Question

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

Answer 1

The correct option is A (-3,5,2)


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 line is $\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}=r,\left(say\right)\therefore x=2r+1,y=-r+3,z=r+4$
So, co-ordinates of Q be (2r+1, -r+3, r+4)
Let R be the mid point of PQ, then co-ordinates of R are $(\frac{2r+1+1}{2},\frac{-r+3+3}{2},\frac{r+4+4}{2})$
i.e. $(r+1,\frac{-r+6}{2},\frac{r+8}{2})$
Since R lies on the plane
$\therefore 2(r+1)-\left(\frac{-r+6}{2}\right)+\left(\frac{r+8}{2}\right)+3=0$ $\Rightarrow r=-2$
So, co-ordinates of Q are (-3,5,2).
Trick: From option(a), mid point of (-3,5,2) and (1,3,4) satisfies the equation of Q are (-3,5,2).