Showing skew lines lie in parallel planes

[mikemichiel]

Im givin these two lines..
L1= x=4+5t y=5+5t z=1-4t
L2= x=4+s y=-6+8s z=7-3s

What i tried doing was taking the directional vector of both lines <5 5 -4> <1 8 -3>, and crossing them to find the normal vector. I have enough information to find 1 equation of a plane, but how can I find the other. Can someone please point me in the right direction. Thanks in advance!

 

[tiny-tim]

Welcome to PF!

Hi mikemichiel! Welcome to PF!

Im givin these two lines..
L1= x=4+5t y=5+5t z=1-4t
L2= x=4+s y=-6+8s z=7-3s

What i tried doing was taking the directional vector of both lines <5 5 -4> <1 8 -3>, and crossing them to find the normal vector. I have enough information to find 1 equation of a plane, but how can I find the other.

You have the normal …

can't you find the plane from that?

And why do you need more than 1 equation for a plane?

 

[HallsofIvy]

To find the equation of a plane, you need a normal vector and a point in the plane. You have two lines that presumably lie in each line in the planes you want. Any point on the line will do.