Tensor product, wedge product, Hodge product, dyad, or what? Suppose I have two vectors $\mathbf{u} = (u_1, u_2, u_3)$ and $\mathbf{v} = (v_1, v_2, v_3)$ in $\mathbb{R}^3$. I can regard $\mathbf{u}$ as a $3 \times 1$ matrix, and $\mathbf{v}$ as a $1 \times 3$ matrix, and then I can form their product to get a $3 \times 3$ matrix, which is sometimes denoted by $\mathbf{u} \otimes \mathbf{v}$: 
$$ 
\mathbf{u} \otimes \mathbf{v} =
\begin{bmatrix}
u_1v_1 & u_1v_2 & u_1v_3 \\ 
u_2v_1 & u_2v_2 & u_2v_3 \\ 
u_3v_1 & u_3v_2 & u_3v_3 \\ 
\end{bmatrix}
$$
It seems like this thing would be very useful in writing vector formulae in 3D geometry, which is what interests me.
So, my specific questions are:
(1) What is this thing called? The names "tensor product", "wedge product", and "Hodge product" all seem related somehow, but none of them seem to fit perfectly. Outer product or dyad, maybe?
(2) Where can I find a systematic account of its algebraic properties? There must be many relationships between this thing and the plain old vector dot and cross products. It's very interesting that $(\mathbf{u} \cdot \mathbf{v})\mathbf{w} = \mathbf{u} (\mathbf{v}\otimes \mathbf{w})$ for example.
(3) Is there any geometric intepretation (as there is with dot and cross products)?
I'm interested in 3D geometry, not in abstractions, so simple concrete answers would be preferable, please. 
 A: This is the tensor product of the vectors $u,v$. The tensor product operation is quite general, but in the case of $2$ vectors the simpler name "outer product" is sometimes used.
The algebraic properties you want to know come from the properties of the tensor product. For example, to see that
$$(u \cdot v)w = u(v \otimes w)$$
Pick a fourth vector $z$. Then
$$u(v \otimes w)z^{T} = (u \cdot v)(w \cdot z) = (u \cdot v)w z^{T}$$
So that both sides define the same linear map, and hence must be the same.
Lastly, regarding geometric intuition, the tensor product is a general algebraic tool that does not always have a geometric explanation. However, one can say that these objects are useful in geometry because a tensor is a linear object that behaves nicely under a change of coordinates, so it is always useful to develop geometric constructions around the framework of tensors and tensor products.
A: As @Elchanan Solomon said, the object $u \otimes v$ is the tensor/outer product of the two vectors $u$ and $v$, also written as $uv^T$ in $\mathbb{R}^n$
With regards to geometric intuition, one can think of the specific case of the outer product of a vector $v$ with itself $v \otimes v$ as a projection operator, that when applied onto another vector $w$, will produce the vector projection of $w$ onto $v$
$$ (v \otimes v)w \propto \vec{\text{Proj}}_{v}(w) $$
The relationship is a proportionality due to the norm of $v$. If $v$ was normalized, we'd have the equality
$$ (v \otimes v)w = \vec{\text{Proj}}_{v}(w) $$
If $v$ was not normalized, we'd end up with an additional factor of the inner product $v \cdot v$ to "cancel out" the norms of $v$ in the outer product,
$$ \frac{(v \otimes v)}{v \cdot v}w = \vec{\text{Proj}_{v}(w)} $$
To see where this comes from, consider the projection operator in quantum mechanics which is constructed as an outer product,
$$ P_i = \vert i \rangle \langle i \vert $$
Where $\vert i \rangle$ is the set of orthonormal basis vectors and $\langle i \vert$ are the corresponding bras from the dual space. Since the bras are the adjoints (basically transpose-conjugates) of the kets, if the ket $\vert i \rangle$ was a column vector, then the bra $\langle i \vert$ is a row vector so $\vert i \rangle \langle i \vert$ is the product of a column vector and a row vector (in that order), the result of which is a matrix (operator).
From this, we have the completeness relation for a finite dimensional Hilbert space:
$$ \sum_{i} P_i = I $$
$$ \sum_{i} \vert i \rangle \langle i \vert = I $$
Which tells us that adding up all the vector projections of a vector along all the basis vectors is equivalent to reconstructing the vector itself, hence the presence of the identity operator.
Now, applying the projection operator to any vector $\vert v \rangle$ from the Hilbert space, we see that,
$$ P_i\vert v \rangle = \vert i \rangle \langle i \vert v \rangle $$
$$ P_i\vert v \rangle = \langle i \vert v \rangle \vert i \rangle $$
Where $\langle i \vert v \rangle$ is the inner product of $\langle i \vert$ with $\vert v \rangle$. If we translate the vector operations and bra-ket notation present here, it's clear that this is the same as the vector projection of $v$ onto $e_i$ in geometric vector spaces such as $\mathbb{R}^n$,
$$ \vec{\text{Proj}}_{e_i}(v) = (e_i \cdot v)e_i$$
When we're not dealing with projections onto normalized vectors, we'd have to account for the norms via
$$ P_{u}\vert v \rangle = \frac{\langle u \vert v \rangle \vert u \rangle}{\langle u \vert u \rangle}$$
And,
$$ \vec{\text{Proj}}_{u}(v) = \frac{(u \cdot v)}{u \cdot u}u $$
Where $ u $ is the vector being projected onto.
Rearranging, we can see that
$$ P_{u}\vert v \rangle = \frac{\vert u \rangle \langle u \vert v \rangle }{\langle u \vert u \rangle} $$
If you allow me to extract out the quantity operating on $ \vert v \rangle $,
$$ P_{u}\vert v \rangle = \left(\frac{\vert u \rangle \langle u \vert }{\langle u \vert u \rangle} \right) \vert v \rangle $$
By comparing,
$$ P_{u} = \frac{\vert u \rangle \langle u \vert }{\langle u \vert u \rangle} $$
Doing the same for $\mathbb{R}^n$,
$$ \vec{\text{Proj}}_{u}(v) = \frac{(u \cdot v)}{u \cdot u}u $$
$$ \vec{\text{Proj}}_{u}(v) = \frac{u(u \cdot v)}{u \cdot u} $$
$$ \vec{\text{Proj}}_{u}(v) = \frac{(u \otimes u)v}{u \cdot u} $$
$$ \vec{\text{Proj}}_{u}(v) = \frac{u \otimes u}{u \cdot u}v $$
Therefore,
$$ \vec{\text{Proj}}_{u} = \frac{u \otimes u}{u \cdot u} $$
With this knowledge in hand, we can understand OP's statement about how the projection and reflection of a vector onto/via a plane involves the outer product.

