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The perimeter of the triangle with vertices at $(1,0,0),(0,1,0)$ and $(0,0,1)$ is
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Verified Answer
The correct answer is:
$3 \sqrt{2}$
Let $A=(1,0,0), B=(0,1,0)$ and $C=(0,0,1)$
Now,
$\begin{aligned}
\text { Now, } A B & =\sqrt{(0-1)^2+(1-0)^2+0^2}=\sqrt{2} \\
B C & =\sqrt{0^2+(0-1)^2+(1-0)^2}=\sqrt{2} \\
\text { and } \quad C A & =\sqrt{(1-0)^2+0^2+(0-1)^2}=\sqrt{2}
\end{aligned}$
$\begin{aligned}
\therefore \quad \text { Perimeter of triangle } & =A B+B C+C A \\
= & \sqrt{2}+\sqrt{2}+\sqrt{2}=3 \sqrt{2}
\end{aligned}$
Now,
$\begin{aligned}
\text { Now, } A B & =\sqrt{(0-1)^2+(1-0)^2+0^2}=\sqrt{2} \\
B C & =\sqrt{0^2+(0-1)^2+(1-0)^2}=\sqrt{2} \\
\text { and } \quad C A & =\sqrt{(1-0)^2+0^2+(0-1)^2}=\sqrt{2}
\end{aligned}$
$\begin{aligned}
\therefore \quad \text { Perimeter of triangle } & =A B+B C+C A \\
= & \sqrt{2}+\sqrt{2}+\sqrt{2}=3 \sqrt{2}
\end{aligned}$
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