Find the equation of the plane through the intersection of the planes:

Asked by  | 25th May, 2009, 12:41: AM

Expert Answer:

Equation of plane passing through the intersection of x+y+z=1 and 2x+3y+4z=5 is
(x+y+z-1)+k(2x+3y+4z-5) =0

i.e (1+2k)x+(1+3k)y+(1+4k)z-1-5k=0......(i)

Also (i) is perpendicular to the plane x-y+z=0.
using the condition of perpendicularity

( 1+2k).1+(1+3k)-1+(1+4k)1=0
this gives k =-1/3
substitute in (i) to get the required equation

Answered by  | 27th May, 2009, 04:14: PM

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