Question

If the sum of the squares of the perpendicular distances in 3D of $P$ from the coordinate planes is $5$, then the locus of $P$ is

• A. $x+y+z=5$
• B. $x+y+z=25$
• C. $x^2+y^2+z^2=\frac{5}{2}$
• D. $x^2+y^2+z^2=5$

Answer 1

The correct option is C $x^2+y^2+z^2=\frac{5}{2}$

Let the point be $P(x,y,z)$

So sum of the squares of the perpendicular distances from the coordinate planes is,

$\left[\right(x-x{)}^2+(y-0{)}^2+(z-0{)}^2]+[(x-0{)}^2+(y-y{)}^2+(z-0{)}^2]+\left[\right(x-0{)}^2+(y-0{)}^2+(z-z{)}^2]=5$(given)

$\Rightarrow x^2+y^2+z^2=\frac{5}{2}$