A plane E is perpendicular to the two planes 2x−2y+z=0 and x−y+2z=4, and passes through the point P(1,−1,1). If the distance of the plane E from the point Q(a,a,2) is \[3 \sqrt{2}\], then (PQ)2 is equal to :
Solutions · 1
Official solution
First plane, P1=2x−2y+z=0, normal vector \[ \equiv {\overline n _1} = (2, - 2,1)\]
Second plane, \[{P_2} \equiv x - y + 2z = 4\], normal vector \[ \equiv {\overline n _2} = (1, - 1,2)\]
Plane perpendicular to P1 and P2 will have normal vector \[{\overline n _3}\]
Where \[{\overline n _3} = \left( {{{\overline n }_1} \times {{\overline n }_2}} \right)\]
Hence, \[{\overline n _3} = ( - 3, - 3,0)\]
Equation of plane E through P(1,−1,1) and \[{\overline n _3}\] as normal vector
\[(x - 1,y + 1,z - 1)\,.\,( - 3, - 3,0) = 0\]
\[ \Rightarrow x + y = 0 \equiv E\]
Distance of PQ(a,a,2) from \[E = \left| {{{2a} \over {\sqrt 2 }}} \right|\]
as given, \[\left| {{{2a} \over {\sqrt 2 }}} \right| = 3\sqrt 2 \Rightarrow a = \, \pm \,3\]
Hence, \[Q \equiv ( \pm \,3, \pm \,3,2)\]
Distance \[7Q = \sqrt {21} \Rightarrow {(PQ)^2} = 21\]