Angle between two lines

The angle between two lines is calculated similarly as the angle between two vectors.

Two lines are given:

$\text{g: } \vec{x} = \vec{u} + r \cdot \vec{a}$

$\text{h: } \vec{x} = \vec{v} + s \cdot \vec{b}$

For determining the angle the direction vectors of the lines are used:

Example

$\text{g: } \vec{x} = \begin{pmatrix} 10 \\ 9 \\ 7 \end{pmatrix} + r \cdot \begin{pmatrix} 9 \\ 8 \\ 7 \end{pmatrix}$

$\text{h: } \vec{x} = \begin{pmatrix} 4 \\ 4 \\ 6 \end{pmatrix} + s \cdot \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$

1. Calculate scalar product

   $\vec{a}\cdot\vec{b}$ $=\begin{pmatrix}9 \\ 8 \\ 7 \end{pmatrix}\cdot\begin{pmatrix}1 \\ 1 \\ 2\end{pmatrix}$ $=9\cdot1+8\cdot1+7\cdot2$ $=31$

2. Calculate magnitude of the direction vectors

   $|\vec{a}|=\sqrt{9^2+8^2+7^2}$ $=\sqrt{194}$
   
   $|\vec{b}|=\sqrt{1^2+1^2+2^2}$ $=\sqrt{6}$

3. Insert results into the formula

   $\cos(\gamma) = \frac{|\vec{a}\cdot\vec{b}|}{|\vec{a}|\cdot|\vec{b}|}$
   
   $\cos(\gamma) = \frac{31}{\sqrt{194}\cdot\sqrt{6}}$
   
   $\gamma = \cos^{-1}\left(\frac{31}{\sqrt{194}\cdot\sqrt{6}}\right)$ $\approx24.68°$