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  2. Find the coordinates of the foot of the perpendicular Q drawn from   to the plane 
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Find the coordinates of the foot of the perpendicular Q drawn from  P\left ( 3,2,1 \right ) to the plane  2x-y+z+1= 0. Also, find the distance PQ and the image of the point P treating this plane as a mirror.

 

 

 

 
 
 
 
 

Answers (1)
R Ravindra Pindel

The d.r's of normal to the plane are 2,-1,1 
\because  PQ is \perp^{r}  to the plane so, its equation is:
\frac{x-3}{2}= \frac{y-2}{-1}= \frac{z-1}{1}= \lambda
The coordinates of any random point on the line PQ
Q\left ( 2\lambda +3,-\lambda +2,\lambda +1 \right )
\because  Q lines on the plane so, 2\left ( 2\lambda +3 \right )-\left ( \; \lambda+2 \right )+\left ( \lambda +1 \right )+1= 0
\Rightarrow 6\lambda +6= 0\; \; \therefore \lambda = -1

\therefore  Foot of the perpendicular : Q\left ( 1,3,0 \right )
Distance  PQ\sqrt{\left ( 3-1 \right )^{2}+\left ( 2-3 \right )^{2}+\left ( 1-0 \right )^{2}}= \sqrt{6}\, units
Let M\left ( \alpha ,\beta ,\gamma \right ) be the image of P in the plane 
so, Q will be mid point of PM
ie  Q\left ( 1,3,0 \right )= Q\left ( \frac{\alpha +3}{2},\frac{\beta +2}{2},\frac{\gamma +1}{2} \right )
On comparing the coordinates, we get  : \alpha = -1,\beta = 4,\gamma = -1
\therefore the image is M\left ( -1,4,-1 \right ).
 

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