TITLE: Area of paralellogram
QUESTION [5 upvotes]: Find the area of parallelogram of the two vectors (6,0,1,3) and (2,1,3,1). The area is magnitude of the cross product of the two vectors. Right now the only way i was taught to do the cross product is getting the determinant of putting the vectors in and i,j,k matrix. I dont know how to do it with 4 points of the vector.

REPLY [3 votes]: The area of your (v,w)-parallelogram is equal to $\sqrt{Gram(v,w)}=\sqrt{\det(AA^T)}$, with the matrix $A=\begin{bmatrix}v & w\end{bmatrix}$.
See, for example,  Why is the volume of a parallelepiped equal to the square root of $\sqrt{det(AA^T)}$ or How do I compute the area of this parallelogram.