Question

Find the coordinates of image of point (1,3,4) in the plane 2x-y+z+3=0

Answer 1

Hey !!

Given equation of plane is

     2x - y + 3 = 0 -----------> (1)

Let Q be the image of the point P (1,3,4) in the plane 2x - y + z + 3 = 0, then PQ is normal to the plane.

Therefore, direction cosines of PQ are proportional to 2 , -1  and 1.

Therefore, equation of PQ is

               $\frac{x-1}{2}= \frac{y-3}{-1}=\frac{z-4}{1}=r$

∴ Co-ordinates of Q are

              x = 2r + 1

              y = -r + 3

              z = r + 4

Q is (2r + 1 , -r + 3 , r + 4)

Let R be the mid-point of PQ which lies on the plane 2x - y + z + 3 = 0, then the co-ordinates of R are

$(\frac{2r + 1 + 1}{2} , \frac{-r + 3 + 3}{2} , \frac{r + 4 + 4}{2} )$

or, R $( \frac{2r + 2}{2} ) - \frac{-r + 6}{2} , \frac{r+8}{2} )$

R lies on the plane 2x - y + z + 3 = 0

∴  $2 (\frac{2r + 2 }{2} ) - (\frac{-r + 6}{2} ) - (\frac{r+8}{2} ) + 3 = 0$

=> 2(2r + 2) - (-r + 6) + (r + 8) + 6 = 0

=> (4r + 4) + (r - 6) + (r + 8) + 6 = 0

=>   6r + 12 = 0

=> $r = \frac{-12}{6}$

=> r = - 2

Now, putting the value of r = - 2 in the co-ordinates of Q

           x = 2 (-2) + 1 = - 3

           y = - (-2) + 3 = 5

           z = - 2 + 4 = 2

∴ Co-ordinates of Q are (-3 , 5, 2 )

Hence, image of P in the plane 2x - y + z + 3 = 0 is Q (-3 , 5 , 2 )

GOOD LUCK !!