\[\vec{r}_1= \vec{r}_{10}+ \vec{u} t, \: \vec{r}_2= \vec{r}_{20}+ \vec{v} t\]
is the angle between their direction vectors. We can find the angle \[\theta\]
using the formula \[cos(\theta ) =\frac{ \vec{u} \cdot \vec{v}}{\| \vec{u} \| \| \vec{v} \| }\]
.Example: Let
\[\vec{r}_1= \begin{pmatrix}1\\0\\2\end{pmatrix} + t \begin{pmatrix}1\\1\\2\end{pmatrix}, \: \vec{r}_2= \begin{pmatrix}0\\0\\4\end{pmatrix} + s \begin{pmatrix}2\\-1\\2\end{pmatrix} \]
.The direction bvectors are
\[\vec{u}= \begin{pmatrix}1\\1\\2\end{pmatrix}, \: \vec{v}= \begin{pmatrix}2\\-1\\2\end{pmatrix} \]
.\[cos \theta=\frac{\begin{pmatrix}1\\1\\2\end{pmatrix} \cdot \begin{pmatrix}2\\-1\\2\end{pmatrix}}{\| \begin{pmatrix}1\\1\\2\end{pmatrix} \| \| \begin{pmatrix}2\\-1\\2\end{pmatrix} \|}=\frac{1 \times 2 + 1 \times -1 +2 \times 2}{\sqrt{1^2+1^2+2^2} \sqrt{2^2+1^2+(-2)^2}}=\frac{5}{\sqrt{6} \sqrt{9}}=0.6804\]
\[\theta=cos^{-1}(0.6804)=47.1^o\]