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. $3\sqrt{2}$

Hint:

The distance between any two standard unit basis vector endpoints $(1,0,0), (0,1,0), (0,0,1)$ is always $\sqrt{2}$.

Answer 1

The Correct Option is D

Step 1: Calculate Side Lengths
Vertices $A(1,0,0), B(0,1,0), C(0,0,1)$. Side $AB = \sqrt{(0-1)^{2}+(1-0)^{2}+0^{2}} = \sqrt{1+1+0} = \sqrt{2}$. Side $BC = \sqrt{0^{2}+(0-1)^{2}+(1-0)^{2}} = \sqrt{0+1+1} = \sqrt{2}$. Side $CA = \sqrt{(1-0)^{2}+0^{2}+(0-1)^{2}} = \sqrt{1+0+1} = \sqrt{2}$.
Step 2: Total Perimeter
Perimeter $= AB + BC + CA = \sqrt{2} + \sqrt{2} + \sqrt{2} = 3\sqrt{2}$.
Final Answer: (d)