Question Number 45259 by rahul 19 last updated on 11/Oct/18 $${Find}\:\:{distance}\:{of}\:{point}\:\left(\mathrm{1},\mathrm{1},\mathrm{1}\right)\:{from} \\ $$$${the}\:{line}\:{passing}\:{through}\:\left(\mathrm{2},\mathrm{3},\mathrm{4}\right)\:\& \\ $$$$\left(−\mathrm{1},\mathrm{2},\mathrm{3}\right)\:? \\ $$ Commented by rahul 19 last updated on 11/Oct/18…
Question Number 44848 by pieroo last updated on 05/Oct/18 $$\left(\mathrm{1}\right)\:\mathrm{If}\:\overset{\rightarrow} {\boldsymbol{\mathrm{a}}}=\mathrm{a}_{\mathrm{1i}} +\mathrm{a}_{\mathrm{2j}} +\mathrm{a}_{\mathrm{3k}} \:\mathrm{and}\:\overset{\rightarrow} {\boldsymbol{\mathrm{b}}}=\mathrm{b}_{\mathrm{1i}} +\mathrm{b}_{\mathrm{2j}} +\mathrm{b}_{\mathrm{3k}} \:\mathrm{show}\:\mathrm{that} \\ $$$$\mathrm{i}.\:\overset{\rightarrow} {\boldsymbol{\mathrm{a}}}×\overset{\rightarrow} {\boldsymbol{\mathrm{b}}}=\begin{vmatrix}{\mathrm{i}}&{\mathrm{j}}&{\mathrm{k}}\\{\mathrm{a}_{\mathrm{1}} }&{\mathrm{a}_{\mathrm{2}} }&{\mathrm{a}_{\mathrm{3}} }\\{\mathrm{b}_{\mathrm{1}}…
Question Number 44035 by rahul 19 last updated on 20/Sep/18 Commented by tanmay.chaudhury50@gmail.com last updated on 20/Sep/18 $$\left(\overset{\rightarrow} {{x}}.\overset{\rightarrow} {{z}}\right)\overset{\rightarrow} {{y}}−\left(\overset{\rightarrow} {{x}}.\overset{\rightarrow} {{y}}\right)\overset{\rightarrow} {{z}}=\overset{\rightarrow} {{a}}…
Question Number 44020 by rahul 19 last updated on 20/Sep/18 Answered by tanmay.chaudhury50@gmail.com last updated on 20/Sep/18 $$\overset{\rightarrow} {{a}}.\overset{\rightarrow} {{d}}=\mid\overset{\rightarrow} {{a}}\mid{cos}\theta_{\mathrm{1}} \:\:\:\:\overset{\rightarrow} {{b}}.\overset{\rightarrow} {{d}}=\mid\overset{\rightarrow} {{b}}\mid{cos}\theta_{\mathrm{2}}…
Question Number 43856 by MASANJA J last updated on 16/Sep/18 $${given}\:{that} \\ $$$${a}×{b}=\mathrm{3}{i}\:+\:{j}\:+{k} \\ $$$${a}×{c}=−{i}\:−\mathrm{2}{j}\:+{k} \\ $$$${find} \\ $$$$\left.{i}\right){c}×{a} \\ $$$$\left.{ii}\right){a}×\left({b}×{c}\right) \\ $$$$\left.{iii}\right)\left({a}×{b}\right)\bullet\left({a}×{c}\right) \\ $$…
Question Number 174489 by pete last updated on 02/Aug/22 $$\mathrm{The}\:\mathrm{points}\:\mathrm{A},\:\mathrm{B}\:\mathrm{and}\:\mathrm{C}\:\mathrm{have}\:\mathrm{position}\:\mathrm{vectors} \\ $$$$\boldsymbol{\mathrm{a}},\:\boldsymbol{\mathrm{b}}\:\mathrm{and}\:\boldsymbol{\mathrm{c}}\:\mathrm{respectively}\:\mathrm{reffrred}\:\mathrm{to}\:\mathrm{an}\:\mathrm{origin}\:\mathrm{O}. \\ $$$$\mathrm{i}.