Vector function of vectors expansion

Question

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?

Answer 1

$\newcommand{\bl}[1]{\boldsymbol{#1}} \newcommand{\e}{\bl=} \newcommand{\p}{\bl+} \newcommand{\m}{\bl-} \newcommand{\gr}{\bl>} \newcommand{\les}{\bl<} \newcommand{\greq}{\bl\ge} \newcommand{\leseq}{\bl\le} \newcommand{\plr}[1]{\left(#1\right)} \newcommand{\blr}[1]{\left[#1\right]} \newcommand{\vlr}[1]{\vert#1\vert} \newcommand{\Vlr}[1]{\Vert#1\Vert} \newcommand{\lara}[1]{\langle#1\rangle} \newcommand{\lav}[1]{\langle#1|} \newcommand{\vra}[1]{|#1\rangle} \newcommand{\lavra}[2]{\langle#1|#2\rangle} \newcommand{\lavvra}[3]{\langle#1|\,#2\,|#3\rangle} \newcommand{\vp}{\vphantom{\dfrac{a}{b}}} \newcommand{\hp}[1]{\hphantom{#1}} \newcommand{\x}{\bl\times} \newcommand{\qqlraqq}{\qquad\bl{-\!\!\!-\!\!\!-\!\!\!\longrightarrow}\qquad} \newcommand{\qqLraqq}{\qquad\boldsymbol{\e\!\e\!\e\!\e\!\Longrightarrow}\qquad} \newcommand{\tl}[1]{\tag{#1}\label{#1}} \newcommand{\hebl}{\bl{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}}$

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}

Answer 2

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 $~\vec r\cdot\vec p\ne 0$