## Thomas' Calculus 13th Edition

$y=1$
Form an equation using the distance formula (distance from (0,0,0) to (x,y,z))=(distance from (0,2,0) to (x,y,z)) $\sqrt{x^{2}+y^{2}+z^{2}}=\sqrt{x^{2}+(y-2)^{2}+z^{2}}\quad$ ... square both sides $x^{2}+y^{2}+z^{2}=x^{2}+(y-2)^{2}+z^{2}\quad$ ... simplify $y^{2}=y^{2}-4y+4$ $4y=4$ $y=1$ (the plane parallel to the xz plane, containing (0,1,0)).