Question

Find the reflection of the point (1, 2, −1) in the plane 3x − 5y + 4z = 5.

 

Answer 1

\[\text{ Let Q be the image of the point P (1, 2, -1) in the plane 3x - 5y + 4z = 5 .} \]
\[\text{ Then PQ  is normal to the plane. So, the direction ratios of PQ are proportional to 3, -5, 4.} \]
\[\text{ Since PQ  passes through P (1, 2, -1) and has direction ratios proportional to 3 , -5, 4, equation of PQ is } \]
\[\frac{x - 1}{3} = \frac{y - 2}{- 5} = \frac{z + 1}{4} = r (\text{ say } )\]
\[\text{ Let the coordinates of Q be } \left( 3r + 1, - 5r + 2, 4r - 1 \right). \text{ Let R be the mid-point of PQ. Then } ,\]
\[R = \left( \frac{3r + 1 + 1}{2}, \frac{- 5r + 2 + 2}{2}, \frac{4r - 1 - 1}{2} \right) = \left( \frac{3r + 2}{2}, \frac{- 5r + 4}{2}, \frac{4r - 2}{2} \right)\]
\[\text{ Since R lies in the plane } 3x - 5y + 4z = 5, \]
\[3 \left( \frac{3r + 2}{2} \right) - 5 \left( \frac{- 5r + 4}{2} \right) + 4 \left( \frac{4r - 2}{2} \right) = 5\]
\[ \Rightarrow 9r + 6 + 25r - 20 + 16r - 8 = 10\]
\[ \Rightarrow 50r - 32 = 0\]
\[ \Rightarrow r = \frac{32}{50} = \frac{16}{25}\]
\[\text{ Substituting the value of r in the coordinates of Q, we get } \]
\[Q = \left( 3r + 1, - 5r + 2, 4r - 1 \right) = \left( 3 \left( \frac{16}{25} \right) + 1, - 5 \left( \frac{16}{25} \right) + 2, 4 \left( \frac{16}{25} \right) - 1 \right) = \left( \frac{73}{25}, \frac{- 6}{5}, \frac{39}{25} \right)\]