Question

If the sum of square of distances of a point $P$ from three mutually perpendicular axes is $8\text{units},$ then the distance(in $\text{units}$) of that point from origin is

Answer 1

Let $P(x,y,z),$ then
Distance from $x-$axis $=\sqrt{y^2+z^2},$
Distance from $y-$axis $=\sqrt{x^2+z^2}$ and
Distance from $z-$axis $=\sqrt{x^2+y^2}$
$\Rightarrow (y^2+z^2)+(x^2+z^2)+(x^2+y^2)=8\Rightarrow 2(x^2+y^2+z^2)=8\Rightarrow x^2+y^2+z^2=4$
So, distance from origin $=\sqrt{x^2+y^2+z^2}=2\text{units}$