Area of paralellogram

Question

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.

Answer 1

For: $$ \vec v=(6,0,1,3)^T \qquad \vec u=(2,1,3,1)^T $$ we have: $$\vec v \cdot \vec u= 18 \qquad |\vec v|=\sqrt{46} \qquad |\vec u|=\sqrt{15} $$

so, the angle $\theta$ between the two vectors is such that: $$ \cos \theta= \frac{18}{\sqrt{46}\sqrt{15}} \qquad \sin \theta= \sqrt{1-\cos^2 \theta} $$

And the area is: $$ A= |\vec v| \cdot |\vec u| \sin \theta= \sqrt{46}\sqrt{15}\sqrt{1-\frac{18^2}{46 \cdot 15}} =\sqrt{46\cdot 15 - 18^2} $$

Answer 2

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.