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?

  • $\begingroup$ It's not clear why you believe that "this entire cone of vectors" also work as solutions. Are you claiming that each vector in the cone is perpendicular to both $v$ and $w$? $\endgroup$ – Ben Grossmann Oct 6 '20 at 2:12
  • $\begingroup$ Here is a mathematical fact: suppose that $v$ and $w$ are two non-parallel vectors. If $p,q$ are (non-zero) vectors such that $p$ is perpendicular to both $v$ and $w$ and $q$ is perpendicular to both $v$ and $w$, then $p$ and $q$ must be parallel. Is this a fact that you were not aware of? Is this the source of your confusion? $\endgroup$ – Ben Grossmann Oct 6 '20 at 2:17
  • $\begingroup$ @BenGrossmann My confusion is that 3blue1brown simply states that one perpendicular vector to v and w satisfies the this equation, when I think I can see a whole range of solutions, being that cone according to that definition he gave. Since the dot product is just |a| * |b| * cos(theta), and the magnitudes of p and xyz can stay the same while p rotates around xyz at the same angle that it was when it was perpendicular to v and w, cos(theta) can remain constant, giving a range of solutions. I feel like i'm missing some simple fact since there very obviously is only one answer algebraically. $\endgroup$ – SpyGuyTBM Oct 6 '20 at 3:04
  • $\begingroup$ I've asked some yes/no questions, it would be very helpful if you could respond to those $\endgroup$ – Ben Grossmann Oct 6 '20 at 15:26
  • $\begingroup$ @BenGrossmann Sorry, yes I am aware of the fact that p and q would be parallel if they were both perpendicular to v and w. And no, i'm claiming that the dot product of each vector in that cone with the white vector is the same. $\endgroup$ – SpyGuyTBM Oct 6 '20 at 16:29

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 $\vec w$.

  • $\begingroup$ Thank you, that made sense $\endgroup$ – SpyGuyTBM Oct 6 '20 at 17:48

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