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Find a Cartesian equation for a plane parallel to a given plane and equidistant from a given point

by RoRi

Consider a plane given by the equation

    \[ 2x - y + 2z + 4 = 0. \]

Find the Cartesian equation for a plane parallel to this one and the same distance as this plane from the point (3,2,-1).

Since the requested plane is parallel to the given plane we know that they must have the same normal vector, N = (2,-1,2). Therefore, the Cartesian equation of the requested plane is of the form

    \[ 2x - y + 2z + d = 0. \]

From the previous exercise (Section 13.17, Exercise #19) we know the distance from (3,2,-1) to a plane ax + by + cz + d = 0 is given by the formula

    \[ \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2+b^2+c^2}}. \]

Therefore, the distance from the given plane to the point (3,2,-1) is

    \[ \frac{|6-2-2+4|}{\sqrt{9}} = 2. \]

Since the distance from the point to the requested plane must be the same we must have

    \[ \frac{|6-2-2+d|}{3} = 2 \quad \implies \quad d = -8. \]

(Since the d = 4 solution belongs to the other plane.)

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