Prove the equation for the point of intersection of two vector lines

Question

This is my first question here and I'm not sure if it's the right place but I'm kinda desperate. I was learning about vector lines and parametric equations. My teacher gave me the assignment below and it is nothing like I saw in my lessons and when I asked him for help he didn't really help (it's an online school so communication is difficult). Here is the question word for word:

*Prove that (if exists) the point of intersection between two lines L1: r→=r01→+tu1→ , t∈R and L2: r→=r02→+su2→ , s∈R is given by the vector formula:

r→=r01→+[( r02→-r01→ )u2→] (u1→×u2→)u1→×u2→2 u1→*

Like I said, nothing similar to that was talked about in the lessons and I really have no idea of what to do or how to prove it so I was hoping someone could help me here. The arrows in the equations above mean that they are vectors.

Answer 1

First, you write your lines into vector form or in homogeneous coordinate. According to the definition of the cross product, if the two lines are non-parallel then the resulting vector $\mathbf{p}=\mathbf{a}\times\mathbf{b}$ is perpendicular to both $\mathbf{a}$ and $\mathbf{b}$. In addition from the definition of dot product, the dot product between the vertical vectors is $0$, so you can get: $$\mathbf{a}^\top\cdot(\mathbf{a}\times\mathbf{b})=\mathbf{a}^\top\cdot\mathbf{p}=0$$ $$\mathbf{b}^\top\cdot(\mathbf{a}\times\mathbf{b})=\mathbf{b}^\top\cdot\mathbf{p}=0$$ Therefore, the point $\mathbf{p}$ lies on the plane of $\mathbf{a}$ and $\mathbf{b}$, and on the line $\mathbf{a}$ and $\mathbf{b}$, and $\mathbf{p}$ is the intersection of the two lines. Note $\mathbf{p}$ is the homogeneous coordinates.

Answer 2

Hint.

Assuming your lines are in $\mathbb{R}^2$ we have

$$ \cases{ L_1\to r = r_{01}+t u_1\\ L_2\to r = r_{02}+s u_2 } $$

the possible intersection happens when $r_{01}+t u_1 = r_{02}+s u_2$ and now multiplying conveniently we have

$$ u_1\cdot r_{01}+t\|u_1\|^2 = u_1\cdot r_{02}+su_1\cdot u_2\\ u_2\cdot r_{01}+t u_2\cdot u_1 = u_2\cdot r_{02}+s\|u_2\|^2\\ $$

and now we can try to solve for $t,s$

$$ \left( \begin{array}{cc} \|u_1\|^2 & u_1\cdot u_2\\ u_1\cdot u_2 & \|u_2\|^2 \end{array} \right) \left( \begin{array}{c} t\\ s \end{array} \right)=\left( \begin{array}{c} u_1\cdot(r_{02}-r_{01})\\ u_2\cdot(r_{02}-r_{01}) \end{array} \right) $$

NOTE

If $u_1 = \lambda u_2$ then $\det\left( \begin{array}{cc} \|u_1\|^2 & u_1\cdot u_2\\ u_1\cdot u_2 & \|u_2\|^2 \end{array} \right) = 0$