Question

Find the image of the point (0, 0, 0) in the plane 3x + 4y − 6z + 1 = 0.

Answer 1

$$\begin{gathered} \text{Let}\mathit{\text{Q}}\text{be the image of the point}\mathit{\text{P}}\text{(0, 0, 0) in the plane}3x+4y-6z+1=0. \\ \text{Then,}\mathit{\text{PQ}}\text{\hspace{0.17em}is normal to the plane. So, the direction ratios of}\mathit{\text{PQ}}\text{are proportional to 3, 4, -6.} \\ \text{Since}\mathit{\text{PQ}}\text{\hspace{0.17em}passes through}\mathit{\text{P}}\text{(0, 0, 0) and has direction ratios proportional to}\mathrm{3,\; 4}\text{and}-6,\text{equation of}\mathit{\text{PQ}}\text{\hspace{0.17em}is} \\ \frac{x-0}{3}=\frac{y-0}{4}=\frac{z-0}{-6}=r\text{(say)} \\ \text{Let the coordiantes of}\mathit{\text{Q}}\text{be}\left(3r,4r,-6r\right)\text{. Let}\mathit{\text{R}}\text{be the mid-point of}\mathit{\text{PQ}}\text{. Then,} \\ R=\left(\frac{0+3r}{2},\frac{0+4r}{2},\frac{0-6r}{2}\right)=\left(\frac{3r}{2},2r,-3r\right) \\ \text{Since}\mathit{\text{R}}\text{lies in the plane}3x+4y-6z+1=0, \\ 3\left(\frac{3r}{2}\right)+4\left(2r\right)-6\left(-3r\right)+1=0 \\ \Rightarrow r=\frac{-2}{61} \\ \text{Substituting this in the coordinates of}\mathit{\text{Q}}\text{, we get} \\ Q=\left(3r,4r,-6r\right)=\left(3\left(\frac{-2}{61}\right),4\left(\frac{-2}{61}\right),-6\left(\frac{-2}{61}\right)\right)=\left(\frac{-6}{61},\frac{-8}{61},\frac{12}{61}\right) \end{gathered}$$