# Cross Products and Determinants Geometrically

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!

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 $$a \times b = \langle a \times b, e_1 \rangle e_1 + \langle a \times b, e_2 \rangle e_2 + \langle a \times b, e_3 \rangle e_3$$
• @JoshuaRonis: I'm not sure it helps, but I have expanded slightly above. Note that $\det M = \det M^T$ so it is immaterial whether you use the row or column version. – copper.hat Jul 5 at 20:44
• Got you! Just to tie up what you're saying for me to refer back to in the future: Let $\vec{h}$ be the vector such that $\forall \vec{x}: (\vec{h} \bullet \vec{x} = det|\vec{a}, \vec{b}, \vec{x}|)$. Through copper.hat's and 3Blue1Brown's explanations, that means that $\left\Vert \vec{h} \right\Vert = (\vec{a}, \vec{b})$, where $(\vec{a}, \vec{b})$ denotes the area of the parallelogram created by $\vec{a}$ & $\vec{b}$. Additionally, $\vec{h}$ must be orthogonal to that parallelogram. That's the first part. Now, for the second part, note that: $(\vec{h} \bullet \hat{e_1})(\hat{e_1})$ is ... – Joshua Ronis Jul 6 at 21:22
• ...the $\hat{e_1}$ component of $\vec{h}$, and so on for the other two basis vectors. That means we can "recover" $\vec{h}$ by writing it as a linear combo of its three components: $\vec{h} = (\vec{h} \bullet \hat{e_1})(\hat{e_1}) + (\vec{h} \bullet \hat{e_2})(\hat{e_2}) + (\vec{h} \bullet \hat{e_3})(\hat{e_3})$. But, from our first rule about $\vec{h}$, that can be rewritten as $\vec{h} = (det|\vec{a}, \vec{b}, \hat{e_1}|)(\hat{e_1}) +(det|\vec{a}, \vec{b}, \hat{e_2}|)(\hat{e_2}) +(det|\vec{a}, \vec{b}, \hat{e_3}|)(\hat{e_3})(\hat{e_3})$, and from there arises the cofactor expansion! – Joshua Ronis Jul 6 at 21:24