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Plane - Plane Passing Through the Intersection of Two Given Planes

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Let `π_1` and `π_2` be two planes with equations `vec r  . hat n _1 = d_1` and `vec r . hat n _2 = d_2` respectively.  The position vector of any point on the line of intersection must satisfy both the equations fig.

If `vec t ` is the position vector of a point on the line , then
`vec t . hat n_1 = d_1` and `vec t . hat n _2 = d_2` 
Therefore , for all real values of  λ, we have
`vec t . (hat n _1 + lambda hat n_2) = d_1 + lambda d_2`
Since `vec t` is arbitrary, it satisfies for any point on the line.
Hence , the equation `vec r . (vec n_1 + lambda vec n_2) = d_1 + lambda d_2`   represents a plane `π_3` which is such  that if any vector ` vec r` satisfies both the equations `π_1` and `π_2`, it also satisfies the equation `π_3` i.e., any plane passing through the intersection of the planes 
`vec r . vec n_1 = d_1` and `vec r . vec n_2 = d_2`
has the equation  `vec r . (vec n_1 + lambda vec n_2) = d_1 + lambda d_2`    ...(1)

Cartesian form:
In Cartesian system, let  `vec n_1 = A_1 hat i + B_2 hat j + C_1 hat k`
`vec n_2 = A_2 hat i + B_2 hat j + C _2 hat k` 
and `vec r = x hat i + y hat j + z hat k`
Then (1) becomes 
`x (A_1 + lambda A_2) + y(B_1 + lambda B_2) + z(C_1 +lambda C_2) = d_1 + lambda d_2`
or `(A_1x +B_1y + C_1z -d_1) + lambda (A_2x + B_2y + C_2z -d_2) = 0`     ..(2)
which is the required Cartesian form of the equation of the plane passing through the intersection of the given planes for each value of λ.

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