Question

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.

Answer 1

Equation of any plane through the intersection of given planes can be written as

$x+y+z-1+\lambda (2x+3y+4z-5)=0\Rightarrow (1+2\lambda )x+(1+3\lambda )y+(1+4\lambda )z-1-5\lambda =0$ ..(i)

The diection ratios, $a_1,b_1,c_1,$ of the plane are $(2\lambda +1),(3\lambda +1)and(4\lambda +1)$ The plane in Eq. (i) is perpendicular to x-y+z=0. Its direction ratios, $a_2,b_2,c_2$ are 1,-1, and 1.

Since, the planes are perpendicular.

$\therefore a_1{a}_2+b_1{b}_2+c_1{c}_2=0\Rightarrow 1(1+2\lambda )-1(1+3\lambda )+1(1+4\lambda )=0\Rightarrow 1+2\lambda -1-3\lambda +1+4\lambda =0\Rightarrow 3\lambda =-1\Rightarrow \lambda =-\frac{1}{3}$

Substituting this value of $\lambda$ in Eq. (i), we obtain the required plane as

$(1-\frac{2}{3})x+(1-\frac{3}{3})y+(1-\frac{4}{3})z-1+\frac{5}{3}=0\Rightarrow \frac{1}{3}x-\frac{1}{3}z+\frac{2}{3}=0\Rightarrow x-z+2=0$

This is the required equation of the plane.