Area of quadrilateral using determinant/cross product

Question

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.

Answer 1

Consider an arbitrary quadrilateral.

Move two vertices parallelly to a diagonal, so that two sides become aligned with the other diagonal. This transformation does not change the area.

Then move a vertex so that one side becomes aligned with the first diagonal. This transformation also preserves the area.

The area is that of a triangle, half the cross-product of the diagonal vectors.

Answer 2

For parallelograms:

Assuming that $\vec a$ and $\vec b$ are the 2 non-parellal vectors of the parallelogram, then the diagonals of this parallelogram are $\vec a +\vec b$ and $\vec a - \vec b$

Now by applying the cross product you get $||(\vec a +\vec b) × (\vec a -\vec b)||=2||(\vec a ×\vec b)||=2A$

($A=$area of the parallelogram).

For the general case:

Consider the quadrilateral $ABCD$ of center $O$ (the intersection of diagonals) and consider the parallelogram formed by drawing 2 parallel lines to $(AC)$ one passing through $D$ and the other through $B$, and also another 2 parallel lines to $(BD)$ one passing through $A$ and the other through $C$. Let this parallelogram be $A'B'C'D'$ (for example name the vertex that is to the right of $A$ by $A'$ and the vertex right to $B$ by $B'$ and so on).

Now the determinant of the cross product of the diagonals of the quadrilateral gives you the area of this parallelogram ($||\vec {AC} ×\vec {BD}||=||\vec {A'B'} ×\vec {A'D'}||= A_{A'B'C'D'}$), but also you can easily show that the triangles $AA'B$ and $AOB$ are congruent (which means that they have the same area), and also the triangles $BB'C$ and $BOC$ are also congruent and so on, this means that the area of $A'B'C'D'$ is just double the area of the quadrilateral $ABCD$, and this is the result we want.

$$A_{ABCD}=\frac{A_{A'B'C'D'}}{2}=\frac{||\vec {AC} ×\vec {BD}||}{2}$$

Answer 3

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 $$ |\vec v\times w| = |{\rm det}[\vec v, \vec w]|$$