Understanding Cross Product Definition Derivation by 3blue1brown I've been following through with 3blue1brown's linear algebra series, and I have a question regarding the definition of the cross product he gives. https://www.youtube.com/watch?v=BaM7OCEm3G0&list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab&index=11
$$\begin{bmatrix} 
p_1
\\p_2 
\\p_3
\end{bmatrix}
\cdot 
\begin{bmatrix} 
x
\\y
\\z
\end{bmatrix}
=
det\left(\begin{bmatrix}
 x & v_1 & w_1\\ 
y & v_2 & w_2\\ 
z & v_3 & w_3
\end{bmatrix}\right)$$
where p is the resultant vector from the cross product of v and w for any x y and z. 3blue1brown essentially says that since the determinant of a matrix is the area of the parallelepiped with side lengths of the column vectors, the determinant is also just the height of that parallelepiped times the base of it. And the dot product of p and xyz is the projection of xyz on p, times the magnitude of p. If p is a vector perpendicular to v and w, then the projection of the final side of the parallelepiped(xyz) onto that perpendicular vector would be the height of the parallelepiped, and then the magnitude of p would the area of the base.
So that makes logical sense, but according to this definition couldn't this entire cone of vectors also be solutions?
The Cone of Vectors
since the projection of xyz on p and the magnitude of p stays the same?

The point is that we're looking for a vector $\vec p$ that has the correct dot-product for every choice of (white) vector $(x,y,z)$.
It is true that any choice of vectors from the cone gives you the correct dot product for the particular input shown in the figure, but the point behind the duality argument is that our $p$ needs to correctly encode the function that takes in a vector $(x,y,z)$ and produces the corresponding area. It is only true that $\vec p$ correctly encodes the function if $(p_1,p_2,p_3) \cdot(x,y,z)$ gives the right output for every possible input.
We can see by plugging in $(x,y,z) = \vec v$ that $\vec p$ needs to be perpendicular to $\vec v$ (since the volume should be $0$). Similarly, $\vec p$ needs to be perpendicular to $\vec w$.
