I have two lines:
$ r_l:\begin{cases}lx-y=1 \\ x+ly-z=1 \end {cases}\ $
and
$s_l:\begin{cases}x+y=0 \\ y-lz=-1 \end {cases}\ $
and $l $ is a variable. As $ l $ varies, I have to find their relative position.
I'm guessing that you need to change them to parametric form, equate the equations and do some algebra manipulation, but the last point is where I'm having difficulties.
Can someone help me?
Thanks!
Best Answer
What do you mean by "relative position"? Whether they are skew, parallel, or intersect? And if they intersect, the angle between them?
You can always use one of the variables themselves as a parameter.
For the first line, from lx- y= 1 we get y= lx- 1. Then x+ ly- z= 1 gives z= x+ ly+ 1= x+ l^2x- l+ 1= (1+ l^2)x. Letting x= t we have the parametric equations x= t, y= lt- 1, z= (1+ l^2)t. A tangent vector to the line is given by <1, l, 1+ l^2>.
For the second line, from x+ y= 0 we get y= -x. Then from y- lz= -1, lz= y+ 1= -x+ 1 so z= -x/l+ 1/l. Letting s= x/l (because I don't like fractions) x= ls, y= -ls, z= -s+ 1/l. A tangent vector to the line is given by .
In order for the two lines to be parallel (or the same line) those tangent vectors would have to be in the same direction- one must be a multiple of the other so we must have 1= al, l= -al, and 1+ l^2= -a for some number a. From l= -al, a= -1. Then 1= al= -l so l= -1. 1+ l^2= 1+ 1= 2 which is NOT equal to -a= 1. So the two lines are not the same line nor are they parallel. The lines either cross or are skew.
To determine whether they cross we need to try to solve the equations x= t= ls, y= lt- 1= -ls, and z= (1+ l^2)t= -s+ 1/l for s and t. Since t= ls, lt- 1= l^2s- 2= -ls so (l^2+l)s= 2, s= 2/(l^2+ 1) and t= ls= 2l/(l^2+ 1). Then the third equation, for z, becomes (1+ l^2)t= (1+ l^2)[2l/(l^2+ 1)= 2l= -s+ 1/l= -2/(l^2+ 1)+ 1/l= (-2l+ l^2+ 1)/(l(l^2+ 1))= (l- 1)^2/(l^3+ l) which is definitely NOT 2l!
The lines do NOT intersect so must be skew lines.