Point On The Globe

Geometry Level 1

Point $P$ is some point on the surface of the sphere ${(x-1)}^{2}+{(y+2)}^{2}+{(z-3)}^{2}=1 .$ What is the shortest possible distance between $P$ and $O=(0, 0, 0)?$

\sqrt{14} -1

\sqrt{12} -1

\sqrt{13} -1

\sqrt{15} -2

1 solution

Constantin Merlusca
Jun 26, 2014

Our sphere has center $C = (1, -2, 3)$ and radius 1. The shortest distance between $P$ and $O$ is the segment $PO$ when $P$ is on the line containing $OC$ , Since $OC = \sqrt{ 1^2+2^2+3^2} = \sqrt{14}$ and $PC = r =1$ , we have that $PO=\sqrt{14} - 1$

True enough. This one wasn't much of a challenge.

Whitney Clark - 5 years, 4 months ago

Point On The Globe

1 solution

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