What is the Intersection and Plane Determination for Parametric Lines?

[knowLittle]

Homework Statement

Find the point of intersection of the lines r(t)=< 2t+1, 3t+2, 4t+3> and
x=s+2
y=2s+4
z=-4s-1
Then, find the plane determined by these lines.

Homework Equations

Intersection is when points meet.
So, just equating x,y, and z variables will yield the point of intersection.

The Attempt at a Solution

The point of intersection is P(1,2,3), but I'm clueless on how to find the plane with two parametric equations. It's very simple to find the plane, when they are not parametric.

Could someone help?

 

[AdkinsJr]

You need to find the normal vector n between the two vectorys r(t) and p(t)=<s+2, 2s+4,-4s-1> by taking their cross product. Then just plug that into the equation of a plane:

n\mdot (r-p) where r and p are the vectors you have...

 

[knowLittle]

That's my problem. How do I find the cross product of two parametric equations?

 

[knowLittle]

Could someone help, please?

I've been working on it and looking for info on the net, but I have been unsuccessful.

Thank you.

 

[Ericv_91]

You will have to change the format of the two equations. An easier way to look at it would be in vector form:

L1(s)= (2,4,-1)+ s(1,2,-4)
L2(t)= (1,2,3)+ t(2,3,4)

then take the cross product of the vector portion of those two equations to find the normal line of the plane. Your cross product should look something like

| i j k |
| 1 2 -4 |= vector for line perpendicular to plane
| 2 3 4 |

Hope that helps a bit

 

[knowLittle]

It does help. It's interesting, the vector directions are use as if they were in a plane to find the "normal" vector passing through them.

 

[Ericv_91]

That's right :)