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  • Find the coordinates of the foot of perpendicular and perpendicular distance from the point    to the plane <img alt="x+2y+3z=2" src="https:/

Find the coordinates of the foot of perpendicular and perpendicular distance from the point  P(4,3,2)  to the plane x+2y+3z=2. Also find the image of P in the plane.

 

 

 

 
 
 
 
 

Answers (1)

Let \pi : x+2y+3z=2

\therefore  The direction of its normal are 1,2,3

Draw  PR \perp plane\; \; \pi

So equation of PR : 

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

Coordinates of any random point on the line PR is R(\lambda +4,2\lambda +3,3\lambda +2)

As R lies on the plane so,

\lambda +4+(2\lambda +3)2+3(3\lambda +2)=2\Rightarrow \lambda =-1

So, R must be the mid point of PQ.

That is R\left ( \frac{h+4}{2},\frac{p+3}{2},\frac{s+2}{2} \right )=R(3,1,-1)

On comparing the coordinates, \frac{h+4}{2}=3,\frac{p+3}{2}=1,\frac{s+2}{2}=-1

\Rightarrow h=2,p=-1,s=-4

Hence the image is Q(2,-1,-4)

 

Posted by

Ravindra Pindel

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