What is the distance between $(- 5, 13, 4)$ $(-5,13,4)$ and $(2, - 16, - 2)$ $(2,-16,-2)$ ?

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What is the distance between $(- 5, 13, 4)$ $(-5,13,4)$ and $(2, - 16, - 2)$ $(2,-16,-2)$ ?

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Answer 1 · 2016-02-05T11:30:44

$d \approx 30.4302$ $d ~~ 30.4302$

Explanation:

We know that the distance formula for cartesian coordinates is as follows:

$\sqrt{{(x_{2} - x_{1})}_{2}^{2} + {(y_{2} - y_{1})}_{2}^{2}}$ $sqrt((x_2-x_1)^2 + (y_2 - y_1)^2)$

In the case where there are three dimensions $(x, y, z)$ $(x, y, z)$ , you just need to add another term to account for the z-coordinates.

$\sqrt{{(x_{2} - x_{1})}_{2}^{2} + {(y_{2} - y_{1})}_{2}^{2} + {(z_{2} - z_{1})}_{2}^{2}}$ $sqrt((x_2-x_1)^2 + (y_2 - y_1)^2 + (z_2 -z_1)^2)$

Plugging in the values from the given...

[Given]
$P_{1} : (- 5, 13, 4)$ $P_1: (-5, 13, 4)$
$P_{2} : (2, - 16, - 2)$ $P_2: (2, -16, -2)$

[Solution]
$d = \sqrt{{(2 - (- 5))}^{2} + {(- 16 - 13)}^{2} + {(- 2 - 4)}^{2}}$ $d = sqrt((2-(-5))^2 + (-16-13)^2 + (-2-4)^2)$
$d = \sqrt{7^{2} + {(- 29)}^{2} + {(- 6)}^{2}}$ $d = sqrt(7^2 + (-29)^2 + (-6)^2)$
$d = \sqrt{49 + 841 + 36}$ $d = sqrt(49 + 841 + 36)$
$d = \sqrt{926}$ $d = sqrt (926)$
$d \approx 30.4302$ $d ~~ 30.4302$

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What is the distance between $(- 5, 13, 4)$ $(-5,13,4)$ and $(2, - 16, - 2)$ $(2,-16,-2)$ ?

1 Answer

Explanation:

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Impact of this question

Science

Math

Humanities

... and beyond

What is the distance between (−5,13,4)(-5,13,4) and (2,−16,−2)(2,-16,-2)?

1 Answer

Explanation:

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Impact of this question

What is the distance between $(- 5, 13, 4)$ $(-5,13,4)$ and $(2, - 16, - 2)$ $(2,-16,-2)$ ?