The proof of the angle formula

We prove the formula

\boxed{\cos(\alpha)=\frac{\vec{a}\bullet \vec{b}}{\vert\vec{a}\vert\cdot \vert\vec{b}\vert} }

where $\vec{a}=\left(\begin{array}{r} a_x\\ a_y\\ a_z \end{array}\right)$ and $\vec{b}=\left(\begin{array}{r} b_x\\ b_y\\ b_z \end{array}\right)$ . It is a simple application of the cosine-law.

Find the vector $\vec{c}$ that runs from the tip of vector $\vec{a}$ to the tip of vector $\vec{b}$ .

As $\vec{a}+\vec{c}=\vec{b}$ (completing the triangle), it follows that

\vec{c}=\vec{b}-\vec{a}=\left(\begin{array}{r} b_x-a_x\\ b_y-a_y\\ b_z-a_z \end{array}\right)

Consider the triangle that is given by the vectors $\vec{a}$ , $\vec{b}$ , and $\vec{c}$ . What are the side lengths of this triangle?

a=\vert\vec{a}\vert=\sqrt{a_x^2+a_y^2+a_z^2}

b=\vert\vec{b}\vert=\sqrt{b_x^2+b_y^2+b_z^2}

c=\vert\vec{c}\vert=\sqrt{(b_x-a_x)^2+(b_y-a_y)^2+(b_z-a_z)^2}

Recall the cosine-law, which states that for a triangle with side lengths $a$ , $b$ , and $c$ it is

c^2=b^2+a^2-2\cdot a\cdot b \cdot \cos(\alpha)

where $\alpha$ is the angle between the sides $a$ and $b$ (follows from SOHCAHTOA). Applying this law to our triangle, we get

\vert\vec{c}\vert^2=\vert\vec{a}\vert^2+\vert\vec{b}\vert^2-2\cdot \vert\vec{a}\vert\cdot \vert\vec{b}\vert \cdot \cos(\alpha)

By bringing $\cos(\alpha)$ on the left side, and everything else on the right side of the equal sign, we get

\cos(\alpha)=\frac{\vert\vec{a}\vert^2+\vert\vec{b}\vert^2-\vert\vec{c}\vert^2}{2\cdot \vert\vec{a}\vert \cdot \vert\vec{b}\vert}

Plug in the components, and simplify as much as possible:

\begin{array}{rll} \cos(\alpha) & =& \frac{a_x^2+a_y^2+a_z^2+b_x^2+b_y^2+b_z^2 - ((b_x-a_x)^2+(b_y-a_y)^2+(b_z-a_z)^2)}{2\cdot \sqrt{a_x^2+a_y^2+a_z^2}\cdot \sqrt{b_x^2+b_y^2+b_z^2}} \\ & = & \frac{a_x^2+a_y^2+a_z^2+b_x^2+b_y^2+b_z^2 - b_x^2+2a_x b_x - a_x^2 - b_y^2+2a_y b_y - a_y^2 - b_z^2+2a_z b_z - a_z^2}{2\cdot \sqrt{a_x^2+a_y^2+a_z^2}\cdot \sqrt{b_x^2+b_y^2+b_z^2}} \\ & = & \frac{2a_x b_x+2a_y b_y+2a_z b_z}{2\cdot \sqrt{a_x^2+a_y^2+a_z^2}\cdot \sqrt{b_x^2+b_y^2+b_z^2}} \\ & = & \frac{a_x b_x+a_y b_y+a_z b_z}{\sqrt{a_x^2+a_y^2+a_z^2}\cdot \sqrt{b_x^2+b_y^2+b_z^2}} = \frac{\vec{a}\bullet \vec{b}}{\vert\vec{a}\vert\cdot \vert\vec{b}\vert} \end{array}

And we are done!