Question

Find the distance between the point (5, 4, -6) and its image in xy-plane.

Answer 1

1 no distance in them

Answer 2

$The \: distance \: between \: (x_{1},y_{1},z_{1})\\and \: (x_{2},y_{2},z_{2})\:is :$

$= \sqrt{ (x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2} + (z_{2}-z_{1})^{2} }$

$Given \: the \: point \:A(5,4,-6)$

$Image \: of \: the \:point \: A \:is \: A'(5,4,6)$

$Distance \: between \:AA' \\= \sqrt{ (5-5)^{2} + (4-4)^{2}+[6+(-6)]^{2} } \\= \sqrt{0+0+(6+6)^{2}} \\= \sqrt{12^{2}} \\= 12\ :units$

Therefore.,

$\red{ Required \: distance }\green {= 12\ :units }$

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