Question

Prove that the plane passes through the points $(1,0,1),(1,1,1)$ and $(-7,-3,-5)$ is perpendicular to the $XZ$ plane.

Answer 1

The plane passing through the point $(1,0,1)$ is $A(x-1)+B(y-0)+C(z-1)=0$ ...(1)
If plane (1) passes through the point $(1,1,1)$
$A(1-1)+B(1-0)+C(1-1)=0$
$0.A+B+0.C=0$ ............(2)
If plane (1) passes through point $(-7,-3,-5)$, then
$A(-7-1)+B(-3-0)+C(-5-1)=0$
$\Rightarrow -8A-3B-6C=0$
$\Rightarrow 8A+3B+6C=0$ ........(3)
Solving Equations (2) and (3)
$\frac{A}{\left(1\right)\left(6\right)-\left(0\right)\left(3\right)}=\frac{B}{\left(8\right)\left(0\right)-\left(0\right)\left(6\right)}=\frac{C}{\left(0\right)\left(3\right)-\left(8\right)\left(1\right)}$
$=\frac{A}{6}=\frac{B}{0}=\frac{C}{-8}$
$\Rightarrow \frac{A}{3}=\frac{B}{0}=\frac{C}{-4}=K$ (Say)
Putting the values of $A,B,C$ the equation of the plane is:
$3K(x-1)+0(y-0)-4k(z-1)=0$
$\Rightarrow 3x-3-4z+4=0$
$\Rightarrow 3x-4z+1=0$ ........(4)
Equation plane $XZ$ will be
$y=0$
$0.x+y+0.Z=0$ ........(5)
The direction ratios of the plane (4) are $3,0,-4$ and the direction ratios of the plane (5) are $0,1,0$.
$\therefore$ The planes (4) and (5) will be perpendicular when their normal also be perpendicular to each other than condition is:
$a_1{a}_2+b_1{b}_2+c_1{c}_2=0$
$\Rightarrow 3\left(0\right)+0\left(1\right)+(-4)\left(0\right)=0$.
Hence, plane (4) and $XZ$ plane are perpendicular.