# Math Help - Equation of a Plane

1. ## Equation of a Plane

Find the equation of the plane containing the line L1 , and parallel to the line L2 , where:

L1: (x,y,z) = (0,0,1) + t (2,-3,1).

L2: is the intersection of the planes -1x -1y +3z = 0 and 2x -2y +1z = 3

Write your answer in the form Ax + By -29z = D , and give the values of A, B and D as your answer.

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I know the first vector in the plane is given in the line L1 as (2, -3, 1). I am not sure how to get the second vector I need to solve the plane equation from the L2 information given.

Does anyone know the correct steps to get a second direction vector created by the line from the two intersecting planes? I know given the line created by the intersection is parallel to the plane that it will give me the second usable diection vector to solve the equation of the plane, just don't know how to go about solving it to get that line.

2. Originally Posted by LostInCalculus
Find the equation of the plane containing the line L1 , and parallel to the line L2 , where:

L1: (x,y,z) = (0,0,1) + t (2,-3,1).

L2: is the intersection of the planes -1x -1y +3z = 0 and 2x -2y +1z = 3

Write your answer in the form Ax + By -29z = D , and give the values of A, B and D as your answer.

************************************************** *

I know the first vector in the plane is given in the line L1 as (2, -3, 1). I am not sure how to get the second vector I need to solve the plane equation from the L2 information given.

Does anyone know the correct steps to get a second direction vector created by the line from the two intersecting planes? I know given the line created by the intersection is parallel to the plane that it will give me the second usable diection vector to solve the equation of the plane, just don't know how to go about solving it to get that line.
The direction vector of the intersection line of the 2 planes must be perpendicular to both normal vectors of the planes. Thus the direction vector is the crossproduct of the 2 normal vectors:

$[-1, -1, 3] \times [2, -2, 1] = [5, 7, 4]$

Therefore the normal vector of the plane is:

$[5, 7, 4] \times [2, -3,1] = [-19, -3, 29]$

The point (0, 0, 1) must lie on the plane.

3. Thanks =)