Interpretation of a proof on inner product space. Let $\mathcal{B} = \{v_1,…,v_n\}$ be a basis for the finite dimensional vector space $V$ ($\dim V = n < +\infty$) such that $\langle v_i,v_j\rangle =0$ if $i \neq j$. If $\mathcal{B}$ is an orthogonal basis and $x \in V$ then
\begin{align*}
x = \sum_{i=1}^n \frac{\langle x,v_i\rangle}{\|v_i\|^2} v_i
\end{align*}
Proof
Since $\displaystyle x = \sum_{i=1}^n c_i v_i$ for some $c_1,\ldots,c_n \in\textbf{F}$, one has that
$$\langle x,v_j\rangle = \Bigg\langle\sum_{i=1}^{n} c_i v_i, v_j\Bigg\rangle =\sum_{i=1}^{n} c_i \langle v_i, v_j\rangle = c_j \|v_j\|^2$$
Thus we have that $\langle x,v_j\rangle = c_j\|v_j\|^2$, so $\displaystyle c_j = \frac{\langle x,v_j\rangle}{\|v_j\|^2}$
My question is: what is the interpretation of $\langle x,v_j\rangle$ on an inner product space?
 A: You can think of $\langle x,y\rangle/\|y\|$ as the signed length of the projection of the vector $x$ on the direction of $y/\|y\|$ (provided that $y\neq 0$): a plus sign corresponds to the same direction and a minus sign corresponds to opposite directions.
Such interpretation comes from $\textbf{R}^{3}$, where the standard inner product satisfies $\langle x,y\rangle = \|x\|\|y\|\cos(\theta)$. Indeed, given $x = (x_{1},x_{2},x_{3})$, $y = (y_{1},y_{2},y_{3})$ and $z = x - y = (x_{1} - y_{1}, x_{2} - y_{2}, x_{3} - y_{3})$, the application of the cosine's law $\|z\|^{2} = \|x\|^{2} + \|y\|^{2} - 2\|x\|\|y\|\cos(\theta)$ results into
\begin{align*}
& (x_{1}-y_{1})^{2} + (x_{2} - y_{2})^{2} + (x_{3}-y_{3})^{2} = x^{2}_{1} + x^{2}_{2} + x^{2}_{3} + y^{2}_{1} + y^{2}_{2} + y^{2}_{3} - 2\|x\|\|y\|\cos(\theta) \Longleftrightarrow\\\\
& -2(x_{1}y_{1} + x_{2}y_{2} + x_{3}y_{3}) = -2\|x\|\|y\|\cos(\theta) \Longleftrightarrow x_{1}y_{1} + x_{2}y_{2} + x_{3}y_{3} = \|x\|\|y\|\cos(\theta)
\end{align*}
where $\langle x,y\rangle = x_{1}y_{1} + x_{2}y_{2} + x_{3}y_{3}$ stands for the standard inner product on $\textbf{R}^{3}$. In fact, this is how we define angles between vectors within euclidean spaces. More precisely, given $x\neq 0$ and $y\neq 0$, one has
\begin{align*}
\cos(\theta) = \frac{\langle x,y\rangle}{\|x\|\|y\|}
\end{align*}
which is between $-1$ and $1$ due to the Cauchy-Schwarz inequality $|\langle x,y\rangle|\leq\|x\|\|y\|$. In order to prove it, you can approach it by considering the expression $\|x + \lambda y\|^{2}\geq 0$, which is quadratic in the variable $\lambda$.
A: In an inner-product space over $\Bbb R,$ if $x\ne 0\ne v$ and if $x$ is not a scalar multiple of $v$ then $<x,v>=\|x\|\cdot \|v\|\cdot \cos A$ where $A$ is the angle at $0$ in the $\triangle$ with vertices $0,x,v.$ The inner-product vector-subspace generated by a linearly independent pair $\{x,v\}$ is isomorphic to $\Bbb R^2.$ If also $\|v\|=1$ then $|<x,v>|=\|y\|$ where $y$ is the foot of the perpendicular from $x$ to the line through $0,v.$
BTW. I found that helpful to remember the Linear Correlation Co-efficient formula from Statistics: Given $x=<x_1,...,x_n>$ and $y=<y_1,...,y_n>,$ first we "normalize" $x$ and $y $ by taking $x'=<x_1-m_x,...,x_n-m_x>$ and $y'=<y_1-m_y,...,y_n-m_y>$ where $m_x=n^{-1}\sum_{j=1}^nx_j$ and $m_y=n^{-1}\sum_{j=1}^ny_j$ are the means (averages). The correlation co-efficient is $\cos \angle x'0y',$ which, if $x'\ne 0\ne y',$  is $\frac {<x',y'>}{\|x'\|\cdot \|y'\|}.$
