Question

The perimeter of triangle with vertices at $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$ is ...P... . Here, $P$ refers to

• 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 $C=(0,0,1)$
Now, $AB=\sqrt{(1-0{)}^2+(0-1{)}^2+0^2}=\sqrt{2}$
$BC=\sqrt{0^2+(1-0{)}^2+(0-1{)}^2}=\sqrt{2}$
and $CA=\sqrt{(0-1{)}^2+0^2+(1-0{)}^2}=\sqrt{2}$
$\therefore$ Perimeter of triangle $=AB+BC+CA$
$=\sqrt{2}+\sqrt{2}+\sqrt{2}$
$=3\sqrt{2}$