Question

Find the image of the point (1, 3, 4) in the plane 2x − y + z + 3 = 0.

 

Answer 1

\[\text{ Let Q be the image of the point P(1, 3, 4) in the plane 2x - y + z + 3 = 0 }  . \]
\[\text{ Then PQ is normal to the plane. So, the direction ratios of PQ are proportional to 2, -1, 1.} \]
\[\text{ Since PQ  passes through P (1, 3, 4) and has direction ratios proportional to 2, -1 and 1 , equation of PQ is } \]
\[\frac{x - 1}{2} = \frac{y - 3}{- 1} = \frac{z - 4}{1} = r (say)\]
\[\text{ Let the coordinates of Q be } \left( 2r + 1, - r + 3, r + 4 \right). \text{ Let R be the mid-point of PQ. Then,} \]
\[R = \left( \frac{2r + 1 + 1}{2}, \frac{- r + 3 + 3}{2}, \frac{r + 4 + 4}{2} \right) = \left( r + 1, \frac{- r + 6}{2}, \frac{r + 8}{2} \right)\]
\[\text{ Since R lies in the plane } 2x - y + z + 3 = 0, \]
\[2 \left( r + 1 \right) - \left( \frac{- r + 6}{2} \right) + \frac{r + 8}{2} + 3 = 0\]
\[ \Rightarrow 4r + 4 + r - 6 + r + 8 + 6 = 0\]
\[ \Rightarrow 6r + 12 = 0\]
\[ \Rightarrow r = - 2\]
\[\text{ Substituting this in the coordinates of Q, we get } \]
\[Q = \left( 2r + 1, - r + 3, r + 4 \right) . = \left( 2 \left( - 2 \right) + 1, 2 + 3, - 2 + 4 \right) = \left( - 3, 5, 2 \right)\]