Equation of a plane through the intersection of two planes

picture in 3D

In the above  figure, thick black and thin black line represents the 2D form of two planes through which a black dotted plane passes . and the equation of the thick black line and thin line plane are a₁x+b₁y+c₁z+d₁=0   and a₂x+b₂y+c₂z+d₂=0 respectively. Here  we have to find the equation of black dotted plane. Let , OP , PQ and OR are constructed so that they are normals of black dottted, thin black and thick black plane  respectively.

Let the coordinate of O, R, P  are (x_3,y₃,z₃), (x_2,y₂,z₂)  and (x_1,y₁,z₁) repectively. Then since by definition of plane , the a₁, b₁ and c₁  represents the direction ratio of the normal of thick black plane. As we assume that OR is also the normal then direction ratio of OR is

           (x₃-x₂, y₃-y_2, z₃-z₂)  so OR is also normal so it is parallel to any normal of that plane so we have following relation

                  

Then we have , x₃-x₂=ka₁, y₃-y₂=kb_1, z₃-z₂=kc₁

Here PQ is the normal of the thin black plane which is also parallel to PR so PR is parallel to any normal of that plane so by similar as above we get

Then x₂-x₁=ma₂, y₂-y₁=mb_2, z₂-z₁=mc₂

Now the equation of black dotted plane since it passes through (x_3,y₃,z₃)

We have a(x-x₃)+b(y-y₃)+c(z-z₃)=0  ………..(1)where a, b, c are the direction ratio of the normal i.e OP of the required plane.

But By figure the direction  ratio of OP is (x₃-x₁, y₃-y_1, z₃-z₁ )

\a= x₃-x₁=x₃-x₂+x₂-x₁=ka₁+ma₂,     similarly b= kb₁+mb₂ and  c= kc₁+mc₂

Now from (1)

 (ka₁+ma₂ )(x-x₃)+ (kb₁+mb₂ )(y-y₃)+ (kc₁+mc₂ )(z-z₃)=0

Or, (a₁+λa₂ )(x-x₃)+ (b₁+λb₂ )(y-y₃)+ (c₁+λc₂ )(z-z₃)=0          where λ=m/k 

Or, a₁x-a₁x₃+λa₂(x-x₃)+b₁y-b₁y₃+λb₂(y-y₃)+c₁z-c₁z₃+λc₂(z-z₃)=0

Or, a₁x+b₁y+c₁z+λ(a₂x+b₂y+c₂z)-{(a₁x₃+b₁y₃+c₁z₃)+λ(a₂x₃+b₂y₃+c₂z₃)}=0……..(2)

Now since (x_3,y₃,z₃) also passes through two planes  a₁x+b₁y+c₁z+d₁=0   and a₂x+b₂y+c₂z+d₂=0 so

               

                                       a₁x₃+b₁y₃+c₁z₃+d₁=0   ………..´1

                                       a₂x₃+b₂y₃+c₂z₃+d₂=0………´ λ

add above we get(a₁x₃+b₁y₃+c₁z₃)+λ(a₂x₃+b₂y₃+c₂z₃)=-(d₁+λd₂)……..(3)

now from (2) and (3) we get

a₁x+b₁y+c₁z+λ(a₂x+b₂y+c₂z)+ (d₁+λd₂)=0

\ a₁x+b₁y+c₁z+d₁+λ(a₂x+b₂y+c₂z+d₂)=0  is the required plane .