## Line-line intersection

Given two lines in 3D defined as rays

$r_1(t_1) = \mathbf{p_1} + t_1 \mathbf{d_1} \\ r_2(t_2) = \mathbf{p_2} + t_2 \mathbf{d_2}$

Where $$t_1, t_2 \in \mathbb{R}$$, the two lines intersect if

$\mathbf{p_1} + t_1 \mathbf{d_1} = \mathbf{p_2} + t_2 \mathbf{d_2}$

We can apply the cross multiplication operation on both sides with $$\mathbf{d_2}$$ and work from there to find the value of $$t_1$$

$t_1 = \frac{\norm{(\mathbf{p_2} - \mathbf{p_1}) \times \mathbf{d_2} }}{ \norm{\mathbf{d_1} \times \mathbf{d_2}} }$

Similarly we can find the value of $$t_2$$ by crossing with $$\mathbf{d_1}$$ and work from there to find the value of $$t_2$$

$t_2 = \frac{\norm{ (\mathbf{p_2} - \mathbf{p_1}) \times \mathbf{d_1}} }{ \norm{\mathbf{d_1} \times \mathbf{d_2}} }$

The proof can be found here

We can actually solve this problem graphically by using triangle similarity, imagine the following situation

The intersection point $$\mathbf{p}$$ is equal to

$$$\label{line-line-intersection-point} \begin{split} \mathbf{p} &= \mathbf{a} + \norm{\mathbf{p - a}} \unit{ \mathbf{b - a} } \\ &= \mathbf{a} + \norm{\mathbf{p - a}} \frac{ \mathbf{b - a} }{ \norm{\mathbf{b - a}} } \end{split}$$$

By triangle similarity we see that

$\frac{ \norm{\mathbf{p - a}} }{ \norm{\mathbf{b - a}} } = \frac{ \norm{\mathbf{n -a}} }{ \norm{\mathbf{m - a}} }$

Multiplying the left side with an identity

$$$\label{line-line-triangle-similarity} \frac{ \norm{\mathbf{p - a}} }{ \norm{\mathbf{b - a}} } = \frac{ \norm{\mathbf{n -a}} }{ \norm{\mathbf{m - a}} } \frac{ \norm{\mathbf{d - c}} }{ \norm{\mathbf{d - c}} }$$$

We see that the quantity $$\norm{ \mathbf{ n - a } } \norm{\mathbf{d - c}}$$ is equal to the equation of the area of a parallelogram, we can skew the parallelogram (in the graphic towards the $$x$$-axis) so that the left side becomes $$\mathbf{c - a}$$ and the bottom side $$\mathbf{d - c}$$ (which is not affected by the skew), note that the area can also be expressed with the cross product of the vectors $$\mathbf{c - a}$$ and $$\mathbf{d - c}$$ therefore

$$$\label{numerator-area} \norm{\mathbf{n - a}} \norm{\mathbf{d - c}} = \norm{(\mathbf{c - a}) \times (\mathbf{d - c})}$$$

A similar equation can be derived for the parallelogram with sides $$\mathbf{m - a}$$ and $$\mathbf{d - c}$$, only this time the skewed side will become $$\mathbf{b - a}$$

$$$\label{denominator-area} \norm{\mathbf{m - a}} \norm{\mathbf{d - c}} = \norm{(\mathbf{b - a}) \times (\mathbf{d - c})}$$$

Replacing \eqref{numerator-area}, \eqref{denominator-area} in \eqref{line-line-triangle-similarity} and \eqref{line-line-intersection-point} we see that the intersection point is equal to

$\mathbf{p} = \mathbf{a} + (\mathbf{b - a}) \frac{ \norm{(\mathbf{c - a}) \times (\mathbf{d - c})} }{ \norm{(\mathbf{b - a}) \times (\mathbf{d - c})} }$

References