Question

find the equation of the plane through the line of intersection of the plane x+y+z=1 and 2x+3y+4z=5 which is perpendicular to the plane x-y+z=0.
then find the distance of the plane thus obtained from the point A(1,3,6).







=

Answer 1

Equation of the plane passing through the intersection of the planes x+y+z=1 and 2x+3y+4z=5 is x+y+z-1+k(2x+3y+4z-5)=0 ie (1+2k)x+(1+3k)y+(1+4k)z=1+5k As this plane is perpendicular to the plane x-y+z=0, their dot product gives zero. ie k= -1/3 Therefore the equation of the plane is x-z+2=0 Equation to find the distance from a point P(x1,y1,z1) to a plane Ax+By+Cz+D=0 is d=|Ax1+By1+Cz1+D| / sqrt(A^2+B^2+C^2) d=|(1+0-6-2)|/sqrt(2) d=7/sqrt(2)