both the given planes r perpendicular to each oder
so image is the same plane x-y+z+6=0
so a=1 b=-1 c=1
so ans 0
Q1. The image of the plane x-y+z+6=0 with respect to the plane mirror 4x-2y-6z+14=0 is ax+by+cz+6=0, then the value of cos-1a + cos-1b+ cos-1c is
both the given planes r perpendicular to each oder
so image is the same plane x-y+z+6=0
so a=1 b=-1 c=1
so ans 0
thx.... (lapse in observation)
But if the two planes were not perpendicular to each other or parallel to each other...
then how to proceed in such type of questions
ya
the in that case wat i think is
firstly find teh line of intersection of the two planes(that u can find i guess)
now take any point on that line suppose(p,q,r)
now that point will lie on the image plane
assume direction ratios of the normal to the image plane be (a,b,c)
so eq of image plane is a(x-p)+b(y-q)+c(z-r)=0
let teh direction ratios of incident plane be(g,h,j)
now find the angle between the incident plane and plane mirror,,,,let that be α
so angle between the incident plane and image palne wiill be 2α....hence angle betwenn the normals form des 2 planes will be 2α
so use cos(2α)=ga+hb+cr√a2+b2+c2√g2+h2+j2
except(a,b,c) everthinmg is known u hav to find (a,b,c)
now use the condition that the teh line of intersection of the two planes lies on the image plane
if (r1,p1,q1,) be direction ratio of line of intersection
so r1a+p1b+q1c=0
u will get two eqautions in a b c
find it and hence the image plane
caan u please elaborate me the process,Please yaar I can't remember how to find intersection line of two planes.Pleaaase help me.