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 $C=(0,0,1)$
Now, $A B=\sqrt{(0-1)^{2}+(1-0)^{2}+0^{2}}=\sqrt{2}$
$B C=\sqrt{0^{2}+(0-1)^{2}+(1-0)^{2}}=\sqrt{2}$
and $C A=\sqrt{(1-0)^{2}+0^{2}+(0-1)^{2}}=\sqrt{2}$
$\therefore $ Perimeter of triangle $=A B+B C+C A$
$=\sqrt{2}+\sqrt{2}+\sqrt{2}=3 \sqrt{2}$