Question

find the equation of the plane through the point (1,1,2)and perpendicular to the planes:x+2y-4z=3;x+2y-3z=5​

Answer 1

Answer:

z -2 = 0

Step-by-step explanation:

using family of lines: (x1 + y1 + z1 +c1 )+ k(x2 + y2 + z2 + c2)=0

now we are given equation of 2 planes so we can find the equation of plan perpendicular to both which will be:

x+2y-4z-3 + k(x+2y-3z-5)=0

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

now point 1,1,2 lies on the reqd plane, therefore it should satisfy the eqn of plane.

putting point in eqn of plane

1(1+k) +1(2+2k) +2(-4-3k) -3-5k=0

1 + k + 2 + 2k -8 -6k -3 -5k =0

8k=-8

k=-1

Therefore, eqn of plane: x(0) + y(0) + z(-1) +2=0

z -2 = 0