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The distance between $(1, 2, 3)$ and $(4, 6, 3)$ in 3D is:
A$3$, taking only the $x$-difference here on chart
B$4$, taking only the $y$-difference here on chart
C$5$, since $\sqrt{9 + 16 + 0} = \sqrt{25} = 5$
D$25$, the squared distance reported as distance on chart
Answer & Solution
Correct answer: C. $5$, since $\sqrt{9 + 16 + 0} = \sqrt{25} = 5$
$\sqrt{(4-1)^2 + (6-2)^2 + (3-3)^2} = \sqrt{9 + 16 + 0} = 5$.
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