1.

Find the equation of theimage of the plane `x-2y+2z-3=0`in plane `x+y+z-1=0.`

Answer» The image of the plane `x-2y+2z-3=0" "` (i)
in the plane `x+y+z-1=0" "`(ii)
passes through the line of intersection of the given planes.
Therefore, the equation of such a plane is
`(x-2y+2z-3) + t(x+y+z-1)=0, t in R`
`(1+t)x+ (-2+t) y + (2+t) z-3 - t =0" "` (iii)
Now plane (ii) makes the same angle with plane (i) and image plane (iii). Thus,
`" "(1-2+2)/(3sqrt(3))=pm(1+t-2+t+2+t)/(sqrt(3)sqrt((t+1)^(2)+ (t-2)^(2)+ (2+t)^(2)))`
`" "(1)/(3)=pm (3t+1)/(sqrt(3t^(2)+2t+9))`
`" "3t^(2)+2t+9=9(9t^(2)+6t+1)`
`" "3t^(2)+2t+9=81t^(2)+54t+9`
`" "78t^(2)+52t=0`
`" "t=0 or t=-(2)/(3)`
For `t=0`, we get plane (i) , hence for image plane,
`t=-(2)/(3)`
The equation of the image plane is
`" "3(x-2y+2z-3)-2(x+y+z-1)=0`
or `" "x-8y+4z-7=0`


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