What is the distance between $(15, 3, - 4)$ $(15,3,-4)$ and $(21, - 6, - 2)$ $(21,-6,-2)$ ?

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Answer 1 · 2016-04-28T13:40:13

$d i s tan c e = 11$ $distance=11$

$A = (15, 3, - 4)$ $A=(15,3,-4)$

$a_{x} = 15$ $a_x=15$
$a_{y} = 3$ $a_y=3$
$a_{z} = - 4$ $a_z=-4$

$B = (21, - 6, - 2)$ $B=(21,-6,-2)$

$B_{x} = 21$ $B_x=21$
$B_{y} = - 6$ $B_y=-6$
$B_{z} = - 2$ $B_z=-2$

$x^{2} = {(B_{x} - A_{x})}_{x}^{2}$ $x^2=(B_x-A_x)^2$

$x^{2} = {(21 - 15)}^{2} x^{2} = 6^{2} x^{2} = 36$ $x^2=(21-15)^2" "x^2=6^2" "x^2=36$

$y^{2} = {(B_{y} - A_{y})}_{y}^{2}$ $y^2=(B_y-A_y)^2$

$y^{2} = {(- 6 - 3)}^{2} b_{y}^{2} = - 9^{2} b_{y}^{2} = 81$ $y^2=(-6-3)^2" "b_y^2=-9^2" "b_y^2=81$

$z^{2} = {(B_{z} - A_{z})}_{z}^{2}$ $z^2=(B_z-A_z)^2$

$z^{2} = {(- 2 + 4)}^{2} z^{2} = 2^{2} z^{2} = 4$ $z^2=(-2+4)^2" "z^2=2^2" "z^2=4$

$d i s tan c e = \sqrt{x^{2} + y^{2} + z^{2}}$ $distance=sqrt(x^2+y^2+z^2)$

$d i s tan c e = \sqrt{36 + 81 + 4}$ $distance=sqrt(36+81+4)$

$d i s tan c e = 11$ $distance=11$

What is the distance between (15,3,−4)(15,3,-4) and (21,−6,−2)(21,-6,-2)?