Question 1011829: "The lines l1 and l2 have equations r=2i+k+s(i-j+2k) and r=3i+8j+t(i+2j-3k) respectively, where s and t are parameters.
The plane p contains l1 and has no common point with l2. Find an equation of p in scalar product form."
I know that if p has no common point with l2, it means l2 is parallel to p. I have read the solution, and it says we need to cross product (i-j+2k) x (i+2j-3k). But how can we cross product (i-j+2k) x (i+2j-3k) ? Doesn't two parallel vector cannot be crossed ? I need your help and further explanation please.
Thank you very much.
Answer by mathmate(429)   (Show Source):
You can put this solution on YOUR website!
Question:
"The lines l1 and l2 have equations r=2i+k+s(i-j+2k) and r=3i+8j+t(i+2j-3k) respectively, where s and t are parameters.
The plane p contains l1 and has no common point with l2. Find an equation of p in scalar product form."
I know that if p has no common point with l2, it means l2 is parallel to p. I have read the solution, and it says we need to cross product (i-j+2k) x (i+2j-3k). But how can we cross product (i-j+2k) x (i+2j-3k) ? Doesn't two parallel vector cannot be crossed ? I need your help and further explanation please.
Solution:
We have
L1=2i+k+s(i-j+2k)=‹s+2,-s,2s+1›
L2=3i+8j+t(i+2j-3k)=‹t+3,2t+8,-3t›
Vectors parallel to L1 and L2 are:
V1=<1,-1,2>
V2=<1,2,-3>
We see clearly that V1 and V2 are not parallel.
The idea is to pass a plane P through L1 and rotate it in such a direction that the plane is parallel to V2.
We need to find the normal vector to the plane, which is perpendicular to both L1 and L2, thus we use cross product of V1 and V2:
n=V1×V2 = <-st,5st,3st>
which can be shortened to
N=<-1,5,3>
We know that L1 passes through (2,0,1) (=L1-sV1)
The plane perpendicular to N and passing through has the equation:
P(x,y,z)=-(x-2)+5y+3(z-1)=0, or
x-5y-3z+1=0........... plane P.
Check: that P is parallel to L2
we need to show that N is orthogonal to L2, or
N.V2=<-1,5,3>.<1,2,-3>=-1+10-9=0 => P is parallel to L2.
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