Dot Product Associates with Scalar Multiplication proof I’m confused and don’t know where to start to prove the following, it’s one of the dot product properties:
$(ca)\cdot b = c(a\cdot b) = a\cdot(cb)$
$a$ and $b$ are vectors. could I say since the product of each is some real number then it has a real number property which is associative? 
I hope I don’t sound dumb.. 
 A: The dot product is commutative and distributive, but not associative! Just use the definition:
\begin{equation}
\vec{c}\cdot\vec{a}=\sum_{i=1}^{n}c_{i}a_{i}
\end{equation}
If you multiply that by $\vec{b}$, that is actually a scalar dotted with $\vec{b}$. 
A: It looks like you want $c$ to be a scalar, so the definition of $ca$ is as the vector whose $i$th entry is $ca_i$, with $a_i$ the $i$th entry of $a$. With implicit summation over $i$ throughout,$$(ca_i)b_i=c(a_ib_i)=a_i(cb_i),$$by the associativity of multiplication on scalars. Or in dot product notation,$$(ca)\cdot b=c(a\cdot b)=a\cdot(cb),$$as required.
A: You can work with two definitions. The first is scalar multiplication:
$$c\vec{a} = c 
    \left(\begin{matrix}
    a_1 \\
    a_2 \\
    \vdots \\
    a_n
    \end{matrix}\right) = \left(\begin{matrix}
    ca_1 \\
    ca_2 \\
    \vdots \\
    ca_n
    \end{matrix}\right)
$$
The second is the inner product:
$$\vec{a} \cdot \vec{b} = \left(\begin{matrix}a_1 a_2 \dots a_n\end{matrix}\right) \left(\begin{matrix}
    b_1 \\
    b_2 \\
    \vdots \\
    b_n
    \end{matrix}\right) = \sum_{j=1}^n a_jb_j$$
To prove the statement, simply write out each in terms of components, and show that they all are the same.
Can you take it home from here?
A: The dot product is NOT associative.  For example, consider
$$\vec{a} \cdot (\vec{b} \cdot \vec{c}) = \vec{a} \cdot (\| b\| \| c \| \cos \theta) $$
The quantity in parenthesis, $\| b\| \| c \| \cos \theta$, is a scalar.  So you have a situation where you're left with
$$\vec{a} \cdot (scalar)$$
which is undefined.
A common mistake people make is they treat $\vec{a} \cdot (scalar)$ as multiplying the vector $\vec{a}$ by a scalar ... which is it not.
