Find the equation of the plane (if it exists) containing the two lines:

A) R1(t)= <1-t, t, 3+2t> ; R2(t)= <t, 1+t, 5-2t>

B) R1(t)= <1+3t, -2t, -1+t> ; R2(t)= <6t,1-4t,3+2t>

C) R1(t)= <t, 1+t, 5+t> ; R2(t)= <-2+t, t, 7-2t>

Thanks!

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- March 11th 2009, 12:37 AMjffyxEquation of The Plane
Find the equation of the plane (if it exists) containing the two lines:

A) R1(t)= <1-t, t, 3+2t> ; R2(t)= <t, 1+t, 5-2t>

B) R1(t)= <1+3t, -2t, -1+t> ; R2(t)= <6t,1-4t,3+2t>

C) R1(t)= <t, 1+t, 5+t> ; R2(t)= <-2+t, t, 7-2t>

Thanks! - March 11th 2009, 07:24 AMmr fantastic
Take the cross product of the vectors in the direction of each line to get a normal to the plane. Get a point in the plane by choosing a point on one of the lines.

You should know how to get the equation of a plane in the form ax + by + cz = d when you have a point in the plane and a normal to the plane. - March 11th 2009, 02:01 PMjffyx
- March 11th 2009, 02:32 PMrunning-gag
Hi

There can be different situations

1) the lines are parallel but not equal => one plane is containing the 2 lines

Take one point on the first line, one point on the second line, form a vector with these 2 points and make the cross product of this vector with the direction vector of the lines

This will give a normal vector to the plane

2) the lines are parallel and equal => an infinite number of planes is containing the line

3) the lines are secant (and not equal) => one plane is containing the 2 lines

Make the cross product of the 2 direction vector of the lines

This will give a normal vector to the plane, giving the form of its Cartesian equation

Define the point which is on the 2 lines, this will give you a complete equation of the plane

4) the lines are not parallel and not secant => no plane is containing the 2 lines

You have first to define the status of the 3 cases you have to treat (1, 2, 3 or 4)

Reminder : if you know a normal vector (a,b,c) to a plane then one Cartesian equation of the plane is ax+by+cz+d=0. To find d you have to chose one point on the plane.