Question

The perimeter of the triangle with vertices at $\left(1,0,0\right),\left(0,1,0\right)$ and $\left(0,0,1\right)$ is

• A. $3$
• B. $2$
• C. $2\sqrt{2}$
• D. $3\sqrt{2}$

Answer 1

The correct option is D

$3\sqrt{2}$

Explanation for the correct option:

Step 1: State the given data and formula used.

The given coordinates of the vertices of a triangle are $\left(1,0,0\right),\left(0,1,0\right)$ and $\left(0,0,1\right)$.

Let, $\left(0,0,1\right),\left(0,1,0\right)$ and $\left(1,0,0\right)$ be the coordinates of the vertices $A,B$ and $C$ respectively of the $△ABC$.

So, $AB,BC$ and $CA$ are the sides of $△ABC$.

Now, we know that in a three-dimensional space ($XYZ$-space), the distance $\left(d\right)$ between two points $A\left(x_1,y_1,z_1\right)$ and $B\left(x_2,y_2,z_2\right)$ can be calculated by the following formula,

$d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2+{\left(z_2-z_1\right)}^2}$

Step 2: Calculate the lengths of each side.

In the $△ABC$, $A\left(0,0,1\right),B\left(0,1,0\right)$ and $C\left(1,0,0\right)$ are the vertices.

The lengths of the sides $AB,BC$ and $CA$ can be calculated using the distance formula as follows,

$\begin{array}{cl}& AB=\sqrt{{\left(0-0\right)}^2+{\left(1-0\right)}^2+{\left(0-1\right)}^2}\\ \Rightarrow & AB=\sqrt{0+1+1}\\ \Rightarrow & AB=\sqrt{2}\end{array}$

And, $BC=\sqrt{{\left(1-0\right)}^2+{\left(0-1\right)}^2+{\left(0-0\right)}^2}$

$\begin{array}{cl}\Rightarrow & BC=\sqrt{1+1+0}\\ \Rightarrow & BC=\sqrt{2}\end{array}$

And, $CA=\sqrt{{\left(0-1\right)}^2+{\left(0-0\right)}^2+{\left(1-0\right)}^2}$

$\begin{array}{cl}\Rightarrow & CA=\sqrt{0+1+1}\\ \Rightarrow & CA=\sqrt{2}\end{array}$

Step 3: Calculate the perimeter of the triangle.

Now, as we know, the perimeter of a triangle is the sum of the lengths of its side. So,

The perimeter of the triangle $△ABC=AB+BC+CA$

$\Rightarrow$ The perimeter of the triangle $△ABC=\sqrt{2}+\sqrt{2}+\sqrt{2}$

$\Rightarrow$ The perimeter of the triangle $△ABC=3\sqrt{2}$

Hence, option $D$ is the correct option.