In this question…
Geometric interpretation of the cofactor expansion theorem
…Grigory explained (beautifully, in my opinion) why the cofactor expansion for calculating determinants worked by breaking it up into the dot product of the vector $\vec{u}$ and the product $\vec{v} \otimes \vec{w}$.
However, I still don't understand the equation for $\vec{v} \otimes \vec{w}.$
Why should… $$\left|\begin{matrix}{1}&{0}&{0}\\v_1&v_2&v_3\\w_1&w_2&w_3\end{matrix}\right| \vec{e_1} – \left|\begin{matrix}{0}&{1}&{0}\\v_1&v_2&v_3\\w_1&w_2&w_3\end{matrix}\right| \vec{e_2} + \left|\begin{matrix}{0}&{0}&{1}\\v_1&v_2&v_3\\w_1&w_2&w_3\end{matrix}\right| \vec{e_3}$$ …or alternatively, $$\left|\begin{matrix}\overrightarrow{e_1}&\overrightarrow{e_2}&\overrightarrow{e_3}\\v_1&v_2&v_3\\w_1&w_2&w_3\end{matrix}\right|$$ …give us a vector orthogonal to $\vec{v}, \vec{w}$ but whose magnitude is equal to the area of the parallelogram they create?
How come we can add the vectors in such a way? What does Grigory mean by "linearity"?
Thanks!
Best Answer
One way of defining the cross product is the unique vector $a \times b$ that satisfies $\langle a \times b, x \rangle = \det \begin{bmatrix} a & b & x\end{bmatrix}$ for all $x$.
Then it is clear that $a \bot (a \times b)$ and similarly for $b$. Furthermore, $\|a \times b \|^2 = \det \begin{bmatrix} a & b & a \times b\end{bmatrix}$, or $\|a \times b \| = \det \begin{bmatrix} a & b & {a \times b \over \|a \times b \|}\end{bmatrix}$, and the latter is the area of the parallogram created by $a,b$.
Note: Observe that $\langle a \times b, x \rangle = \sum_k x_k \langle a \times b, e_k \rangle $, so you can 'recover' the components of $a \times b$ using this formula (by choosing $x=e_1,e_2,e_3$). In particular, this gives $a \times b = \langle a \times b, e_1 \rangle e_1 + \langle a \times b, e_2 \rangle e_2 + \langle a \times b, e_3 \rangle e_3$