Components of vector In the notation:
Vector(v) = ai + bj + ck

Aren't a, b, c the components of vector? I looked on Wikipedia which says they "are the magnitudes of the components of v." 
Does it mean the same?
If not then how can I convert the components of vector given in set notation:
$v = (c_x, c_y, c_z)$.
Where, $c_x, c_y, c_z$ are the components into the $i, j, k$ notation?

Consider a vector $\mathbf v \in V$ where $V$ is a $3$-dimensional real vector space.  There is a theorem which tells us that every vector space has at least one basis.  That allows us to say there exists a basis $S=\{\mathbf s_1, \mathbf s_2, \mathbf s_3\}$ of $V$.  By the definition of a basis, we can write $\mathbf v$ as a linear combination of $S$:
$$\mathbf v = v_1\mathbf s_1 + v_2\mathbf s_2 + v_3\mathbf s_3$$
After choosing a basis for $V$ we can make use of another very useful theorem in linear algebra: every $n$-dimensional real vector space is isomorphic to $\Bbb R^n$, where $\Bbb R^n$ is the space of all ordered "tuples" of real numbers: for instance $\Bbb R^3 = \{(a,b,c) \mid a,b,c \in \Bbb R\}$.  What that means for us is that we can freely convert any vectors in our abstract real vector space $V$ to vectors in $\Bbb R^3$ and vice versa.  The "natural" way to do so is $$\mathbf v = v_1\mathbf s_1 + v_2\mathbf s_2 + v_3\mathbf s_3 \leftrightarrow (v_1, v_2, v_3) = (\mathbf v)_S$$
The fact that $V$ and $\Bbb R^3$ are isomorphic as vector spaces means that if we'd like to perform a linear transformation $T$ of a vector $\mathbf v\in V$ we can instead first convert $\mathbf v \to (\mathbf v)_S$, perform the transformation in $\Bbb R^3$, and then convert back $T((\mathbf v)_S) \to T(\mathbf v)$.  Either way we'll get the same answer.
Here's an example: let $\mathbf v= 2\mathbf s_1 + 3\mathbf s_3$ and $\mathbf w=\mathbf s_1 -2\mathbf s_2 + \mathbf s_3$.  Consider the linear transformation $T(\mathbf x) = 2\mathbf x + \mathbf w$.  Then $$\begin{align}T(\mathbf v) &= 2\mathbf v + \mathbf w \\ &= 2(2\mathbf s_1 + 3\mathbf s_3) + (\mathbf s_1 -2\mathbf s_2 + \mathbf s_3) \\ &= (4+1)\mathbf s_1 +(0-2)\mathbf s_2 + (6+1)\mathbf s_3 \\ &= 5\mathbf s_1 -2\mathbf s_2 + 7\mathbf s_3\end{align}$$
vs
$$\begin{align}T(\mathbf v) \rightarrow T([\mathbf v]_S) &= 2(\mathbf v)_S + (\mathbf w)_S \\ &= 2(2,0,3) + (1,-2,1) \\ &= (4+1,0-2,6+1) \\ &= (5,-2,7) \\ &\rightarrow 5\mathbf s_1 -2\mathbf s_2 + 7\mathbf s_3\end{align}$$
Note: Technically $T: V\to V$ and $T:\Bbb R^3\to \Bbb R^3$ are different transformations -- and thus should probably have different names -- but usually we're not that observant.
As for the nomenclature, apparently some people use the term "components of a vector $\mathbf v$ (with respect to $S$)" to mean the scalars $v_1$, $v_2$, and $v_3$ whereas others call the components $v_1\mathbf s_1$, $v_2\mathbf s_2$, and $v_3\mathbf s_3$ -- distinguishing $v_1$, $v_2$, and $v_3$ as the coordinates of $\mathbf v$.
And for the notation, for historical reasons (look up the quaternions for more info) the "standard" basis of $\Bbb R^3$ is often labeled $\mathcal E = \{\mathbf i, \mathbf j, \mathbf k\}$ where $\mathbf i=(1,0,0)$, $\mathbf j = (0,1,0)$, and $\mathbf k= (0,0,1)$ by definition.
Thus $$(\mathbf v)_S = (v_1, v_2, v_3) = v_1\mathbf i + v_2\mathbf j + v_3\mathbf k$$
