# Thread: Cartesian equation containing the two lines

1. ## Cartesian equation containing the two lines

The questions asks to find the Cartesian equation of the lines r=<1,1,0>+t<1,-2,2> and r=<2,0,2>+t<-1,1,0>

Just trying to work out how to solve. Is it okay to take the point vector and direction vector of each line equation above and do a cross product on them to get a vector normal ie <1,1,0> x <1,-1,2> = <2,-2,-2>= n

than a dot product <2,-2,-2> . <x-1, y-1, z-0>=0
to get the Cartesian equation of the first line
and similarly do the same for the second line

Than elimate x from first equation, to get expression for y etc

2. Originally Posted by Craka
The questions asks to find the Cartesian equation of the lines r=<1,1,0>+t<1,-2,2> and r=<2,0,2>+t<-1,1,0>

Just trying to work out how to solve. Is it okay to take the point vector and direction vector of each line equation above and do a cross product on them to get a vector normal ie <1,1,0> x <1,-1,2> = <2,-2,-2>= n Where came these vectors from?

than a dot product <2,-2,-2> . <x-1, y-1, z-0>=0 This is the equation of a plane passing through (1,1,0)
to get the Cartesian equation of the first line
and similarly do the same for the second line

Than elimate x from first equation, to get expression for y etc
Then you'll get maybe the intersection of 2 planes which could be a line, but not the equation of the given lines.
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3. Originally Posted by Craka
The questions asks to find the Cartesian equation of the lines r=<1,1,0>+t<1,-2,2> and r=<2,0,2>+t<-1,1,0>

Just trying to work out how to solve. Is it okay to take the point vector and direction vector of each line equation above and do a cross product on them to get a vector normal ie <1,1,0> x <1,-1,2> = <2,-2,-2>= n

than a dot product <2,-2,-2> . <x-1, y-1, z-0>=0
to get the Cartesian equation of the first line
and similarly do the same for the second line

Than elimate x from first equation, to get expression for y etc
Exactly what does the question ask? The Cartesian equations of the lines are simply
1) x= 1+ t, y= 1- 2t, z= 2t and
2) x= 2- t, y= t, z= 2.

If you are asking for the equation of the plane containing both lines, take the cross product of <1, -2, 2> and <-1, 1, 0> to find a vector perpendicular to the plane and, having determined that the lines are not skew so that such a plane does exist, do as you say.