Prove that $(\mathbf{1}\times\textbf{u})\cdot\mathbf{v}=\mathbf{u}\cdot(\mathbf{1}\times\textbf{v})=\textbf{u}\times\textbf{v} $

Question

As the title, $\textbf{u}$ and $\textbf{v}$ be two vectors. The problem didn't explain, but I think $\textbf{1}$ is a dyadic?
I did try using index notation, and here is the following steps I did.

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Proof:

$$(\mathbf{1}\times\textbf{u})\cdot\mathbf{v} =(\mathbf{e}_{i}\mathbf{e}_{i}\times{u}_{j}\mathbf{e}_{j})\cdot{v}_{k}\mathbf{e}_{k}={u}_{j}{v}_{k}[\mathbf{e}_{i}(\mathbf{e}_{i}\times\mathbf{e}_{j})\cdot\mathbf{e}_{k}]$$

$$={u}_{j}{v}_{k}[\mathbf{e}_{i}(\mathbf{e}_{k}\times\mathbf{e}_{i})\cdot\mathbf{e}_{j}] =-{u}_{j}{v}_{k}[\mathbf{e}_{i}(\mathbf{e}_{i}\times\mathbf{e}_{k})\cdot\mathbf{e}_{j}] =-(\mathbf{e}_{i}\mathbf{e}_{i}\times{v}_{k}\mathbf{e}_{k})\cdot{u}_{j}\mathbf{e}_{j} $$

$$=-(\mathbf{1}\times\textbf{v})\cdot\mathbf{u}=-\mathbf{u}\cdot(\mathbf{1}\times\textbf{v})$$

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Finally, I got $(\mathbf{1}\times\textbf{u})\cdot\mathbf{v} =-\mathbf{u}\cdot(\mathbf{1}\times\textbf{v})$ which is different from answer $\mathbf{u}\cdot(\mathbf{1}\times\textbf{v})$
Were there anything I'm missing or logical misunderstanding that make outcome wrong?
Any advice would be greatly appreciated!!!

Answer 1

If the course handout defines the dyadic $\mathbf{1}$ as $\mathbf{e}_i\mathbf{e}_i$ (with summation over $i=1,2,3$) it is extremely likely, see [1], that it also defines

• $\mathbf{1}\times\mathbf{u}$ as $\mathbf{e}_i(\mathbf{e}_i\times\mathbf{u})$ where the cross product between vectors $\mathbf{e}_i$ and $\mathbf{u}$ is the ordinary one.

Since $$\mathbf{e}_1\times\mathbf{u}=\begin{pmatrix}0\\-u_3\\u_2\end{pmatrix},\quad\mathbf{e}_2\times\mathbf{u}=\begin{pmatrix}u_3\\0\\-u_1\end{pmatrix}\,,\quad\mathbf{e}_3\times\mathbf{u}=\begin{pmatrix}-u_2\\u_1\\0\end{pmatrix} $$ and since $\mathbf{e}_i(\mathbf{e}_i\times\mathbf{u})$ is most likely a dyadic (tensor product) we get (with summation over $i=1,2,3$) $$\tag{1} \mathbf{e}_i(\mathbf{e}_i\times\mathbf{u})=\begin{pmatrix}0&-u_3&u_2\\u_3&0&-u_1\\-u_2&u_1&0\end{pmatrix}\,. $$ It is obvious and well-known that the multiplication of this anti-symmetric matrix with a vector $\mathbf{v}$ equals the ordinary cross product $\mathbf{u}\times\mathbf{v}$.

Remarks:

• By [1] all products $\mathbf{a}\mathbf{b}$ are tensor (outer products) and thus nothing else than matrices: $$ \mathbf{a}\mathbf{b}=\mathbf{a}\otimes\mathbf{b}=\mathbf{a}\mathbf{b}^\top= \begin{pmatrix}a_1b_1&a_1b_2&a_1b_3\\a_2b_1&a_2b_2&a_2b_3\\ a_3b_1&a_3b_2&a_3b_3\end{pmatrix}\,. $$

• For the unit basis vectors $\mathbf{e}_i$ we therefore have $$ \mathbf{1}=\mathbf{e}_1\mathbf{e}_1+\mathbf{e}_2\mathbf{e}_2+\mathbf{e}_3\mathbf{e}_3=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}\,. $$

• The cross product of a dyad and a vector, $(\mathbf{a}\mathbf{b})\times\mathbf{c}$, is defined as the dyad $\mathbf{a}(\mathbf{b}\times\mathbf{c})$ where $\mathbf{b}\times\mathbf{c}$ is an ordinary cross product.

• Applying this to the individual summands of the dyad $\mathbf{1}$ I get from linearity $$ \mathbf{1}\times\mathbf{u}=\mathbf{e}_1(\mathbf{e}_1\times\mathbf{u})+\mathbf{e}_2(\mathbf{e}_2\times\mathbf{u})+\mathbf{e}_3(\mathbf{e}_3\times\mathbf{u}) $$ which leads to (1) above.