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:

In 3D, distance between \((x_1,y_1,z_1)\) and \((x_2,y_2,z_2)\) is \(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}\).

Answer 1

The Correct Option is D

Step 1: Compute side lengths using 3D distance formula.
Between \(A(1,0,0)\) and \(B(0,1,0)\):
\[ AB=\sqrt{(1-0)^2+(0-1)^2+(0-0)^2} =\sqrt{1+1}=\sqrt{2} \]
Between \(B(0,1,0)\) and \(C(0,0,1)\):
\[ BC=\sqrt{(0-0)^2+(1-0)^2+(0-1)^2} =\sqrt{1+1}=\sqrt{2} \]
Between \(C(0,0,1)\) and \(A(1,0,0)\):
\[ CA=\sqrt{(0-1)^2+(0-0)^2+(1-0)^2} =\sqrt{1+1}=\sqrt{2} \]
Step 2: Add all sides.
\[ P=AB+BC+CA=\sqrt{2}+\sqrt{2}+\sqrt{2}=3\sqrt{2} \]
Final Answer:
\[ \boxed{3\sqrt{2}} \]