Question

To find the equation of a plane perpendicular to the plane $2x-3y+2z+5=0$ and $2y-4z+1=0$ and at a distance of $4$ units from the origin.

Answer 1

Let the required equation of the plane be

$Ax+By+Cz+d=0$ ----- $\left(1\right)$

Since, it is perpendicular to the two given lines
So,

$2\times A-3\times B+2C=0$

$2A-3B+2C=0$ ---- $\left(2\right)$

And,

$0\times A+2\times B+(-4)\times C=0$

$B=2C$

Putting $B=2C$ in equation $\left(2\right)$

$2A-3\times 2C+2C=0$

$A=2C$

Putting $A=2C$ and $B=2C$ in equation $\left(1\right)$

$2C\times x+2C\times y+Cz+d=0$

$x+y+\frac{z}{2}+\frac{d}{2C}=0$ ---- $\left(3\right)$

Since,

Perpendicular distance of this plane from the origin is $4$ units,

So,

$|\frac{0+0+\frac{0}{4}+\frac{d}{2c}}{\sqrt{1^2+1^2+(\frac{1}{2}{)}^2}}|$

$d=12C$

Putting $d=12C$ in equation $\left(3\right)$, we get,

$x+y+\frac{z}{2}+\frac{12C}{2C}=0$

$2x+2y+z+12=0$

This the required equation of the plane.