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Find the equation of the plane through the line of intersection of the planes $ x+y+z=1$ and $2x+3y+4z=5$ which is perpendicular to the plane $x-y+z=0$. Also find the distance of the plane obtained above, from the origin.

1 Answer

The equation of the plane passing through the equation of the plane $ x + y + z = 1$ and $ 2x+3y+4z=5$ is
$( x+y+z-1 )+x (2x+3y+4z-5) = 0$
$ \therefore (1+2\lambda)x + (1+3\lambda)y +(1+4\lambda)y$
$-5\lambda+1=0$ ---------(1)
The direction ratio a, b, c of the planes are
$(1+2\lambda), ) (1+3\lambda), (1+4\lambda)$
The above plane is perpendicular to $x-y+z=0$
The direction ratio of this plane is (1, -1, 1)
Since the two planes are perpendicular
$a_1a_2 + b_1b_2 +c_1c_2=0$
$ \therefore (1+2\lambda)(1)+(1+3\lambda)(-1) + (1+4\lambda)(1)=0$
$ \Rightarrow 1+2\lambda -1 - 3\lambda+1+4\lambda=0$
$ \Rightarrow \lambda = - \large\frac{1}{3}$
answered Mar 25, 2014 by thanvigandhi_1
edited Apr 1, 2014 by thanvigandhi_1

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