dripcima24

2022-10-08

What is the distance between (-9,3,1) and (1,5,2)?

Joaquin Cisneros

Beginner2022-10-09Added 4 answers

Distance

$=\sqrt{{({x}_{2}-{x}_{1})}^{2}+{({y}_{2}\_{y}_{1})}^{2}+{({z}_{2}-{z}_{1})}^{2}}=\sqrt{{(1+9)}^{2}+{(5-3)}^{2}+{(2-1)}^{2}}=\sqrt{105}=10.25\left(2dp\right)$

$=\sqrt{{({x}_{2}-{x}_{1})}^{2}+{({y}_{2}\_{y}_{1})}^{2}+{({z}_{2}-{z}_{1})}^{2}}=\sqrt{{(1+9)}^{2}+{(5-3)}^{2}+{(2-1)}^{2}}=\sqrt{105}=10.25\left(2dp\right)$ units

$=\sqrt{{({x}_{2}-{x}_{1})}^{2}+{({y}_{2}\_{y}_{1})}^{2}+{({z}_{2}-{z}_{1})}^{2}}=\sqrt{{(1+9)}^{2}+{(5-3)}^{2}+{(2-1)}^{2}}=\sqrt{105}=10.25\left(2dp\right)$

$=\sqrt{{({x}_{2}-{x}_{1})}^{2}+{({y}_{2}\_{y}_{1})}^{2}+{({z}_{2}-{z}_{1})}^{2}}=\sqrt{{(1+9)}^{2}+{(5-3)}^{2}+{(2-1)}^{2}}=\sqrt{105}=10.25\left(2dp\right)$ units

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