# Vectors Please find the equation of a plane that passes the point (0,1,1) and is equidistant from plane X+y+z=4.

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### 1 Answer

The question is really asking for a plane parallel to the plane:

x+y+z=4

which passes through the point (0,1,1).

To get the equation of a plane, you want a vector normal to the plane, and a point on the plane. We already have a point on the plane (0,1,1).

To get the normal vector of the plane, look at the coefficients of the x, y, and z, in the equation. Your normal vector is <1,1,1>

(If your plane were 2x+3y+4z=5, the normal vector would be <2,3,4> )

The equation of the plane, then, is:

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

The outer multipliers are the normal vector, the subtracted pieces are the point. The right side is always 0.

This simplifies to:

`x+y+z=2`