• # question_answer Image point of $(1,\,3,4)$ in the plane $2x-y+z+3=0$ is A)            (? 3, 5, 2) B)            (3, 5, ? 2) C)            (3, ? 5, 3) D)            None of these

Let Q be image of the point $P(1,3,\,4)$ in the given plane, then PQ is normal to the plane.                    The d.r.'s of PQ  are 2, ?1,1                    Since PQ passes through (1, 3, 4) and has d.r's 2, 1, ?1; therefore, equation of plane is $\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}=r$, (say)                    $\therefore$ $x=2r+1,\,y=-r+3,\,z=r+4$                    So, co-ordinates of Q be                    $(2r+1,\,-r+3,\,r+4)$                                       Let R be the mid point of PQ, then co-ordiantes of R are $\left( \,\frac{2r+1+1}{2},\,\frac{-r+3+3}{2},\,\frac{r+4+4}{2} \right)$                    i.e., $\left( r+1,\,\frac{-r+6}{2},\,\frac{r+8}{2} \right)$                    Since R lies on the plane                    $\therefore$ $2(r+1)-\left( \frac{-r+6}{2} \right)+\left( \frac{r+8}{2} \right)+3=0$ Þ $r=-2$                    So, co-ordinates of Q are (?3, 5, 2).                    Trick : From option , mid  point of (?3, 5, 2) and   (1,3,4) satisfies the equation of plane $2x-y+z+3=0$.