Question Number 120801 by TITA last updated on 02/Nov/20 Commented by liberty last updated on 02/Nov/20 $$\mathrm{let}\:\begin{cases}{\mathrm{i}=\left(\mathrm{1},\mathrm{0},\mathrm{0}\right)}\\{\mathrm{j}=\left(\mathrm{0},\mathrm{1},\mathrm{0}\right)}\\{\mathrm{k}=\left(\mathrm{0},\mathrm{0},\mathrm{1}\right)}\end{cases}\rightarrow\begin{cases}{\mathrm{i}.\mathrm{i}=\mathrm{1}.\mathrm{1}+\mathrm{0}.\mathrm{0}+\mathrm{0}.\mathrm{0}=\mathrm{1}}\\{\mathrm{j}.\mathrm{j}=\mathrm{0}.\mathrm{0}+\mathrm{1}.\mathrm{1}+\mathrm{0}.\mathrm{0}=\mathrm{1}}\\{\mathrm{k}.\mathrm{k}=\mathrm{0}.\mathrm{0}+\mathrm{0}.\mathrm{0}+\mathrm{1}.\mathrm{1}=\mathrm{1}}\end{cases} \\ $$ Commented by TITA last updated on…
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Question Number 54955 by gunawan last updated on 15/Feb/19 $$\mathrm{Let}\:\mathrm{vector}\:\mathrm{set}\:\left\{{u}_{\mathrm{1}} ,\:{u}_{\mathrm{2}} ,\:{u}_{\mathrm{3}} ,\:{u}_{\mathrm{4}} \right\}\:\mathrm{in}\:\mathbb{C}^{{n}} \\ $$$$\mathrm{free}\:\mathrm{linear}.\:\mathrm{So}\:\mathrm{that}\: \\ $$$$\left\{{u}_{\mathrm{1}} +\alpha{u}_{\mathrm{2}} ,\:{u}_{\mathrm{2}} +\alpha{u}_{\mathrm{3}} ,\:{u}_{\mathrm{3}} +\alpha{u}_{\mathrm{4}} ,\:{u}_{\mathrm{4}} +\alpha{u}_{\mathrm{1}}…
Question Number 120258 by bramlexs22 last updated on 30/Oct/20 Answered by TANMAY PANACEA last updated on 30/Oct/20 $$\overset{\rightarrow} {{p}}={pcosa}\:{i}+{psina}\:{j} \\ $$$$\overset{\rightarrow} {{q}}={qcosb}\:{i}−{qsinb}\:{j} \\ $$$$\overset{\rightarrow} {{p}}.\overset{\rightarrow}…
Question Number 185752 by Rupesh123 last updated on 26/Jan/23 Commented by Rupesh123 last updated on 26/Jan/23 Denote [XYZ] area of the Triangle XYZ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 54578 by pieroo last updated on 07/Feb/19 $$\mathrm{A}\:\mathrm{particle}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{1}.\mathrm{5kg}\:\mathrm{rests}\:\mathrm{on}\:\mathrm{a}\:\mathrm{rough}\: \\ $$$$\mathrm{plane}\:\mathrm{inclined}\:\mathrm{at}\:\mathrm{45}°\:\mathrm{to}\:\mathrm{the}\:\mathrm{horizontal}. \\ $$$$\mathrm{It}\:\mathrm{is}\:\mathrm{maintained}\:\mathrm{in}\:\mathrm{equilibrium}\:\mathrm{by}\:\mathrm{a}\: \\ $$$$\mathrm{horizontal}\:\mathrm{force}\:\mathrm{of}\:{p}\:\mathrm{newtons}.\:\mathrm{Given} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{friction}\:\mathrm{between} \\ $$$$\mathrm{the}\:\mathrm{particle}\:\mathrm{and}\:\mathrm{the}\:\mathrm{plane}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{4}},\:\mathrm{calculate} \\ $$$$\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{p}\:\mathrm{when}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{is}\:\mathrm{on} \\ $$$$\mathrm{the}\:\mathrm{point}\:\mathrm{of}\:\mathrm{moving} \\…
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Question Number 54061 by behi83417@gmail.com last updated on 28/Jan/19 Commented by mr W last updated on 29/Jan/19 $${C}\left({t},\frac{\mathrm{1}}{{t}}\right) \\ $$$${D}\left(\frac{\mathrm{1}}{{t}},{t}\right) \\ $$$${CD}=\sqrt{\mathrm{2}}\left({t}−\frac{\mathrm{1}}{{t}}\right)=\mathrm{2} \\ $$$${t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}−\mathrm{1}=\mathrm{0}…
Question Number 53468 by ajfour last updated on 22/Jan/19 Commented by ajfour last updated on 22/Jan/19 $${Find}\:\theta\:{in}\:{terms}\:{of}\:\boldsymbol{{h}},\boldsymbol{{r}},\:{and}\:\boldsymbol{{R}}. \\ $$$$\left({ring}\:{rests}\:{inside}\:{cylinder}\right).\:\:\: \\ $$ Commented by mr W…
Question Number 53450 by ajfour last updated on 22/Jan/19 Commented by ajfour last updated on 22/Jan/19 $${Find}\:{radius}\:{of}\:{inscribed}\:{sphere}\:{in} \\ $$$${box}. \\ $$ Answered by ajfour last…