The problem is :
Find parametric equations of the line L intersecting the given lines L1 and L2 and parallel to the given line L3.
L1: x = 1 + (t1), y = 2 + 2(t1), z = -2 + (t1)
L2: x = 2 + (t2), y = 1 + 2(t2), z = 3 + 3(t2)
L3: x = 1 + 2(t3), y = 1 + 7(t3), z = 1 + 3(t3)
(t1), (t2) and (t3) are in R
Thank you
Best Answer
If two lines intersect, then they are coplanar. In particular, $L$ and $L_1$ must lie on the same plane, as must $L$ and $L_2$, therefore $L$ lies on the intersection of these two planes. The cross product of the direction vectors of each of these pairs of lines is normal to the plane in which they lie, and you can extract a known point on each of $L_1$ and $L_2$ from their parametric equations, so you can easily construct an equation for each of these planes. Once you have the two equations, finding their intersection is also straightforward. Indeed, you only need to find a point in the intersection since you already know what the direction vector of the line must be, but it’s worth going through the entire calculation to ensure that the planes do in fact intersect.
Specifically, we have for the first plane’s normal $(2,7,3)\times(1,2,1)=(1,1,-3)$ and so an equation for it is $$x+y-3z=9.$$ For the second plane, we have the normal $(2,7,3)\times(1,2,3)=(15,-3,-3)$, giving the equation $$5x-y-z=6.$$ These two normals are not parallel, so the planes do intersect in a line. I’ll leave the rest to you.