What is the distance between $(- 2, 1, - 13)$ $(-2,1,-13)$ and $(4, 3, - 12)$ $(4,3,-12)$ ?

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What is the distance between $(- 2, 1, - 13)$ $(-2,1,-13)$ and $(4, 3, - 12)$ $(4,3,-12)$ ?

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Answer 1 · 2016-05-03T19:33:02

$s = \sqrt{41}$ $s=sqrt41$

Explanation:

$A = (- 2, 1, - 13)$ $A=(-2,1,-13)$
$B = (4, 3, - 12)$ $B=(4,3,-12)$

$A_{x} = - 2$ $A_x=-2$
$A_{y} = 1$ $A_y=1$
$A_{z} = - 13$ $A_z=-13$

$B_{x} = 4$ $B_x=4$
$B_{y} = 3$ $B_y=3$
$B_{z} = - 12$ $B_z=-12$

$distance between A and B can be calculated by formula:$ $"distance between A and B can be calculated by formula:"$
$s = \sqrt{{(B_{x} - A_{x})}_{x}^{2} + {(B_{y} - A_{y})}_{y}^{2} + {(B_{z} - A_{z})}_{z}^{2}}$ $s=sqrt((B_x-A_x)^2+(B_y-A_y)^2+(B_z-A_z)^2)$

$s = \sqrt{{(4 + 2)}^{2} + {(3 - 1)}^{2} + {(- 12 + 13)}^{2}}$ $s=sqrt((4+2)^2+(3-1)^2+(-12+13)^2)$

$s = \sqrt{6^{2} + 2^{2} + 1^{2}} = \sqrt{36 + 4 + 1} = \sqrt{41}$ $s=sqrt(6^2+2^2+1^2)=sqrt(36+4+1)=sqrt(41)$

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What is the distance between $(- 2, 1, - 13)$ $(-2,1,-13)$ and $(4, 3, - 12)$ $(4,3,-12)$ ?

1 Answer

Explanation:

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Impact of this question

Science

Math

Humanities

... and beyond

What is the distance between (−2,1,−13)(-2,1,-13) and (4,3,−12)(4,3,-12)?

1 Answer

Explanation:

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Impact of this question

What is the distance between $(- 2, 1, - 13)$ $(-2,1,-13)$ and $(4, 3, - 12)$ $(4,3,-12)$ ?