This question was asked in a quiz.I've proved that these points are coplanar,all lying in $x+y-z=0$.I tried it by using the formula for the area of tringles in coordinate geometry(taking points $A,B,C$ & points $A,B,D$),but arranging the points in a correct manner so that there is no area common to both the triangles seems cumbersome.
Is there any general formula for the computation of area of the given quadrilateral?
EDIT:After H.H.Rugh's answer
- $\vec{a} \times \vec{b}=\begin{vmatrix} i & j & k \\ 1 & 0 & 1 \\ 1 & 3 & 4 \\ \end{vmatrix} =i(0-3)-j(4-1)+k(3-0)=-3i-3j+3k$
- $\vec{b} \times \vec{c}=\begin{vmatrix} i & j & k \\ 1 & 3 & 4 \\ 0 & 2 & 2 \\ \end{vmatrix} =i(6-8)-j(2-0)+k(2-0)=-2i-2j+2k$
$\vec{c} \times \vec{d}=\begin{vmatrix} i & j & k \\ 0 & 2 & 2 \\ 3 & 2 & 5 \\ \end{vmatrix} =i(10-4)-j(0-6)+k(0-6)=6i+6j-6k$
$\vec{d} \times \vec{a}=\begin{vmatrix} i & j & k \\ 3 & 2 & 5 \\ 1 & 0 & 1 \\ \end{vmatrix} =i(2-0)-j(3-5)+k(0-2)=2i+2j-2k$
Putting above values in,$ \vec{S} = \frac12 \left( \vec{a}\times \vec{b}+ \vec{b}\times \vec{c} +\vec{c}\times \vec{d} +\vec{d}\times \vec{a}\right) $,we get $\vec{S}=3i+3j-3k \implies \vert \vec{S} \vert=\frac12 \vert 3i+3j-3k \vert=\frac32 \sqrt 3$.
But,its correct answer is $\frac92 \sqrt 3$
You may compute the signed area of $OAB$ by: $$ \frac12 \vec{OA}\times \vec{OB} = \frac12 \vec{a}\times \vec{b}$$ In your case the order of points which produce a genuine quadrilatere is: $A,D,B,C$ (look e.g. at the (x,y) projection) so the formula is: Now you simple add all the contributions by cyclically permuting $ADBC$: $$ \vec{S} = \frac12 \left( \vec{a}\times \vec{d}+ \vec{d}\times \vec{b} +\vec{b}\times \vec{c} +\vec{c}\times \vec{a}\right) $$ Cancellation will occur for the part which is outside the quadrilateral. The area is the length of $S$.
If you know that the points form a convex quadrilatere then you may add the absolute value of the areas of $ABC$ and $ABD$. Complications may arise, however, when it is not convex.