Find Equation of Plane: Cartesian & Parametric Form

[forty]

Find the equation of the following plane in both Cartesian and (vector) parametric form.

the plane through the point (6,5,-2) and parallel to the plane x+y-z+1=0




Cartesian form is just -x-y+z=d as both planes share the same normal. solving for d using (6,5,-2) gives -x-y+z=-13.

as for the vector form..

r = (6,5,-2) + s(x,y,z) + t(a,b,c) (s,t are real numbers)

How do i work out the (x,y,z) and (a,b,c)? Can i just take the dot product of each of them with the normal vector and find values that give 0. meaning they are at right angles (i hope that's right). and then just make sure that (x,y,z) and (a,b,c) aren't parallel?

e.g.

(-1,-1,1) . (x,y,z) = 0
-x-y+z=0

so (1,0,1) would work? then do the same with the (a,b,c) and say take (0,1,1)??

is this the right thing to do? is there an easier way?

 

[Mark44]

You should be able to find two more points in the plane - you have its equation, so for each point, pick values of x and y and solve for z. Let's call these points A and B.

Now form a vector from P(6, 5, -2) to A and another vector from P to B. These two vectors and your first point will determine the plane. As a check each vector should be perpendicular to the plane's normal.

 

[HallsofIvy]

Find the equation of the following plane in both Cartesian and (vector) parametric form.

the plane through the point (6,5,-2) and parallel to the plane x+y-z+1=0




Cartesian form is just -x-y+z=d as both planes share the same normal. solving for d using (6,5,-2) gives -x-y+z=-13.

Given the form of the original plane, I would have been inclined to write either x+ y-z+ d1= 0 or x+ y- z= d2, but they give the same result as yours: x+y-z-13= 0 or x+y-z= 13.

as for the vector form..

r = (6,5,-2) + s(x,y,z) + t(a,b,c) (s,t are real numbers)

How do i work out the (x,y,z) and (a,b,c)? Can i just take the dot product of each of them with the normal vector and find values that give 0. meaning they are at right angles (i hope that's right). and then just make sure that (x,y,z) and (a,b,c) aren't parallel?

e.g.

(-1,-1,1) . (x,y,z) = 0
-x-y+z=0

so (1,0,1) would work? then do the same with the (a,b,c) and say take (0,1,1)??

is this the right thing to do? is there an easier way?

Simplest thing to do: since -x- y+ z= -13 (your form), solve the equation for z: z= x+ y- 13. Taking x= 0, y= 0, z= -13. Taking x= 1, y= 0, z= -12, so <1-0,0-0,-12-(-13)>= <1, 0, 1> is a vector in the plane. Taking x= 0, y= 1, z= -12 so <0-0, 1-0, -12-13>= <0, 1, 1> is a vector in the plane. The plane can be written r= (0, 0, -13)+ s(1, 0, 1)+ t(0, 1, 1).

 

[forty]

Always helpful!

Just one thing, seeing as in the question they give you a point in the plane "the plane through the point (6,5,-2) and parallel to the plane x+y-z+1=0" wouldn't it be more logical to just use that point then solving for another, ie (0, 0, -13). The only reason I bring this up is to make sure you can use that original point they give in the question!

Thanks again! :-D

 

[Mark44]

Always helpful!

Just one thing, seeing as in the question they give you a point in the plane "the plane through the point (6,5,-2) and parallel to the plane x+y-z+1=0" wouldn't it be more logical to just use that point then solving for another, ie (0, 0, -13). The only reason I bring this up is to make sure you can use that original point they give in the question!

Thanks again! :-D

It's probably no more logical, but it's a lot quicker!