Proving area of plane quadrilateral is $\frac{\|\vec{p}\times\vec{q}\|}{2}$, where $p, q$ are diagonals

Question

$$\text{Area}=\frac{\|\vec{p}\times\vec{q}\|}{2}$$

We have to show the above formula for area of a planer quadrilateral, where $\vec{p}, \vec{q}$ are diagonals of quadrilateral.

I consider following quadrilatrel:

$$\vec{a}+\vec{b}+\vec{c}+\vec{d} = 0 \\ \vec{a}+\vec{d}=\vec{p}\\ \vec{a}+\vec{b}=\vec{q}$$

Also, $$\frac{\|\vec{a}\times\vec{d}\|}{2} + \frac{\|\vec{b}\times \vec{c}\|}{2} = \text{Area} $$

Now I cannot get into the required form. Any hint (big one) will be helpful!

Answer 1

First note that if we arrange the order of the cross products in the right way we have,

$$ \| \vec{a}\times \vec{d} \| + \| \vec{c} \times \vec{b} \| = \| \vec{a}\times \vec{d} + \vec{c} \times \vec{b} \|, $$

then note the following,

$$\vec{a} \times \vec{d} = (\vec{a} + \vec{d}) \times (\vec{d}+\vec{c}-\vec{c}) = \vec{p} \times(\vec{q}-\vec{c}) = \vec{p} \times \vec{q} - \vec{p} \times \vec{c},$$

and,

$$ \vec{c} \times \vec{b} = \vec{c} \times ( \vec{b} + c ) = \vec{c} \times (-\vec{p}) = \vec{p}\times \vec{c}, $$

from which we get,

$$\vec{a} \times \vec{d} + \vec{c} \times \vec{b} = \vec{p} \times \vec{q} $$