Question Number 193809 by byaw last updated on 20/Jun/23 A(-1, 2), B(3, 5) and C(4, 8) are the vertices of triangle ABC. Forces whose magnitudes…
Question Number 130889 by mnjuly1970 last updated on 30/Jan/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\:\:\:{advanced}\:\:{mathematcs}\:\:… \\ $$$$\:{prove}\:{that}:: \\ $$$$\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{1}+{n}^{\mathrm{2}} }\:=\frac{{csch}\left(\pi\right)−\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$ Answered by mindispower…
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Question Number 130537 by benjo_mathlover last updated on 26/Jan/21 Answered by EDWIN88 last updated on 26/Jan/21 $$\Leftrightarrow\:\overset{\rightarrow} {{a}}×\left(\overset{\rightarrow} {{b}}×\overset{\rightarrow} {{c}}\right)=\left(\overset{\rightarrow} {{a}}.\overset{\rightarrow} {{c}}\right)\overset{\rightarrow} {{b}}−\left(\overset{\rightarrow} {{a}}.\overset{\rightarrow} {{b}}\right)\overset{\rightarrow}…
Question Number 130523 by benjo_mathlover last updated on 26/Jan/21 Answered by TheSupreme last updated on 26/Jan/21 $${P}\left(\mathrm{4},\mathrm{6},\mathrm{2}\right) \\ $$$$\begin{cases}{\mathrm{2}{x}−\mathrm{3}{y}=\mathrm{2}}\\{\mathrm{7}{x}−\mathrm{3}{z}=\mathrm{10}}\\{{x}+{y}−{z}=\mathrm{8}}\end{cases} \\ $$$$\begin{bmatrix}{\mathrm{2}}&{−\mathrm{3}}&{\mathrm{0}}\\{\mathrm{7}}&{\mathrm{0}}&{−\mathrm{3}}\\{\mathrm{1}}&{\mathrm{1}}&{−\mathrm{1}}\end{bmatrix}\begin{pmatrix}{{x}}\\{{y}}\\{{z}}\end{pmatrix}=\begin{pmatrix}{\mathrm{2}}\\{\mathrm{10}}\\{\mathrm{8}}\end{pmatrix} \\ $$$${n}_{\mathrm{1}} −{n}_{\mathrm{2}} +\mathrm{3}{n}_{\mathrm{3}}…
Question Number 64348 by Rio Michael last updated on 16/Jul/19 $${the}\:{vectors}\:\boldsymbol{{a}}\:{and}\:\boldsymbol{{b}}\:{are}\:{such}\:{that}\:\mid\boldsymbol{{a}}\mid\:=\mathrm{3}\:,\:\mid\boldsymbol{{b}}\mid=\mathrm{5}\:{and}\:\boldsymbol{{a}}.\boldsymbol{{b}}=−\mathrm{14} \\ $$$${find}\:\mid\boldsymbol{{a}}−\boldsymbol{{b}}\mid \\ $$ Answered by Tanmay chaudhury last updated on 17/Jul/19 $$\mid\boldsymbol{{a}}−\boldsymbol{{b}}\mid^{\mathrm{2}} =\left(\boldsymbol{{a}}−\boldsymbol{{b}}\right).\left(\boldsymbol{{a}}−\boldsymbol{{b}}\right)…
Question Number 129804 by bramlexs22 last updated on 19/Jan/21 $$\mathrm{If}\:\overset{\rightarrow} {{p}}=\mathrm{2}\hat {{i}}+\mathrm{5}\hat {{j}}+\mathrm{6}\hat {{k}} \\ $$$$\:\:\:\overset{\rightarrow} {{q}}=\mathrm{3}\hat {{i}}+\mathrm{6}\hat {{j}}+\mathrm{8}\hat {{k}} \\ $$$$\:\:\:\overset{\rightarrow} {{r}}=\mathrm{2}\hat {{i}}+\mathrm{6}\hat {{j}}+\mathrm{10}\hat…
Question Number 64085 by Rio Michael last updated on 12/Jul/19 $${please}\:{just}\:{read}\:{equation}\:{of}\:{a}\:{line}\:{and}\:{a}\:{plane}\:{in}\:{vectors}. \\ $$$${i}\:{don}'{t}\:{understand}\: \\ $$$$\:\:\left(\boldsymbol{{r}}−\boldsymbol{{a}}\right)×\boldsymbol{{b}}=\mathrm{0}\:\:?? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 63427 by Rio Michael last updated on 04/Jul/19 $$\left(\mathrm{1}\right)\:{A}\:{plane}\:{contains}\:{the}\:{lines}\:\frac{{x}+\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{4}−{y}}{\mathrm{2}}=\frac{{z}−\mathrm{2}}{\mathrm{3}}\:{and}\: \\ $$$${r}=\:\left(\mathrm{2}{i}+\mathrm{2}{j}\:+\:\mathrm{12}{k}\right)+{t}\left(−{i}+\mathrm{2}{j}\:+\mathrm{4}{k}\right).\:{find} \\ $$$$\left({a}\right)\:{the}\:{angle}\:{between}\:{these}\:{lines}. \\ $$$$\left({b}\right)\:{A}\:{cartesian}\:{equation}\:{of}\:{the}\:{plane}. \\ $$$$\left(\mathrm{2}\right)\:{Given}\:{the}\:{lines}\:\boldsymbol{{l}}_{\mathrm{1}} :\frac{{x}−\mathrm{10}}{\mathrm{3}}=\frac{{y}−\mathrm{1}}{\mathrm{1}}=\frac{{z}−\mathrm{9}}{\mathrm{4}}\:\:\boldsymbol{{l}}_{\mathrm{2}} :{r}=\left(−\mathrm{9}{j}+\mathrm{13}{k}\right)+\mu\left({i}+\mathrm{2}{j}−\mathrm{3}{k}\right) \\ $$$${where}\:\mu\:{is}\:{a}\:{parameter};\:\boldsymbol{{l}}_{\mathrm{3}} :\frac{{x}+\mathrm{10}}{\mathrm{4}}=\frac{{y}+\mathrm{5}}{\mathrm{3}}=\frac{{z}+\mathrm{4}}{\mathrm{1}}. \\…
Question Number 128830 by bemath last updated on 10/Jan/21 Answered by liberty last updated on 10/Jan/21 $$\:\mathrm{let}\:\mid\overset{\rightarrow} {{a}}\mid\:=\:\mid\overset{\rightarrow} {{b}}\mid\:=\:\mathrm{1}\:;\:\mathrm{also}\:\mid\:\overset{\rightarrow} {{a}}+\overset{\rightarrow} {{b}}\mid\:=\:\mathrm{1}\: \\ $$$$\:\mathrm{we}\:\mathrm{want}\:\mathrm{to}\:\mathrm{compute}\:\mid\overset{\rightarrow} {{a}}−\overset{\rightarrow} {{b}}\:\mid\:.…