Question

The perimeter of the triangle with vertices at $(1, 0, 0), (0, 1, 0)$ and $(0, 0, 1)$ is

• A. $3$
• B. $2$
• C. $2\sqrt{2}$
• D. $3\sqrt{2}$

Answer 1

Answer: (D)

Let $A = (1, 0, 0), B = (0, 1, 0)$ and $C = (0, 0, 1)$
so, $AB=\sqrt{\left(0-1\right)^{2}+\left(1-0\right)^{2}+0^{2}}=\sqrt{2}$
$BC=\sqrt{O^{2}+\left(0-1\right)^{2}+\left(1-0^{2}\right)}=\sqrt{2}$
and $CA=\sqrt{\left(1-0\right)^{2}+0^{2}+\left(0-1^{2}\right)}=\sqrt{2}$
$\therefore $ Perimeter of triangle $=AB+BC+CA$
$=\sqrt{2}+\sqrt{2}+\sqrt{2}$
$=3\sqrt{2}$