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 palne $x-y+z=0$. Also,find the distance of the plane obtained above,form the origin.

Answer 1

$\text{Any plane through the intersection of given planes}$
$(x+y+z-1=0and2x+3y+4z-5=0isx+y+z-1+\lambda (2x+3y+4z-5)=0$

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

$\text{Also, plane (i) is perpendicular to the plane}$
$x-y+z=0$

$\Rightarrow (1+2\lambda )-(1+3\lambda )+(1+4\lambda )=0$

$\Rightarrow 3\lambda +1=0,then,\lambda =-\frac{1}{3}$

$Put\lambda =-\frac{1}{3}in\left(i\right),weobtain$

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

$\Rightarrow x-z+2=0,\text{which is the required plane.}$
$\text{The perpendicular distance of this plane from origin(0,0,0)}$

$=\frac{|0-0+2|}{\sqrt{1^2+(-1{)}^2}}=\frac{2}{\sqrt{2}}=\sqrt{2}units.$