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. $2\sqrt2$
• D. $3\sqrt2$

Answer 1

Answer: (D)

$3\sqrt2$

Answer 2

Let $A(1, 0, 0), B (0, 1, 0), C (0, 0, 1)$ be the vertices of $\Delta ABC$ $\therefore\, AB = \sqrt{(0 - 1)^2 + (1 -0)^2 + (0 - 0)^2} = \sqrt{2}$ BC = $\sqrt{2}$ [Similarly], CA = $\sqrt{2}$ $\therefore$ Perimeter = $\sqrt{2} +\sqrt{2} + \sqrt{2} = 3\sqrt{2}$