= Vector area of a polygonal boundary
{title2=$\mathbf A=\tfrac12\sum_iB_i\times B_{i+1}$}
For an oriented closed polygon, triangulate with a common vertex $P$. Its total <vector area> is $\tfrac12\sum_i(B_i-P)\times(B_{i+1}-P)$; the terms involving $P$ telescope, giving the displayed boundary-only formula. Its signed projected <area> along a unit vector $V$ is $V\cdot\mathbf A$. The maximum is $|\mathbf A|$, in direction $\mathbf A/|\mathbf A|$ if nonzero, and zero projections have directions perpendicular to $\mathbf A$. If $\mathbf A=0$, every direction has zero signed projection despite possible nonzero unsigned surface area.
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