Projection of skew lines on plane of common perpendicular

Question

Suppose $L_1$ and $L_2$ are two skew lines in the $\mathbb{R}^3$. Draw the common perpendicular line of them $L_3$. The plane $P$ is parallel to the $L_3$. What is the relative position of orthogonal projection of $L_1$ and $L_2$ on the plane $P$? I think the answer is parallel but I can't prove it. I'm looking for an analytic answer to this problem.

Answer 1

Let consider

• $v_1$ direction vector of $L_1$

• $v_2$ direction vector of $L_2$

then

• $v_3=v_1\times v_2$ is such that $v_3$ is a direction vector of $L_3$.

Let consider $P: ax+by+cz=0$ then $n=(a,b,c)$ is such that $n\cdot v_3=0$ then $n \in$ plane spanned by $v_1$ and $v_2$.

Now consider the ortogonal basis $\{n,v_3,n\times v_3\}$ therefore in that basis for some coefficients $a_1,a_2,b_1,b_2$ we have

• $v_1=a_1n+b_1(n\times v_3)$

• $v_2=a_2n+b_2(n\times v_3)$

therefore the orthogonal projection of $L_1$ and $L_2$ onto $P$ have direction vectors parallel to $n\times v_3$ and then they are parallel.