Question

Ex 4. Find the equation of the plane through (4.4.0) and perpendicular to the plane
x+2y+2:=5 and 3x+3y +2:-8=0.
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Answer 1

Answer:

Let ax+by+cz+d=0 is the required plane

Plugging 4,4,0 we get

4a+4b+d=0…………………..(1)

Normal of this plane has direction vector ai+bJ+ck

Normal of other two planes are direction vectors 2i+j+2k and 3i+3j-2k

As this is perpendicular to both, dot products are zero

2a+b+2c=0……………(2)

3a+3b-2c=0…………(3)

Solving these equations,

b=-5a/4; c=-3a/8;d=a

hence required equation is

ax+by+cz+d=0

ax-5a/4*y-3a/8*z+a=0

8x-10y-3z+8=0