Let Q be the mirror image of the point P(1, 0, 1) with respect to the plane S : x + y + z = 5. If a line L passing through (1, -1, -1), parallel to the line PQ meets the plane S at R, then QR2 is equal to :
Solutions · 1
Official solution
As L is parallel to PQ d.r.s of S is
\[\therefore\] \[L \equiv {{x - 1} \over 1} = {{y + 1} \over 1} = {{z + 1} \over 1}\]
Point of intersection of L and S be \[\lambda\]
\[ \Rightarrow (\lambda + 1) + (\lambda - 1) + (\lambda - 1) = S\]
\[ \Rightarrow \lambda = 2\]
\[\therefore\] \[R \equiv (3,1,1)\]
Let \[Q(\alpha ,\beta ,\gamma )\]
\[ \Rightarrow {{\alpha - 1} \over 1} = {\beta \over 1} = {{\gamma - 1} \over 1} = {{ - 2( - 3)} \over 3}\]
\[ \Rightarrow \alpha = 3,\,\beta = 2,\,\gamma = 3\]
\[ \Rightarrow Q \equiv (3,2,3)\]
(QR)2=02+(1)2+(2)2=5