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The symmetric form of the equation of the line of intersection of the planes x+y-z = 1 and 2x-3y+z=2 is
Dear deepika
planes x+y-z = 1 and 2x-3y+z=2
let the direction ratio of the line are a,b,c
so a+b-c =0 and 2a-3b+c =0
so from these two equation
a/2 = b/3 =c/5
now we have to find out the point of intersection of these two planes.
let x=0
so y-z =1 and -3y + z =2
so y=-3/2
z = -5/2
so equation of line become
(x-0)/2 = (y +3/2)/3 = (z+5/2)/5
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