What is the distance between $(-7,6,10)$ and $(7,-4,9)$ ?

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Answer 1 · 2018-03-29T16:38:53

distance $=3sqrt(33) ~~ 17.2$ square units

We seek the distance $d$ , say, between the coordinates $(-7,6,10)$ and $(7,-4,9)$ ? in euclidean space.

Applying Pythagoras theorem in $3$ -Dimensions we have:

$d^2 = (-7-7)^2 + (6-(-4))^2 + (10-9)^2$
$\ \ \ \ = (-14)^2 + (10)^2 + (1)^2$
$\ \ \ \ = 196 + 100 + 1$
$\ \ \ \ = 297$

Thus:

$d = sqrt(297) \ \ \$ (NB - we seek the positive solution)
$\ \ = sqrt(9*33)$
$\ \ = 3sqrt(33)$
$\ \ ~~ 17.2$

What is the distance between (-7,6,10)(-7,6,10) and (7,-4,9)(7,-4,9)?