Question Number 127566 by Ar Brandon last updated on 30/Dec/20 Answered by bramlexs22 last updated on 30/Dec/20 $$\left(\overset{\rightarrow} {\mathrm{a}}+\overset{\rightarrow} {\mathrm{b}}\right)×\left(\overset{\rightarrow} {\mathrm{a}}−\overset{\rightarrow} {\mathrm{b}}\right)=\overset{\rightarrow} {\mathrm{a}}×\overset{\rightarrow} {\mathrm{a}}−\overset{\rightarrow} {\mathrm{a}}×\overset{\rightarrow}…
Question Number 127568 by Ar Brandon last updated on 30/Dec/20 Commented by bramlexs22 last updated on 30/Dec/20 $$\mathrm{R}_{\mathrm{max}} \:=\:\sqrt{\mid\overset{\rightarrow} {\mathrm{a}}\mid^{\mathrm{2}} +\mid\overset{\rightarrow} {\mathrm{b}}\mid^{\mathrm{2}} +\mathrm{2}\mid\overset{\rightarrow} {\mathrm{a}}\mid\mid\overset{\rightarrow} {\mathrm{b}}\mid}\:…
Question Number 61823 by necx1 last updated on 09/Jun/19 $${If}\:{D},{E}\:{and}\:{F}\:{are}\:{midpoints}\:{of}\:{the}\:{sides} \\ $$$${BC},{CA}\:{and}\:{AB}\:{respectively}\:{of}\:{the}\:\bigtriangleup{ABC} \\ $$$${and}\:{O}\:{be}\:{any}\:{point}.{Prove}\:{that} \\ $$$${O}\overset{\rightarrow} {{A}}\:+\:{O}\overset{\rightarrow} {{B}}\:+{O}\overset{\rightarrow} {{C}}={O}\overset{\rightarrow} {{D}}+{O}\overset{\rightarrow} {{E}}+{O}\overset{\rightarrow} {{F}} \\ $$ Answered…
Question Number 126786 by Ar Brandon last updated on 24/Dec/20 Answered by Olaf last updated on 26/Dec/20 $$\mathrm{C}_{\mathrm{1}} \:=\:\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:\left(\mathrm{first}\:\mathrm{city}\right) \\ $$$$\mathrm{C}_{\mathrm{2}} \:=\:\begin{pmatrix}{\mathrm{60}\sqrt{\mathrm{2}}−\mathrm{20}}\\{\mathrm{75}+\mathrm{60}\sqrt{\mathrm{2}}}\end{pmatrix}\:\left(\mathrm{second}\:\mathrm{city}\right) \\ $$$$\left(\mathrm{C}_{\mathrm{1}} \mathrm{C}_{\mathrm{2}}…
Question Number 126778 by Ar Brandon last updated on 24/Dec/20 Answered by Dwaipayan Shikari last updated on 24/Dec/20 $$\sqrt{\left(\overset{\rightarrow} {{A}}_{\mathrm{1}} \right)^{\mathrm{2}} +\left(\overset{\rightarrow} {{A}}_{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{2}\mid\overset{\rightarrow}…
Question Number 126716 by sdfg last updated on 23/Dec/20 Answered by liberty last updated on 23/Dec/20 $$\overset{\rightarrow} {{a}}=\left(\mathrm{3},\mathrm{4}\right)\:;\:\overset{\rightarrow} {{b}}=\left({x},{y}\right)\:\rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{5}…\left({i}\right) \\ $$$$\:\overset{\rightarrow} {{a}}\:\bot\:\overset{\rightarrow} {{b}}\:\rightarrow\:\overset{\rightarrow}…
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Question Number 126567 by sdfg last updated on 21/Dec/20 Answered by physicstutes last updated on 22/Dec/20 $$\:\boldsymbol{\mathrm{A}}\:=\:\left({k},{k}\right)\:\mathrm{and}\:\boldsymbol{\mathrm{B}}\:=\:\left(−\mathrm{3},\mathrm{4}\right) \\ $$$$\boldsymbol{\mathrm{A}}.\boldsymbol{\mathrm{B}}\:=\:\mid\boldsymbol{\mathrm{A}}\mid\mid\boldsymbol{\mathrm{B}}\mid\:\mathrm{cos}\:\theta \\ $$$$\Rightarrow\:−\mathrm{3}{k}+\mathrm{4}{k}\:=\:{k} \\ $$$$\Rightarrow\:{k}\:=\:\sqrt{{k}^{\mathrm{2}} +{k}^{\mathrm{2}} }\:.\sqrt{\left(−\mathrm{3}\right)^{\mathrm{2}}…
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