Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The perpendicular distance of a point $(x_1,y_1,z_1)$ from the plane $ax+by+cz+d=0$ is given by $p=\frac{|a{x}_1+b{y}_1+c{z}_1+d|}{\sqrt{a^2+b^2+c^2}}$.

Here, the point is the origin $(0,0,0)$ and the equation of the plane is $2x+y-2z-18=0$.

Substituting the values, we get:

$p=\frac{|2(0)+1(0)-2(0)-18|}{\sqrt{2^2+1^2+(-2{)}^2}}$

$p=\frac{|-18|}{\sqrt{4+1+4}}$

$p=\frac{18}{\sqrt{9}}$

$p=\frac{18}{3}=6$ units.

Answer: 6 units