Question Number 139055 by EnterUsername last updated on 21/Apr/21
$$\mathrm{Let}\:\mathrm{a}\:\mathrm{and}\:\mathrm{b}\:\mathrm{be}\:\mathrm{complex}\:\mathrm{numbers}\:\mathrm{representing}\:\mathrm{the}\:\mathrm{points} \\ $$$$\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{respectively}\:\mathrm{in}\:\mathrm{the}\:\mathrm{complex}\:\mathrm{plane}. \\ $$$$\mathrm{If}\:\frac{\mathrm{a}}{\mathrm{b}}+\frac{\mathrm{b}}{\mathrm{a}}=\mathrm{1}\:\mathrm{and}\:\mathrm{O}\:\mathrm{is}\:\mathrm{the}\:\mathrm{origin}.\:\mathrm{Then}\:\Delta\mathrm{OAB}\:\mathrm{is}\:? \\ $$
Answered by MJS_new last updated on 22/Apr/21
$$\frac{{a}}{{b}}+\frac{{b}}{{a}}=\mathrm{1}\:\Rightarrow\:{b}={a}\left(\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right) \\ $$$${a}={p}+{q}\mathrm{i}\:\Rightarrow\:{b}=\frac{{p}\pm\sqrt{\mathrm{3}}{q}}{\mathrm{2}}+\frac{{q}\mp\sqrt{\mathrm{3}}{p}}{\mathrm{2}}\mathrm{i} \\ $$$${A}=\begin{pmatrix}{{p}}\\{{q}}\end{pmatrix}\:\wedge{B}=\begin{pmatrix}{\frac{{p}\pm\sqrt{\mathrm{3}}{q}}{\mathrm{2}}}\\{\frac{{q}\mp\sqrt{\mathrm{3}}{p}}{\mathrm{2}}}\end{pmatrix} \\ $$$${AO}=\begin{pmatrix}{−{p}}\\{−{q}}\end{pmatrix}\:\wedge\:{AB}=\begin{pmatrix}{−\frac{{p}\pm\sqrt{\mathrm{3}}{q}}{\mathrm{2}}}\\{−\frac{{q}\mp\sqrt{\mathrm{3}}{p}}{\mathrm{2}}}\end{pmatrix} \\ $$$$\mathrm{cos}\:\alpha\:=\frac{\mid{AO}\bullet{AB}\mid}{\mid{AO}\mid×\mid{AB}\mid}=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:\alpha=\frac{\pi}{\mathrm{3}} \\ $$
Commented by EnterUsername last updated on 22/Apr/21
$${A}=\begin{pmatrix}{{p}}\\{{q}}\end{pmatrix}\:\wedge{B}.\:{I}\:{don}'{t}\:{understand} \\ $$
Commented by mr W last updated on 22/Apr/21
$${it}\:{means} \\ $$$${A}=\begin{pmatrix}{{p}}\\{{q}}\end{pmatrix}\:\:{and}\:{B}=\begin{pmatrix}{\frac{{p}\pm\sqrt{\mathrm{3}}{q}}{\mathrm{2}}}\\{\frac{{q}\mp\sqrt{\mathrm{3}}{p}}{\mathrm{2}}}\end{pmatrix} \\ $$$$\wedge\:{means}\:{and} \\ $$$$\vee\:{means}\:{or} \\ $$
Commented by EnterUsername last updated on 23/Apr/21
$${Ok}\:{thanks}.\:{I}\:{mistakened}\:{it}\:{for}\:“{the}\:{vector}\:{perpendicular}\:{to}…'' \\ $$
Commented by MJS_new last updated on 23/Apr/21
$$\mathrm{we}\:\mathrm{use}\:\overset{\rightharpoonup} {{a}}\bot\overset{\rightharpoonup} {{b}}\:\mathrm{for}\:\mathrm{this} \\ $$
Answered by mr W last updated on 22/Apr/21
$${say}\:{a}={Ae}^{\alpha{i}} \\ $$$${say}\:{b}={Be}^{\beta{i}} \\ $$$${let}\:{k}=\frac{{A}}{{B}},\:\theta=\alpha−\beta \\ $$$$\frac{{A}}{{B}}{e}^{\left(\alpha−\beta\right){i}} +\frac{{B}}{{A}}{e}^{−\left(\alpha−\beta\right){i}} =\mathrm{1} \\ $$$${ke}^{\theta{i}} +\frac{\mathrm{1}}{{k}}{e}^{−\theta{i}} =\mathrm{1} \\ $$$$\left({k}+\frac{\mathrm{1}}{{k}}\right)\mathrm{cos}\:\theta+\left({k}−\frac{\mathrm{1}}{{k}}\right)\mathrm{sin}\:\theta\:{i}=\mathrm{1} \\ $$$$\Rightarrow\left({k}+\frac{\mathrm{1}}{{k}}\right)\mathrm{cos}\:\theta=\mathrm{1} \\ $$$$\Rightarrow\left({k}−\frac{\mathrm{1}}{{k}}\right)\mathrm{sin}\:\theta=\mathrm{0}\:\Rightarrow\begin{cases}{{k}−\frac{\mathrm{1}}{{k}}=\mathrm{0}\:\Rightarrow{k}=\mathrm{1}}\\{\mathrm{sin}\:\theta=\mathrm{0}\:\Rightarrow\theta={n}\pi}\end{cases} \\ $$$$ \\ $$$${with}\:{k}=\mathrm{1}: \\ $$$$\frac{{A}}{{B}}=\mathrm{1} \\ $$$$\mathrm{cos}\:\theta=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\theta=\mathrm{2}{n}\pi\pm\frac{\pi}{\mathrm{3}} \\ $$$$\Rightarrow\Delta{AOB}\:{is}\:{equilateral} \\ $$$$ \\ $$$${with}\:\mathrm{sin}\:\theta=\mathrm{0}: \\ $$$$\mathrm{cos}\:\theta=\pm\mathrm{1} \\ $$$${k}+\frac{\mathrm{1}}{{k}}=\pm\mathrm{1}\:\Rightarrow{no}\:{solution}\:{for}\:{k}>\mathrm{0} \\ $$$$ \\ $$$${therefore}\:{only}\:{possibility}\:{is} \\ $$$$\Delta{AOB}\:{is}\:{equilateral}\:{triangle}. \\ $$
Commented by EnterUsername last updated on 22/Apr/21
$${Thanks}\:{Sir} \\ $$