# Vector/Geometry toughie

Find parametric equations of a line that intersects both l1 and l2 at right angles:

l1= [x,y,z]=[4,8,-1] + t[2,3,-4]
l2= x-7/-6 = y-2/1 = z+1/2

I found the cross product of l1 and l2 to get a normal vector perpendicular to both, which is [10, 20, 20] or n=[1,2,1]

Now I am not sure how to get the parametric equation for the line because I cannot just use one of the vectors from above, like [4,8,-1] because it is not necessarily on l3.

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Find parametric equations of a line that intersects both l1 and l2 at right angles:

l1= [x,y,z]=[4,8,-1] + t[2,3,-4]
l2= x-7/-6 = y-2/1 = z+1/2

I found the cross product of l1 and l2 to get a normal vector perpendicular to both, which is [10, 20, 20] or n=[1,2,1]
How about this? You already know now that a line L3 parallel to the vector <1,2,1> must intersect both L1 and L2. Picture your vector sticking up from, say, L1 at a point (x,y,z). For any value of the parameter t, the prospective line L3 will have the equation

(x,y,z) + <1,2,1>·v = [ (4,8,-1) + <2,3,-4>·t ] + <1,2,1>·v ,

where v is the parameter on line L3. Now, for some value of t, as we slide our candidate L3 along, it's supposed to intersect line L2. If we call its parameter s, this intersection will be given by

[ (4,8,-1) + <2,3,-4>·t ] + <1,2,1>·v = (7,2,-1) + <-6,1,2>·s ,

after converting L2's symmetric equation to parametric equations. We now have a vector equation corresponding to a system of three equations in three unknowns. (Not much worse than what you have to do for the skewness/intersection test for two lines in three-dimensional space.) Once you have
t, s, and v , you will know the points where each pair L1, L3 and L2, L3 meet and the value of parameter v where L3 intersects L2. This will give a complete description of the situation, including information for writing the parametric form for L3.

It also occurred to me that if L3 meets L1 and L2 at right angles, it must meet those lines at the points on L1 and L2 that are at the minimal separation between those lines, since the segment of L3 linking them marks the perpendicular distance between L1 and L2. But the reckoning for that looks hideous...

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