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Asked by  | 19th Mar, 2010, 06:47: AM

Expert Answer:

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


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


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.

Answered by  | 21st Mar, 2010, 11:20: PM

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