TITLE: Area of quadrilateral using determinant/cross product
QUESTION [6 upvotes]: Consider a quadrilateral whose sides are given by the vectors $\vec{a}, \vec{b}, \vec{c}$ and $\vec{d}$, such that $\vec{a}+\vec{b}+\vec{c}+\vec{d}=0$.
I'm told that the area of the quadrilateral can be calculated by half of the determinant of the matrix with columns given by the diagonals of the quadrilateral. 
Why is this the case?
My initial approach was to use the cross product to find the area of the parallelogram spanned by two sides of the quadrilateral but this didn't seem yield anything meaningful.

REPLY [2 votes]: Consider the plane embeded in 3-diomensional space, so that we can use vector product.
Area of a triangle with sides $\vec{a}$, $\vec{b}$ and $\vec{c}$ ($\vec a+\vec b+\vec c= 0$) is given by $$ S_\triangle = \frac12|\vec a||\vec b|\sin (\angle ab) = \frac12|\vec a \times \vec b|$$
Notice that you can divide the quadriteral into two triangles, one with sides $\vec a, \vec b, \vec c+\vec d$, and the other with sides $\vec a+\vec b, \vec c, \vec d$ ($\vec a+\vec b = -\vec c-\vec d$). We have
$$S_1 = \frac12|(\vec c+ \vec d) \times \vec a| $$
$$ S_2 = \frac12|\vec d \times (\vec a+\vec b)|= \frac12|(\vec c +\vec d) \times \vec d| $$
The total area is 
$$ S = \frac12 |(\vec c+ \vec d) \times (\vec d+\vec a)| $$.
Note that $\vec c+\vec d$ and $\vec d+\vec a$ are the two diagonals, and you have $$ |\vec v\times w| = |{\rm det}[\vec v, \vec w]|$$