Vector function of vectors expansion I am reading Landau's Mechanics. In the solution to the problem 4 on page 138, section 42, it is stated that an arbitrary vector function $\vec f(\vec r,\vec p)$ may be written as $\vec f=\vec r\phi_1+\vec p\phi_2+\vec r\times\vec p\phi_3$, where $\phi_1$, $\phi_2$, $\phi_3$ are scalar functions. Why so?
 A: $\newcommand{\bl}[1]{\boldsymbol{#1}} 
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I think this Figure answers your question. It's up to you to find expressions for $\phi_1,\phi_2,\phi_3$ for given $\mathbf r,\mathbf p,\mathbf f$ with $\mathbf r\x\mathbf p \bl\ne \bl0$.
If
\begin{equation}
\mathbf f \e \phi_1\mathbf r\p\phi_2\mathbf p\p \phi_3\plr{\mathbf r\x\mathbf p}
\tl{01}
\end{equation}
try to prove that
\begin{align}
\phi_1 & \e \dfrac{\plr{\mathbf f\bl\cdot\mathbf r}\Vlr{\mathbf p}^2\m\plr{\mathbf f\bl\cdot\mathbf p}\plr{\mathbf r\bl\cdot\mathbf p}}{\Vlr{\mathbf r\x\mathbf p}^2}
\tl{02.1}\\
\phi_2 & \e \dfrac{\plr{\mathbf f\bl\cdot\mathbf p}\Vlr{\mathbf r}^2\m\plr{\mathbf f\bl\cdot\mathbf r}\plr{\mathbf r\bl\cdot\mathbf p}}{\Vlr{\mathbf r\x\mathbf p}^2}
\tl{02.2}\\
\phi_3 & \e \dfrac{\mathbf f\bl\cdot\plr{\mathbf r\x\mathbf p}}{\Vlr{\mathbf r\x\mathbf p}^2}
\tl{02.3}
\end{align}
Note that
\begin{equation}
\Vlr{\mathbf r\x\mathbf p}^2\e \Vlr{\mathbf r}^2\Vlr{\mathbf p}^2\m\plr{\mathbf r\bl\cdot\mathbf p}^2\e \Vlr{\mathbf r}^2\Vlr{\mathbf p}^2\sin^2\theta
\tl{03}
\end{equation}
A: take  arbitrary vector in 3D space
$$\vec R=c_1\hat g_1+c_2\hat g_2+c_3\hat g_3\tag 1$$
where $~\hat g_i~$ are the independent  basis vectors, and  $~c_i~$ are the scalar components of the vector R.
now your case
$$\vec f=\phi_1\,r\,\hat r+\phi_2\,p\,\hat p+\phi_3\,|r\times p|\,\hat{r}_p\\
r_p=\vec r\times \vec p$$
comparing with equation (1)
$$c_1=\phi_1\,r\,,\hat g_1=\hat r\\
c_2=\phi_2\,p\,,\hat g_2=\hat p\\
c_3=\phi_3\,|r\times p|\,,\hat g_3=\hat r_p$$
thus there is noting unusual in this vector description
remarks:
the basis vectors must not be perpendicular to each other  $~\vec r\cdot\vec p\ne 0$
