Question Number 24622 by ajfour last updated on 22/Nov/17
Commented by ajfour last updated on 22/Nov/17
$${If}\:\bar {{b}}\:{is}\:{the}\:{result}\:{of}\:{rotating}\:\bar {{a}} \\ $$$${about}\:\bar {{r}}\:{by}\:{an}\:{angle}\:\theta\:{in}\:{the} \\ $$$${manner}\:{shown},\:{then},\:{express} \\ $$$$\:\bar {\boldsymbol{{b}}}\:{in}\:{terms}\:{of}\:\bar {\boldsymbol{{a}}},\:\bar {\boldsymbol{{r}}}\:,\:{and}\:\theta. \\ $$
Commented by jota last updated on 23/Nov/17
$$\:{PA}=\boldsymbol{{a}}\:\:{PB}=\boldsymbol{{b}} \\ $$
Commented by jota last updated on 23/Nov/17
![O centro del circulo Asumiendo [r]=1 Definiendo i=OA/[OA] ⇒j=((r×a)/( [r×a] )) i=j×r Luego b=a+AB =a+x(θ)i+y(θ)j](https://www.tinkutara.com/question/Q24649.png)
$$\:{O}\:{centro}\:{del}\:{circulo} \\ $$$${Asumiendo}\:\left[\mathrm{r}\right]=\mathrm{1} \\ $$$${Definiendo} \\ $$$$\mathrm{i}={OA}/\left[{OA}\right] \\ $$$$\Rightarrow{j}=\frac{{r}×{a}}{\:\left[{r}×{a}\right]\:\:} \\ $$$$\:\:\:\:\:{i}={j}×{r} \\ $$$${Luego}\:\boldsymbol{{b}}=\boldsymbol{{a}}+{AB} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\boldsymbol{{a}}+{x}\left(\theta\right)\boldsymbol{\mathrm{i}}+{y}\left(\theta\right)\boldsymbol{{j}} \\ $$
Commented by jota+ last updated on 23/Nov/17
$${OA}=\boldsymbol{{a}}−\left(\boldsymbol{{a}}\bullet\boldsymbol{{r}}\right)\boldsymbol{{r}} \\ $$