The dot product is the special case of a more general concept, the inner product. If you have a vector space $ V $ over the reals or the complex numbers, then an inner product is a map $ f : V \times V \to \mathbb{C} $ or $ f : V \times V \to \mathbb{R} $ which is conjugate symmetric, positive definite, and linear in its first argument. We usually write $ f(u, v) = \langle u, v \rangle $, in which case these properties can be summed up as follows:
- Conjugate symmetry: $ \overline{\langle u, v \rangle} = \langle v, u \rangle $, where $ \bar{z} $ denotes complex conjugation. Note that this implies $ \langle u, u \rangle $ is always real for any vector $ u $.
- Positive definiteness: $ \langle v, v \rangle \geq 0 $ for any $ v \in V $, with equality holding iff $ v = 0 $.
- Linearity in the first argument: $ \langle \alpha u + \beta v, w \rangle = \alpha \langle u, w \rangle + \beta \langle v, w \rangle $ where $ u, v, w \in V $ and $ \alpha, \beta $ are in the field of scalars.
If $ V = \mathbb{R}^n $, then we can fix a basis $ B = \{ b_i \in \mathbb{R}, 1 \leq i \leq n \} $ and define $ \langle b_i, b_i \rangle = 1 $ and $ \langle b_i, b_j \rangle = 0 $ for $ i \neq j $. Extending this to all of $ \mathbb{R}^n $ by linearity gives us
$$ \left \langle \sum_{k=1}^{n} c_k b_k, \sum_{j=1}^{n} d_j b_j \right \rangle = \sum_{1 \leq k, j \leq n} d_k c_j \langle b_i, b_j \rangle = \sum_{i=1}^{n} c_i d_i $$
where positive definiteness is readily verified. You will recognize this expression as the definition of the dot product. Indeed, if we take our basis $ B $ to be the standard basis of $ \mathbb{R}^n $, then this inner product is the dot product.
Why is this formalism more powerful? A result about the inner product is the Cauchy-Schwarz inequality, which says that $ |\langle u, v \rangle| \leq |u| |v| $ where $ |u| = \sqrt{\langle u, u \rangle} $. This tells us that
$$ -1 \leq \frac{\langle u, v \rangle}{|u| |v|} \leq 1 $$
assuming that our field of scalars is $ \mathbb{R} $. We then see that the arccosine of this expression is well-defined, so we can define the angle between nonzero vectors $ u $ and $ v $ as
$$ \theta = \arccos \left( \frac{\langle u, v \rangle}{|u| |v|} \right) $$
The properties we expect to be true are then easily verified. This notion extends to infinite dimensional vector spaces over $ \mathbb{R} $, where defining angle is not at all obvious. It is then trivially true that we have $ \langle u, v \rangle = |u| |v| \cos(\theta) $, since that is how $ \theta $ was defined.
The cross product is an entirely separate concept which allows us to find a vector orthogonal to two given vectors in $ \mathbb{R}^3 $. In addition, its magnitude also gives the area of the parallelogram spanned by the vectors. These properties can be taken as the definition of the cross product (with appropriate care for orientation), or they can be derived as theorems starting from the algebraic definition.
The area vector of a flat surface has apparently in your text been defined to be perpendicular to the surface. If we want to measure flow through a surface, we want to know the amount of velocity vector is parallel to the area vector (i.e., perpendicular to the surface).
The diagram is perhaps a little confusing. The area $A$ is a parallelogram drawn on the floor. Its area vector points up, perpendicular to the floor. The velocity vector points somewhere in the halfspace above the floor. The empty parallelogram is parallel transported along $v$ in one unit of time (whatever time units you are using). The volume of flow extruded through the parallelogram on the floor is the area of that parallelogram times the height (measured perpendicularly to the floor) of the parallelepiped standing on it (bounded by the two parallelograms and the four dotted lines).
If $v$ is parallel to the area vector, then the height is as large as it can be and also, the flow through the parallelogram is maximized. If $v$ is parallel to the floor, there is no flow through the parallelogram (all flow is along the parallelogram), so the volume of the parallelepiped is zero. The dot product is maximized when parallel and zero when perpendicular, which reproduces the desired behaviour. The cross product is zero when parallel and maximized when perpendicular, which is exactly the opposite of the desired behaviour.
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No.
What you found was that $v_1\cdot v_2 = \Vert v_1 \times v_r\Vert$, where $v_r$ is a 90-degree rotation of $v_2$ in the xy-plane, right? This is expected. We can show it by working out both sides out from their coordinates. Or we can see geometrically that $v_1\cdot v_2 = \cos\theta \Vert v_1 \Vert \Vert v_2 \Vert$, while the area of the parallelogram is $|\sin(\theta+\pi/2)|\Vert v_1 \Vert \Vert v_r \Vert = |\cos\theta|\Vert v_1 \Vert \Vert v_2 \Vert$. So the dot product is $\pm$ the length of the cross product.
However, when we transpose in the expression $v_1\cdot v_2 = v_1^Tv_2$, that is not a rotation of $v_1$. Note how it doesn't say $v_1^T\times v_2$, i.e. it is not a cross product. Rather, it is to be read as a matrix product, where $v_2$ is a column vector, and $v_1^T$ has been changed into a row vector by transposing. It still represents the same vector in space, but is notated differently (horizontally).