# Thread: Vestors - finding an equaiton of a plane

1. ## Vestors - finding an equaiton of a plane

HI,

I have the plane equation that is

2x-y+4z =2

and have to find the equation of the plane that is parallel to this equation and passes through the points (2, 1, -1)

To work this out do i just cross multiply the (2,-1,4)x(2,1,-1) and this is the equation of parallel plane?

2. Originally Posted by kayne
I have the plane equation that is 2x-y+4z =2
and have to find the equation of the plane that is parallel to this equation and passes through the points (2, 1, -1)
For any k the plane $2x-y+4z=k$ will be parallel to the given plane.
To find k subsititute the point.

3. plane equation $2x-y+4z =2\ normal\ vector\ \begin{bmatrix} 2\\-1\\4 \end{bmatrix}$
Any plane with this normal vector will parallel with it.
So new plane equation : $2(x-2)-(y-1)+4(z+1) =0 \rightarrow 2x-y+4z=-1$