Line-line intersection

Given two lines in 3D, defined as rays:

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

We can apply the cross multiplication operation on both sides with ${\mathbf{d}}_{\mathbf{2}}$ and work from there to find the value of $t_1$.

Similarly, we can find the value of $t_2$ by crossing with ${\mathbf{d}}_{\mathbf{1}}$ and work from there to find the value of $t_2$.

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:

By triangle similarity, we see that:

Multiplying the left side with an identity

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

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

Replacing $\text{(3)}$ and $\text{(4)}$ in $\text{(2)}$ and $\text{(1)}$, we see that the intersection point is equal to: