Question

FIND THE COORDINATES OF THE FOOT OF THE PERPENDICULAR AND THE PERPENDICULAR DISTANCE OF THE POINT P(3,2,1) FROM THE PLANE 2X-Y+Z+1=0. FIND ALSO, THE IMAGE OF THE POINT IN THE PLANE

Answer 1

Let Q=foot of the perpendicular from P to the given plane.
Then the dr's of line PQ will be same as Dr's of the given plane.
So, dr's of line = 2,-1,1
So the equation of line having dr's 2,-1,1 and passing through the point (3,2,1) is,
x−32=y−2−1=z−11=r (say)x−3=2r; y−2=−r; z−1=rx=3+2r, y=2−r; z= 1+r
So, Q= (3+2r,2-r,1+r)
Since Q lies on the given plane,
2x-y+z+1=0
2(3+2r)-(2-r)+1+r+1=0
6+4r-2+r+1+r+1=0
6r+6=0
r=-1
So Q= (3+2r,2-r,1+r)=(3-2,2+1,1-1)=(1,3,0)

Now, let P' is the image of P with respect to the given plane.
Let P'=(x,y,z)
So, Q= mid point of PP'.
(1,3,0)=(3+x2,2+y2,1+z2)3+x2=1; 2+y2=3; 1+z2=03+x=2; 2+y=6; 1+z=0x=−1; y=4; z=−1So, image P'=(−1,4,−1)

Now the perpendicular distance from (xo,yo, zo) to the plane ax+by+cz+d=0 is∣∣ax0+by0+cz0+d∣∣a2+b2+c2√So the required perpendicular distance =∣∣2(3)−2+1+1∣∣22+(−1)2+12√=66√=6√units
keshav yaduvanshi
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