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Question
Find the image of the point (0, 0, 0) in the plane 3x + 4y − 6z + 1 = 0.
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Solution
\[\text{ Let Q be the image of the point P (0, 0, 0) in the plane 3x + 4y - 6z + 1 = 0 } . \]
\[\text{ Then,PQ is normal to the plane. So, the direction ratios of PQ are proportional to 3, 4, -6. } \]
\[\text{ Since PQ passes through P (0, 0, 0) and has direction ratios proportional to 3, 4 and -6 , equation of PQ is } \]
\[\frac{x - 0}{3} = \frac{y - 0}{4} = \frac{z - 0}{- 6} = r (\text{ say } )\]
\[\text{ Let the coordiantes of Q be} \left( 3r, 4r, - 6r \right). \text{ Let R be the mid-point of PQ. Then },\]
\[R = \left( \frac{0 + 3r}{2}, \frac{0 + 4r}{2}, \frac{0 - 6r}{2} \right) = \left( \frac{3r}{2}, 2r, - 3r \right)\]
\[\text{ Since R lies in the plane 3x + 4y - 6z + 1 = 0,} \]
\[3 \left( \frac{3r}{2} \right) + 4 \left( 2r \right) - 6 \left( - 3r \right) + 1 = 0\]
\[ \Rightarrow r = \frac{- 2}{61}\]
\[\text{ Substituting this in the coordinates of Q, we get } \]
\[Q = \left( 3r, 4r, - 6r \right) = \left( 3 \left( \frac{- 2}{61} \right), 4 \left( \frac{- 2}{61} \right), - 6 \left( \frac{- 2}{61} \right) \right) = \left( \frac{- 6}{61}, \frac{- 8}{61}, \frac{12}{61} \right)\]
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