Question

A point $P$ in space is such that the sum of square of its distances from $x-$axis, $y-$axis and $z-$axis is $9\text{units}$ more than the sum of square of its distances from $xy-$plane, $yz-$plane and $zx-$plane. Then the distance( in $\text{units}$) of point $P$ from origin is equal to

• A. 3
• B. 3.00
• C. 3.0

Answer 1

Answer: (A), (B), (C)

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 $=\sqrt{x^2+y^2}$
Also,
Distance from $xy-$ plane $=|z|,$
Distance from $yz-$plane $=|x|$ and
Distance from $zx-$plane $=|y|$
Given,
$(y^2+z^2)+(x^2+z^2)+(x^2+y^2)=z^2+x^2+y^2+9\Rightarrow x^2+y^2+z^2=9$
So, distance from origin $=\sqrt{x^2+y^2+z^2}=3$