Find parametric equations of a line that intersects line 1 and line 2 at right angles.
Line 1: $[x,y,z] = [4,8,-1] + t[2,3,-4]$ and
Line 2: $[x,y,z] = [7,2,-1] + k[-6,1,2]$.
I've tried solving this problem by crossing the direction vectors to get the direction of the normal, but I don't know how to find a point on the normal to get the overall equation of it. I originally thought I could just find the point of intersection of the 2 lines to get the intersection point, but the 2 lines turned out to be skew. Please help.
This question relates to a particular case of a general case that is dealt with here for example.
Let $\color {grey}{\mathcal P}=\mathbb R \color {green}{(2,3,-4)}\oplus \mathbb R \color{red}{(-6,1,2)}, X=(4,8,-1), Y=(7,2,-1)$ $$(2,3,-4)\times (-6,1,2)=(10,20,20)=10(1,2,2)\text{(cross product)}$$
The numbers offered are well chosen to avoid excessively complicated calculations. We obtain $$X-Y=(4,8,-1)-(7,2,-1)=(-3,6,0)=\color{green}{(+1)}(2,3,-4)+(-6,1,2)+(1,2,2)$$ So, In accordance with here, we have with the notations in the figure$$x_1=X\color{green}{-1}(2,3,-4)=(2,5,3)$$
And parametric equation of the line $l $ that intersects line 1 and line 2 at right angles is $$\boxed{l:(2,5,3)+t(1,2,2), t\in \mathbb R}$$
In addition, we have $$y_1=Y+(-6,1,2)=(1,3,1)$$ And we check that $$x_1-y_1=1.(1,2,2)$$