Why are vectors in physics assumed to be represented as column vectors? I was reading over the definition of a dot product, and the stated equivalency was that $a\cdot b = a^Tb$. I always assumed that vectors were represented as row vectors, not as column vectors, but it seems the opposite is generally accepted. Is there a particular reason for this?
 A: Why to use one notation over the other is a matter of convention really. Given a vector space $V$ over $\mathbb{R}$ for example, it is convention that when we introduce a basis $\{e_i\}$ of $V$ we represent
$$v = \sum v^i e_i$$
as either the element $(v^1,\dots, v^n)$ of $\mathbb{R}^n$ or the column vector
$$[v]=\begin{bmatrix}v^1\\ \vdots\\v^n\end{bmatrix}$$
sitting on $M_{n\times 1}(\mathbb{R})$ the space of matrices of $n$ lines and one column.
When the second option is chosen it is then also a convention that given the dual space $V^\ast$ (the vector space of all linear functions $f : V\to \mathbb{R})$ and the dual basis $\{\varepsilon^i\}$ defined by requring $\varepsilon^i(e_j)=\delta^i_j$ one represents the covector $\omega\in V^\ast$ with
$$\omega = \sum \omega_i \varepsilon^i$$
by the row vector
$$[\omega]=\begin{bmatrix} \omega_1 & \cdots & \omega_n\end{bmatrix}$$
living on $M_{1\times n}(\mathbb{R})$ so that the action of $\omega$ on $v$ coincides with the action of the row vector on the column vector.
Of course you could represent vectors on $M_{1\times n}(\mathbb{R})$ and covectors on $M_{n\times 1}(\mathbb{R})$. But then if $[v]\in M_{1\times n}(\mathbb{R})$ and $[\omega]\in M_{n\times 1}(\mathbb{R})$ you would have
$$\omega(v)=[\omega]^T [v].$$
While with the usual choice you have just
$$\omega(v)=[\omega][v].$$
As for the dot product, the thing is that when there is one inner product, every vector yields a covector. So if $a,b\in V$ the covector associated to $a$ is $\omega_a$ given by
$$\omega_a(v) = a\cdot v.$$
In particular $a\cdot b = \omega_a(b)$. Now if $\{e_i\}$ is an orthonormal basis that meaning $e_i\cdot e_j = \delta_{ij}$ we have that $[\omega_a]=[a]^T$ which yields your formula
$$a\cdot b = [a]^T [b].$$
