Question 1134088
</pre>L1: {{{(x-3)/2 = (y-4)/(-1) = (z+1)/1}}} 
L2: {{{(x-5)/4 = (y-1)/3 = (z-1)/2}}}

L1 is parallel to the vector < 2,-1,1 > and passes thru (3,4,-1)
L2 is parallel to the vector < 4,3,2 > and passes thru (5,1,1)

find Cartesian equation of the plane P which contain L1 and parallel to L2. 
<pre>
A vector perpendicular to both those lines will be perpendicular (normal) to the
desired plane.

So we find a vector perpendicular to both vectors by crossing them:

< 2,-1,1 > × < 4,3,2 > =
{{{abs(matrix(3,3,i,j,k,2,-1,1,4,3,2))=(-2-3)i-(4-4)j+(6+4)k=-5i+10k}}} = 
< -5,0,10 >

So we want a plane containing the point (5,1,1) and having normal vector
< -5,0,10 >, which is

-5(x - 5)+0(y - 1)+10(z - 1) = 0
-5x + 25 + 10z - 10 = 0
-5x + 10z + 15 = 0
Divide through by -5
x - 2z - 3 = 0
x - 2z = 3

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L3 pass through point A(-3,-2,-1) and meets P at B(-1,2,1). 
Find Cartesian equation of L3 

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I have to stop here because something is wrong because B(-1,2,1) is not
on plane P because substituting B(-1,2,1) into

     x - 2z = 3
(-1) - 2(1) = 3
       -1-2 = 3 
         -3 = 3 is false. 

Could it be that B(-1,2,1) could have supposed to have been B(1,2,-1)?

Check everything again.  If you answer below, I'll get back to you 
by email.

Edwin</pre>