Question

The image of the point (1, 3, 4) in the plane 2x − y + z + 3 = 0 is
(a) (3, 5, 2)
(b) (−3, 5, 2)
(c) (3, 5, −2)
(d) (3, −5, 2)

Answer 1

(b) (−3, 5, 2)

$$\begin{gathered} \text{Let}\mathit{\text{Q}}\text{be the image of the point}\mathit{\text{P}}\text{(1, 3, 4) in the plane}2x-y+z+3=0 \\ \text{Then}\mathit{\text{PQ}}\text{\hspace{0.17em}is normal to the plane. So, the direction ratios of}\mathit{\text{PQ}}\text{are proportional to 2, -1, 1.} \\ \text{Since}\mathit{\text{PQ}}\text{\hspace{0.17em}passes through}\mathit{\text{P}}(1,\; 3,\; 4)\text{and has the direction ratios proportional to}\mathrm{2,\; -1,\; 1}.,\text{equation of}\mathit{\text{PQ}}\text{\hspace{0.17em}is} \\ \frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}=r\text{(say)} \\ \text{Let the coordinates of}\mathit{\text{Q}}\text{be}\left(2r+1,-r+3,r+4\right)\text{. Let}\mathit{\text{R}}\text{be the mid point of}\mathit{\text{PQ}}\text{. Then,} \\ R=\left(\frac{2r+1+1}{2},\frac{-r+3+3}{2},\frac{r+4+4}{2}\right)=\left(r+1,\frac{-r+6}{2},\frac{r+8}{2}\right) \\ \text{Since}\mathit{\text{R}}\text{lies in the plane}2x-y+z+3=0, \\ 2\left(r+1\right)-\left(\frac{-r+6}{2}\right)+\frac{r+8}{2}+3=0 \\ \Rightarrow 4r+4+r-6+r+8+6=0 \\ \Rightarrow 6r+12=0 \\ \Rightarrow r=-2 \\ \text{Substituting this in the coordinates of}\mathit{\text{Q}}\text{, we get} \\ Q=\left(2r+1,-r+3,r+4\right).=\left(2\left(-2\right)+1,2+3,-2+4\right)=\left(-3,5,2\right) \end{gathered}$$
So, the answer is (b).