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Find the distance between the the points $(2,-1,3)$ and $(-2,1,3)$.

$\begin{array}{1 1}2 \sqrt{5} \\ 6 \\ 2 \\ 2 \sqrt{14}\end{array} $

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  • Distance between two points $(x_1,y_1,z_1)\:\;and \:\:(x_2,y_2,z_2)$ is given by $\sqrt {(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$
Given two points are $A(2,-1,3)$ and $B(-2,1,3)$.
We know that the distance between two points $(x_1,y_1,z_1)\:\;and \:\:(x_2,y_2,z_2)$ is given by $\sqrt {(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$
Here $x_1=22,\:y_1=-1,\:z_1=3$ and
$x_2=-2,\:y_2=1,\:z_2=3.$
$\therefore$ The distance between the give two points $AB$ is
$\sqrt {(-2-2)^2+(1+1)^2+3-3)^2}=\sqrt {16+4+0}=\sqrt {20}=2\sqrt {5}$
answered Apr 16, 2014 by rvidyagovindarajan_1
edited Apr 16, 2014 by rvidyagovindarajan_1
 

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