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

Answer 1

Equation of given planes are

$P_1\Rightarrow x+y+z-1=0$

$P_2\Rightarrow 2x+3y+4z-5=0$

Equation of plane through the line of intersection of planes $P_1,P_2$ is

$P_1+\lambda P_2=0$

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

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

Given that plane represented by equation (1) is perpendicular to plane

$x-y+z=0$

So we use formula

$a_1{a}_2+b_1{b}_2+c_1{c}_2=0$

So, $(1+2\lambda ).1+(1+3\lambda ).(-1)+(1+4\lambda ).1+(-1-5\lambda )=0$

$1+2\lambda -1-3\lambda +1+4\lambda =0$

$3\lambda +1=0$

$\lambda =\frac{-1}{3}$

Put $\lambda =\frac{-1}{3}$ in equation (1), we get

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

$\frac{x}{3}-\frac{z}{3}+\frac{2}{3}=0$

$x-z+2=0$

General points on the line:

$x=2+3\lambda ,y=-4+4\lambda ,z=2+2\lambda$

The equation of plane:

$\overrightarrow{r}\cdot (\hat{i}-2\hat{j}+\hat{k})=0$

The point of intersection of the line and the plane :
Substituting general point of the line in the equation of plane and finding the particular value of $\lambda$.

$\left[\right(2+3\lambda )\hat{i}+(-4+4\lambda )\hat{j}+(2+2\lambda \left)\hat{k}\right]\cdot (\hat{i}-2\hat{j}+\hat{k})=0$

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

$\Rightarrow 12-3\lambda =0$ or $\lambda =4$

The point of intersection is :

$(2+3(4),-4+4(4),(2+2\left(4\right))=(14,12,10)$

Distance of this point from $(2,12,5)$ is

$=\sqrt{(14-2{)}^2+(12-12{)}^2+(10-5{)}^2}=\sqrt{{12}^2+5^2}=13$ units.