Question

find the equation of the plane through the point 2,3,5 and perpendicular to the plane 2x-3y+z=2,4x+y-3z+1=0

Answer 1

Answer:

x(-2/3) + y(-11/3) + z(3) +-8/3=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:

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

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

now point 2,3,5 lies on the reqd. plane, therefore it should satisfy the eqn. of plane.

putting point in eqn. of plane

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

4 +8k - 9 + 3k + 5 - 15k -2+k =0

3k=-2

k=-2/3

Therefore, eqn of plane: x(-2/3) + y(-11/3) + z(3) +-8/3=0