Find the locus of a point whose sum of the squares of its perpendicular distances from the planes $x+y+z=0$, $x-z=0$, and $x-2y+z=0$ equals 19.

Question

Accepted Answer

$x^2+y^2+z^2=9$. For a point $(x,y,z)$, the squared distances from the planes are proportional to the squares of the corresponding linear forms. Adding them and simplifying gives a constant times $x^2+y^2+z^2$. Equating this to 19 yields the locus $x^2+y^2+z^2=9$.

Find the locus of a point whose sum of the squares of its perpendicular distances from the planes \(x+y+z=0\), \(x-z=0\), and \(x-2y+z=0\) equals 19.

Solution

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