Question

Find the equation of planes parallel to the plane $2x-4y+4z=7$ and which are at a distance of five units from the point $(3,-1,2).$

Answer 1

The given plane is $2x-4y+4z=7$ ..... (i)

Equation of any plane parallel to (i) is
$2x-4y+4z+k=0$ ...... (ii)

Now (ii) is at a distance of 5 units from the point (3, -1, 2) if

$\frac{|2\times 3-4.(-1)+4\times 2+k|}{\sqrt{(2{)}^2+(-4{)}^2+(4{)}^2}}=5$

$\Rightarrow \frac{|18+k|}{6}=5$

$\Rightarrow |18+k|=30$

$\Rightarrow 18+k=30$

or $18+k=30$

$\Rightarrow k=12$

or $k=-18-30=-48$

Substituting these values of k in (ii), the equations of the required planes are

$2x-4y+4z+12=0$

and $2x-4y+4z-48=0$

or $x-2y+2z+6=0$

and $x-2y+2z-24=0$