If the image of the point P(1, --2, 3) in the plane, 2x + 3y -- 4z + 22 = 0 measured parallel to the line,
\[{x \over 1} = {y \over 4} = {z \over 5}\] is Q, then PQ is equal to:
Solutions · 1
Official solution
Equation of line PQ is \[{{x - 1} \over 1} = {{y + 2} \over 4} = {{z - 3} \over 5}\]
Let F be (\[\lambda \] + 1, 4\[\lambda \] - 2, 5\[\lambda \] + 3)
Since F lies on the plane
\[ \therefore \] 2(\[\lambda \] + 1) + 3(4\[\lambda \] - 2) - 4(5\[\lambda \] + 3) + 22 = 0
\[ \Rightarrow \] - 6\[\lambda \] + 6 = 0 \[
\Rightarrow \] \[\lambda \] = 1
\[ \therefore \] F is (2, 2, 8)
PQ = 2 PF = 2\[\sqrt {{1^2} + {4^2} + {5^2}} \] = 2\[\sqrt {42} \]