\:\mathrm{Given}\:\mathrm{that}\:\mathrm{the}\:\mathrm{point}\:\mathrm{X}\:\mathrm{lie}\:\mathrm{on}\:\mathrm{AB}\:\mathrm{produced} \\ $$$$\mathrm{so}\:\mathrm{that}\:\mathrm{AB}\::\:\mathrm{BX}=\mathrm{2}:\mathrm{1},\:\mathrm{find}\:{x},\:\mathrm{the}\:\mathrm{position} \\ $$$$\mathrm{vector}\:\mathrm{of}\:\mathrm{X}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\boldsymbol{\mathrm{b}}\:\mathrm{and}\:\boldsymbol{\mathrm{c}}. \\ $$$$\mathrm{ii}.\:\mathrm{if}\:\mathrm{Y}\:\mathrm{lies}\:\mathrm{on}\:\mathrm{BC},\:\mathrm{between}\:\mathrm{B}\:\mathrm{and}\:\mathrm{C}\:\mathrm{so}\:\mathrm{that} \\ $$$$\mathrm{BY}\::\:\mathrm{YC}\:=\:\mathrm{1}:\mathrm{3},\:\mathrm{find}\:{y},\:\mathrm{the}\:\mathrm{position}\:\mathrm{vector} \\ $$$$\mathrm{of}\:\mathrm{Y}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\boldsymbol{\mathrm{b}}\:\mathrm{and}\:\boldsymbol{\mathrm{c}}. \\…
Question Number 108776 by bobhans last updated on 19/Aug/20 $$\:\overset{{bobhans}} {\vdots} \\ $$$$\int\:\frac{{dx}}{\:\sqrt{{x}\sqrt{{x}}\:−{x}^{\mathrm{2}} }}\:=\:? \\ $$ Answered by john santu last updated on 19/Aug/20 $$\:\:\:\:\:\frac{\multimap\mathcal{JS}\multimap}{\heartsuit}…
Question Number 42711 by rahul 19 last updated on 01/Sep/18 $$\mathrm{If}\:\mathrm{2}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{are}\:\hat {\mathrm{i}}+\mathrm{2}\hat {\mathrm{j}}\:\mathrm{and} \\ $$$$\hat {\mathrm{i}}+\hat {\mathrm{k}}\:,\:\mathrm{then}\:\mathrm{find}\:\mathrm{all}\:\mathrm{possible}\:\mathrm{third}\:\mathrm{side}\:? \\ $$ Answered by MJS last updated on…
Question Number 42528 by rahul 19 last updated on 27/Aug/18 $$\mathrm{Solve}\:: \\ $$$$\frac{\mathrm{p}+\mathrm{2}}{\mathrm{2p}+\mathrm{1}}\:=\:\frac{\mathrm{q}+\mathrm{2p}}{\mathrm{2q}+\mathrm{p}}\:=\:\frac{\mathrm{1}+\mathrm{2q}}{\mathrm{2}+\mathrm{q}}\:=\:\lambda. \\ $$$$\mathrm{Find}\:\left(\mathrm{p},\mathrm{q}\right)\:? \\ $$ Answered by math1967 last updated on 27/Aug/18 $$\lambda=\frac{{p}+\mathrm{2}+{q}+\mathrm{2}{p}+\mathrm{1}+\mathrm{2}{q}}{\mathrm{2}{p}+\mathrm{1}+\mathrm{2}{q}+{p}+\mathrm{2}+{q}}=\frac{\mathrm{3}\left({p}+{q}+\mathrm{1}\right)}{\mathrm{3}\left({p}+{q}+\mathrm{1}\right)}=\mathrm{1}…
Question Number 42521 by rahul 19 last updated on 27/Aug/18 $$\mathrm{Let}\:\overset{\rightarrow\:} {\mathrm{a}},\:\overset{\rightarrow} {\mathrm{b}}\:,\:\overset{\rightarrow} {\mathrm{c}}\:\mathrm{be}\:\mathrm{three}\:\mathrm{unit}\:\mathrm{vectors} \\ $$$$\mathrm{such}\:\mathrm{that}\:\mathrm{3}\overset{\rightarrow} {\mathrm{a}}+\mathrm{4}\overset{\rightarrow} {\mathrm{b}}+\mathrm{5}\overset{\rightarrow} {\mathrm{c}}\:=\:\mathrm{0}.\:\mathrm{Then}\:\mathrm{prove} \\ $$$$\mathrm{that}\:\overset{\rightarrow\:} {\mathrm{a}},\:\overset{\rightarrow} {\mathrm{b}},\overset{\rightarrow} {\mathrm{c}}\:\mathrm{are}\:\mathrm{coplanar}. \\…