Problem on the area of a triangle in 3-space
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Find the area of the triangle in 3-space between $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$.
Solution
We begin by noting that the area of a triangle spanned by two vectors is half the area of the parallelogram spanned by those vectors:
We recall
Cross product and area of parallelograms
The area of the parallelogram spanned by the vectors $\mathbf{x}$ and $\mathbf{y}$ is $| \mathbf{x} \times \mathbf{y}|$.
We proceed by identifying two vectors that span the triangle.
Component view of vector subtraction
These two vectors are: \begin{align}\mathbf{x} &= ( 0,1,0 ) - ( 1,0,0 ) = ( -1, 1, 0 )\\ \mathbf{y} &= ( 0,0,1 ) - ( 1,0,0 ) = ( -1, 0, 1 ).
\end{align}
We now compute
Cross product definition
$$\begin{align}\mathbf{x} \times \mathbf{y} &= (-1,1, 0) \times (-1, 0, 1) \\ &= (1,1,1). \end{align}$$
Length of a vector
The area of the spanned parallelogram is $\left| \mathbf{x} \times \mathbf{y} \right| = \sqrt{3}$.
Thus, the area of the triangle between these vectors is $\sqrt{3}/2$.
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