## Angle Between Two Lines

The angle between two lines
$\vec{r}_1= \vec{r}_{10}+ \vec{u} t, \: \vec{r}_2= \vec{r}_{20}+ \vec{v} t$
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
$\vec{u}= \begin{pmatrix}1\\1\\2\end{pmatrix}, \: \vec{v}= \begin{pmatrix}2\\-1\\2\end{pmatrix}$
.
$cos \theta=\frac{\begin{pmatrix}1\\1\\2\end{pmatrix} \cdot \begin{pmatrix}2\\-1\\2\end{pmatrix}}{\| \begin{pmatrix}1\\1\\2\end{pmatrix} \| \| \begin{pmatrix}2\\-1\\2\end{pmatrix} \|}=\frac{1 \times 2 + 1 \times -1 +2 \times 2}{\sqrt{1^2+1^2+2^2} \sqrt{2^2+1^2+(-2)^2}}=\frac{5}{\sqrt{6} \sqrt{9}}=0.6804$

$\theta=cos^{-1}(0.6804)=47.1^o$

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