tag{3}\label{na3}$$ and $$\vec{b} = ( b_x , b_y ) = ( r_b \cos\theta_b , r_b \sin\theta_b ) \tag{4}\label{na4}$$ where the rightmost sides are simply the same exact vectors expressed in polar coordinates.

For $\eqref{na2}$ to make sense, we need to define $\lvert\vec{a}\rvert$ and $\lvert\vec{b}\rvert$: $$\lvert\vec{a}\rvert = \sqrt{a_x^2 + a_y^2} = r_a \tag{5}\label{na5}$$ $$\lvert\vec{b}\rvert = \sqrt{b_x^2 + b_y^2} = r_b \tag{6}\label{na6}$$

Because $\theta$ is the angle between the two vectors, it must be $$\pm\theta = \theta_a - \theta_b$$ The sign depends on the order of the vectors (because the angle is always measured the "shorter way"). Fortunately, that does not matter, because $\cos(-\theta) = \cos(\theta)$, i.e. $$\cos\theta = \cos(\theta_a - \theta_b) = \cos(\theta_b - \theta_a)$$ If we apply the cosine addition formula, we see that $$\cos\theta = \cos\theta_a \cos\theta_b + \sin\theta_a \sin\theta_b \tag{7}\label{na7}$$

Substituting $\eqref{na5}$, $\eqref{na6}$, and $\eqref{na7}$ into $\eqref{na2}$ we get $$\vec{a} \cdot \vec{b} = r_a r_b \cos \theta = r_a r_b (\cos \theta_a \cos \theta_b + \sin \theta_a \sin \theta_b ) \tag{I}\label{nai}$$

Substituting $a_x = r_a \cos \theta_a$ and $a_y = r_a \sin \theta_a$ from $\eqref{na3}$, and $b_x = r_b \cos \theta_b$ and $b_y = r_b \sin \theta_b$ from $\eqref{na4}$, to $\eqref{na1}$, we get $$\vec{a} \cdot \vec{b} = r_a \cos \theta_a \, r_b \cos \theta_b + r_a \sin \theta_a \, r_b \sin \theta_b \tag{II}\label{naii}$$

Because $\eqref{nai}$ and $\eqref{naii}$ are equivalent, $\eqref{na1}$ and $\eqref{na2}$ must be equivalent also.

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I would not consider this a proof myself, actually; it's more like my own notes on how I would explore the question (whether the two are equivalent or not), if I was unsure.

Answer 2

You can proceed as follows:

Let's say that $(\vec{a}, \vec{b}) = |\vec{a}| |\vec{b}| \cos(\vec{a},\vec{b})$, where I use $(\vec{a}, \vec{b})$ to denote scalar product. Note that $\vec{a} = a_x \vec{e_x}+a_y \vec{e_y}$, where $\vec{e_x}$ and $\vec{e_y}$ are orthonormal units vectors (such that $\cos(\vec{e_x},\vec{e_y}) = 0$ and $|\vec{e_x}|=|\vec{e_y}|=1$).

Now $$(\vec{a}, \vec{b}) = (a_x \vec{e_x}+a_y \vec{e_y}, b_x \vec{e_x}+b_y \vec{e_y}) = (a_x \vec{e_x},b_x \vec{e_x})+(a_y \vec{e_y},b_y \vec{e_y})+(a_x \vec{e_x},b_y \vec{e_y})+(a_y \vec{e_y},b_x \vec{e_x}) = a_x b_x |\vec{e_x}|^2+a_y b_y |\vec{e_y}|^2 = a_x b_x + a_y b_y,$$

just as you wished to show.