## Equation of a Plane Parallel to a Given Plane

Suppose a plane
$P$
? passes through the point
$(1,3,4)$
and is parallel to the plane
$2x-3y+z=7$
.
How do we find the equation of
$P$
?
Because the planes are parallel they have the same normal. The normal to a plane given in Cartesian form is equal to the coefficients of the equation of the plane, written as a vector, in this case
$\mathbf{n}= \begin{pmatrix}2\\-3\\\end{pmatrix}$
, so both have the same coefficients in the Cartesian form of the plane equation, and the plane
$P$
will have equation
$2x-3y+z=d$
.
$d$
can be found by substituting the point above into the equation.
$d=1(1)-3(3)+4=-4$
.
The equation of the plane is
$2x-3y+z=-4$
.