Question

If a point moves so that the sum of the squares of its distances from the six faces of a cube given by $x=\pm 1,y=\pm 1,z=\pm 1$ is $10$ units, then the locus of the point is

• A. $x^2+y^2+z^2=1$
• B. $x^2+y^2+z^2=2$
• C. $x+y+z=1$
• D. $x+y+z=2$

Answer 1

Answer: (B)

$x^2+y^2+z^2=2$

Let $P(x,y,z)$ be any point on the locus, then the distances from the six faces are
$|x+1|,|x-1|,|y-1|,|z-1|$
According to the given condition, we have
${|x+1|}^2+{|x-1|}^2+{|y+1|}^2+{|y-1|}^2+{|z+1|}^2+{|z-1|}^2=10$

$\Rightarrow 2(x^2+y^2+z^2)=10-6=4$

$\Rightarrow x^2+y^2+z^2=2$

Hence, option B.