Question
Fri March 19, 2010 By:

# plss ans this ?????????????///////

Sun March 21, 2010

u have given 2-3 questions of the same type. so here's the method.

Assume that the perpendicular from the given point  A to the given plane meets the plane at say pt  P(a,b,c)

then P(a,b,c)  is a point of the plane also. so it must satisfy the equation of the plane.

so we'll get

2a-b+c+3=0...(i)

since the perpendicular to the plane must be  parallel to the normal to the plane

so the d.r.s of the line joining  AP have to be 2,-1,1[since the plane has the eqn 2 x-1 y+1 z+3=0

let the reqd pt be Q then P is the mid point of AQ. as Q is the reflection of A in the plane and has to be lying along the line  AP.

using this, you can find the coordiantes of P.

Take eqn of line AP in the form

a-1/1=b-2/2=c-3/3=k

find a,b,c in terms of k, substitute in eqn of plane.

once you have k,

take the point P as mid point of AQ and get coordinates of Q.

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