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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…
Question Number 53034 by gopikrishnan005@gmail.com last updated on 16/Jan/19 $$\mathrm{Angle}\:\mathrm{between}\:\mathrm{the}\:\mathrm{lines}\:\frac{\mathrm{x}−\mathrm{1}}{\mathrm{1}}=\frac{\mathrm{y}−\mathrm{1}}{\mathrm{1}}=\frac{\mathrm{z}−\mathrm{1}}{\mathrm{2}}\mathrm{and}\:\frac{\mathrm{x}−\mathrm{1}}{−\sqrt{\mathrm{3}−\mathrm{1}}}=\frac{\mathrm{y}−\mathrm{1}}{\:\sqrt{\mathrm{3}−\mathrm{1}}}=\frac{\mathrm{z}−\mathrm{1}}{\mathrm{4}}\:\mathrm{is} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 16/Jan/19 $${cos}\theta=\frac{{a}_{\mathrm{1}} {a}_{\mathrm{2}} +{b}_{\mathrm{1}} {b}_{\mathrm{2}} +{c}_{\mathrm{1}} {c}_{\mathrm{2}}…
Question Number 53025 by ajfour last updated on 16/Jan/19 Commented by ajfour last updated on 16/Jan/19 $${Find}\:{minimum}\:{distance}\:{of}\:{line} \\ $$$${AB}\:{from}\:{cylinder}\:{axis}. \\ $$$$\left({correction}:\:{the}\:{shown}\:{segment}\right. \\ $$$${with}\:{which}\:{AB}\:{makes}\:\angle\:\alpha\:{should} \\ $$$$\left.\:{end}\:{in}\:{foot}\:{of}\:{vertical}\:{from}\:{A}.\right)…
Question Number 118411 by bramlexs22 last updated on 17/Oct/20 $${Write}\:{the}\:{vector}\:{v}=\left(\mathrm{1},−\mathrm{2},\mathrm{3}\right)\:{as}\:{a} \\ $$$${linear}\:{combination}\:{of}\:{vectors} \\ $$$${u}_{\mathrm{1}} =\left(\mathrm{1},\mathrm{1},\mathrm{1}\right)\:,{u}_{\mathrm{2}} =\left(\mathrm{1},\mathrm{2},\mathrm{3}\right)\:{and}\:{u}_{\mathrm{3}} =\left(\mathrm{2},−\mathrm{1},\mathrm{1}\right) \\ $$ Answered by benjo_mathlover last updated on…
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Question Number 118083 by pramod1977 last updated on 15/Oct/20 Answered by bemath last updated on 15/Oct/20 $$\mathrm{let}\:\overset{\rightarrow} {\mathrm{a}}=\left(\mathrm{2},\mathrm{1},−\mathrm{2}\right)=\mathrm{diagonal}\:\mathrm{and}\:\overset{\rightarrow} {\mathrm{b}}=\left(\mathrm{3},\mathrm{1},−\mathrm{1}\right)=\mathrm{side} \\ $$$$\mathrm{The}\:\mathrm{area}\:\mathrm{of}\:\mathrm{parallelogram}\:=\:\mid\overset{\rightarrow} {\mathrm{a}}\:×\overset{\rightarrow} {\mathrm{b}}\:\mid \\ $$$$\mathrm{where}\:\overset{\rightarrow}…
Question Number 52520 by byaw last updated on 10/Jan/19 $$\mathrm{Please}\:\mathrm{he}{lp} \\ $$$$\:\mathrm{John}\:\mathrm{travels}\:\mathrm{a}\:\mathrm{distance}\:\mathrm{of}\:\mathrm{24km} \\ $$$$\mathrm{from}\:\mathrm{twon}\:\mathrm{A}\:\mathrm{on}\:\mathrm{a}\:\mathrm{bearing}\:\mathrm{of}\:\mathrm{060}°\:\mathrm{to} \\ $$$$\mathrm{town}\:\mathrm{B}.\:\mathrm{He}\:\mathrm{then}\:\mathrm{travels}\:\mathrm{a}\:\mathrm{distance} \\ $$$$\mathrm{of}\:\mathrm{18km}\:\mathrm{to}\:\mathrm{town}\:\mathrm{C},\:\mathrm{which}\:\mathrm{is}\:\mathrm{30km} \\ $$$$\mathrm{east}\:\mathrm{of}\:\mathrm{town}\:\mathrm{A}. \\ $$$$\left(\mathrm{i}\right)\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{bearing}\:\mathrm{of}\:\mathrm{town}\:\mathrm{C}\:\mathrm{from} \\ $$$$\:\:\:\:\:\:\:\mathrm{town}\:\mathrm{B}? \\…