Solving a system of Vector Equations involving dot and cross products

Question

I have a following question on vector and would like to verify if my solution is correct or not .

Suppose $\vec{a}$ be non zero and $\vec{b}$ and a scalar $c$ be given, if point P with position vector $\vec{r}$ satisfies the equation

$\vec{a}.\vec{r} = c \ldots (1)$

and $\vec{a} \times \vec{r} = \vec{b} \ldots(2)$, find $\vec{r}$

Now, let $\displaystyle\vec{r_{p}}$ denote the vector component of $\vec{r}$ along $\vec{a}$ and

$\vec{r_{n}}$ denote vector component of $\vec{r}$ perpendicular to $\vec{a}$

So, $\vec{r_{p}}$ = $\dfrac{c\vec{a}}{\vec{a}.\vec{a}}$

$\vec{r_{n}}$ = $\vec{r}$ - $\vec{r_{p}}$,

Now $\vec{r_n}$ is perpendicular to $\vec{a}$ so by equation (2)

$\vec{r_{n}}$ is parallel to $\vec{b}$ , or

$\vec{r_{n}}$ = $\lambda \vec{b}$ , using this I get

$\vec{r}$ = $\dfrac{c\vec{a}}{\vec{a}.\vec{a}} + \lambda \vec{b}$,

Now using equation (2) gives me $\lambda = \frac{1}{\left|a\right|}$, hence

$\vec{r} = \dfrac{c\vec{a}}{\vec{a}.\vec{a}} + \dfrac{\vec{b}}{\left|a\right|}$

Is my solution correct? Can someone please check??

Thank you!!

Answer 1

Let$$r_1=\frac{(r.a)a}{\|a\|^2}\quad\text{and}\quad r_2=r-r_1.$$Then $r_1\perp r_2$ and $r=r_1+r_2$. Since you alredy know $r_1$, it remains to find $r_2$. Just take$$r_2=\frac{b\times a}{\|a\|^2}.$$This will work, because then\begin{align}a\times r&=a\times r_2\text{ (since $a$ and $r_1$ are parallel)}\\&=a\times\frac{b\times a}{\|a\|^2}\\&=\frac1{\|a\|^2}a\times(b\times a)\\&=\frac1{\|a\|^2}\bigl((a.a)b-(a.b)a\bigr)\\&=b.\end{align}

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ In an above comment, user ${\tt @sato91}$ said that $\ds{\bbx{\vec{a} \perp \vec{b}}}$. Then, $\ds{\vec{r} \equiv \alpha\vec{a} + \beta\vec{b} + \gamma\vec{a}\times\vec{b}}$ and \begin{align} &\left\{\begin{array}{lclcl} \ds{c} & \ds{=} & \ds{\vec{a}\cdot\vec{r}} & \ds{=} & \ds{\alpha\,a^{2}} \\ \ds{\vec{b}} & \ds{=} & \ds{\vec{a}\times\vec{r}} & \ds{=} & \ds{\beta\,\vec{a}\times\vec{b} + \gamma\vec{a}\times\pars{\vec{a}\times\vec{b}} = \beta\,\vec{a}\times\vec{b} - \gamma\,a^{2}\,\vec{b}} \end{array}\right. \end{align}

• The first one $\ds{\implies \bbx{\alpha = {c \over a^{2}}}}$.
• The second one $$ \left\{\begin{array}{l} \ds{\vec{b}\cdot\vec{a}\times\vec{b} = \beta\pars{\vec{a}\times\vec{b}}^{2} \implies \bbx{\beta = 0}} \\ \ds{\vec{b}\cdot\vec{b} = b^{2} = -\gamma a^{2}b^{2} \implies \bbx{\gamma = -\,{1 \over a^{2}}}} \\ \mbox{} \end{array}\right. $$ $$ \mbox{Then,}\quad \bbx{\vec{r} = {c \over a^{2}}\vec{a} - {1 \over a^{2}}\vec{a}\times\vec{b}} \\ $$