Answer to Question #104249 in Analytic Geometry for Deepak Rana

Question

Find the equation of the plane passing through the line of intersection of the planes
2x+3y+z = 4 and x+y+z = 2, and which is perpendicular to the plane
2x+3y−z = 3.

Answer 1

Let us assume that the equation of the first plane is P and that of the second is Q

P= 2x+3y+z-4=0

Q= x+y+z-2=0

The equation of plane passing through the intersection of these two planes is given by

P+øQ=0,

where ø is an arbitrary constant.

Let A be the new plane, then

A=(2x+3y+z-4)+ø(x+y+z-2)=0

Equating the above equation

A=x(ø+2)+y(3+ø)+z(1+ø)-(4+2ø)

If two planes Ax+By+Cz+D=0 and ax+by+cz+d=0 are perpendicular then Aa+Bb+Cc=0

The planes 2x+3y-z-3=0 and

x(ø+2)+y(ø+3)+z(ø+1)-(4+2ø)=0 are perpendicular, then

(ø+2)2+(3+ø)3+(1+ø)(-1)=0

4ø+12=0 ,then ø=-3

E: x(-3+2)+y(3-3)+z(-3+1)-(4-6)=0

E: -x+0-2z+2=0

E: x+2z-2=0