A) (? 3, 5, 2)
B) (3, 5, ? 2)
C) (3, ? 5, 3)
D) None of these
Correct Answer: A
Solution :
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 \[2x-y+z+3=0\].You need to login to perform this action.
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