Cartesian equation of the plane

1. Jan 17, 2013

yy205001

1. The problem statement, all variables and given/known data
If the line "l" is given by the equations 2x-y+z=0, x+z-1=0, and if M is the point (1,3,-2), find a Cartesian equation of the plane.
a) passing through M and l
b) passing through M and orthogonal to l
2. Relevant equations
(r-r0)n=0

3. The attempt at a solution
I expanded the equation above, so i got rn=r0n
then, (x,y,z)n=(1,3,-2)n.
And i don't know how to get the normal vector from those two equations

Last edited: Jan 17, 2013
2. Jan 17, 2013

Staff: Mentor

can you be more specific? does the plane contain the line or the point? or does it contain both?

3. Jan 17, 2013

yy205001

sorry, i forgot some of them.

a) passing through M and l
b) passing through M and orthogonal to l

4. Jan 17, 2013

Staff: Mentor

and what course? vector algebra?

also you really need to show more work before anyone here at PF will help.

5. Jan 17, 2013

HallsofIvy

Staff Emeritus
I presume you know: If <A, B, C> is a vector normal to a plane and $(x_0, y_0, z_0)$ is a point in the plane, then the Cartesian equation of the plane is $A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0$.

2x- y+ z= 0, x+ z=1. Subtract the second equation from the first to eliminate z: x- y= -1 or x= y- 1. If you let y= 0, x= -1 so x+ z- 1= -1+ z- 1= 0 gives z= 2. (1, 0, 2) is a point on the given line. If you let y= 1 in x= y- 1, x= 0 so x+ z- 1= z- 1= 0 gives z= 1. (0, 1, 1) is a point on the given line. Those, together with (1, 3, -2) give you three points in the plane. <1- 1, 3- 0, -2- 2>= <0, 3, -4> is a vector in the plane. <1- 0, 3- 1, -2- 1>= <1, 2, -3> is another vector in the plane. Their cross product is normal to the plane.

<2, 1, 1> is a vector normal to 2x- y+ z= 0. <1, 0, 1> is a vector normal to x+ z= 1. Their cross product is normal to both so normal to any plane perpendicular to the line. You have a normal vector and a point in the plane.

Last edited: Jan 17, 2013