dot product and law of cosines

Dot product and Law of cosines

Since the euclidean norm is defined as

$$|\mathbf{v}|=\sqrt{\sum _{i=1}^n{\mathbf{v}}_i^2}$$

and the dot product as

$$\delta =\sum _{i=1}^n{\mathbf{a}}_i{\mathbf{b}}_i$$

the dot product of a vector with itself is the squared norm of that vector

$$\mathbf{v}\cdot \mathbf{v}=\sum _{i=1}^n{\mathbf{v}}_i{\mathbf{v}}_i=\sum _{i=1}^n{\mathbf{v}}_i^2=|\mathbf{v}{|}^2\text{\hspace{1em}(1)}$$

Then, I consider the not immediately related fact that two vectors $\mathbf{a}$ and $\mathbf{b}$, and their subtraction $\mathbf{c}$ actually form a triangle having sides $\mathbf{a},\mathbf{b},\mathbf{c}$

$$\mathbf{c}=\mathbf{b}-\mathbf{a}$$

with $\mathbf{c}$ thus having length

$$|\mathbf{c}{|}^2=|\mathbf{b}-\mathbf{a}{|}^2$$

expanding, that means

$$=(\mathbf{b}-\mathbf{a})\cdot (\mathbf{b}-\mathbf{a})=|\mathbf{b}{|}^2+|\mathbf{a}{|}^2-2(\mathbf{a}\cdot \mathbf{b})$$

Given the fact that $c=|\mathbf{c}|$—and consequently $c^2=|\mathbf{c}{|}^2$, that is to say the vector norm is the length of the triangle side $c$ in vector interpretation, the Law of Cosines

$$c^2=a^2+b^2-2ab\cos ({\theta }_{a,b})$$

can thus also have a vector interpretation

$$|\mathbf{c}{|}^2=|\mathbf{a}{|}^2+|\mathbf{b}{|}^2-2|\mathbf{a}||\mathbf{b}|\cos ({\theta }_{\mathbf{a},\mathbf{b}})$$

I have therefore to conclude that

$$|\mathbf{b}{|}^2+|\mathbf{a}{|}^2-2(\mathbf{a}\cdot \mathbf{b})=|\mathbf{a}{|}^2+|\mathbf{b}{|}^2-2|\mathbf{a}||\mathbf{b}|\cos ({\theta }_{\mathbf{a},\mathbf{b}})$$

that is to say

$$\mathbf{a}\cdot \mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos ({\theta }_{\mathbf{a},\mathbf{b}})\text{\hspace{1em}(2)}$$

the dot product is the product of the vectors norms scaled by the cosine of the angle between the two vectors.