Discussion Forum

Let P be a point in the first octant , whose image Q in the plane x+y= 3 (that is, the line segment PQ is perpendicular to the plane x+y=3 and the mid- point of PQ lies in the plane x+y=3) lies on the Z-axis. Let the distance of P from the X-axis be 5. If R is the image of P  in the XY-plane, then the length of PR is .....

Answer:

Explanation:

Let $P\left( \alpha, \beta, \gamma\right)$  and R  is the image of P in the XY - plane.

$\therefore$  $R\left( \alpha, \beta, \gamma\right)$ 

   Also, Q is the image of P in the plane  x+y=3

$\therefore$   $\frac{x-\alpha}{1}= \frac{y-\beta}{1}=\frac{z-\gamma}{0}$

                     = $\frac{-2(\alpha+\beta-3)}{2}$

           $x=3-\beta ,y=3-\alpha ,z=\gamma$

 Since , Q is lies on Z-axis

  $\therefore$  $\beta =3, \alpha =3 , z=\gamma$

   $\therefore$  $P (3,3, \gamma )$

Given , distance of  P  from  X-axis  be 5

  $\therefore$            $5= \sqrt{3^{2}+\gamma^{2}}$

                       $25-9=\gamma^{2}$

$\Rightarrow$     $\gamma \pm 4$

Then  , PR = $\mid2\gamma \mid =\mid 2\times 4\mid=8$