Question 1134088: line L1 and L2 has equation (x-3)/2 = (y-4)/-1 = (z+1)/1 and (x-5)/4 = (y-1)/3 = (z-1)/2. find Cartesian equation of the plane P which contain L1 and parallel to L2. L3 pass through point A(-3,-2,-1) and meets P at B(-1,2,1). find Cartesian equation of L3 and acute angle between P and L3 (in nearest degree). Another line L4 with Cartesian equation (x-2)/1 = (y+3)/2 = (z+2)/1 pass through plane P. find intersection point.
Answer by Edwin McCravy(20054)   (Show Source):
You can put this solution on YOUR website! L1: 
L2: 
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.
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 > =
 =
< -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
-----------------------
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
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