Question Number 56899 by Tawa1 last updated on 26/Mar/19 $$\mathrm{Let}\:\:\mathrm{W}\:\:\mathrm{be}\:\mathrm{the}\:\mathrm{subspace}\:\mathrm{of}\:\:\mathbb{R}^{\mathrm{4}} \:\:\mathrm{generated}\:\mathrm{by}\:\mathrm{vector}\: \\ $$$$\left(\mathrm{1},\:−\:\mathrm{2},\:\mathrm{5},\:−\:\mathrm{3}\right),\:\:\:\:\left(\mathrm{2},\:\mathrm{3},\:\mathrm{1},\:−\:\mathrm{4}\right),\:\:\:\left(\mathrm{3},\:\mathrm{8},\:−\:\mathrm{3},\:−\:\mathrm{5}\right)\:\:\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{basis}\:\mathrm{and}\:\mathrm{dimension}\:\mathrm{of}\:\:\mathrm{W}. \\ $$ Answered by kaivan.ahmadi last updated on 26/Mar/19 $$\begin{vmatrix}{\mathrm{1}\:\:\:\:\:\:\:−\mathrm{2}\:\:\:\:\:\:\:\:\:\mathrm{5}\:\:\:\:\:\:−\mathrm{3}}\\{\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:−\mathrm{4}\:\:}\\{\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\mathrm{8}\:\:\:\:\:−\mathrm{3}\:\:\:\:\:\:\:−\mathrm{5}}\end{vmatrix}\underset{−\mathrm{3}{R}_{\mathrm{1}}…
Question Number 56803 by Tawa1 last updated on 24/Mar/19 $$\mathrm{Evaluate}:\:\:\:\int_{\:\mathrm{1}} ^{\:\mathrm{2}} \:\left(\mathrm{A}\centerdot\mathrm{B}\:×\:\mathrm{C}\right)\:\mathrm{dt}\:\:\:\:\mathrm{and}\:\:\:\int_{\:\mathrm{1}} ^{\:\mathrm{2}} \:\mathrm{A}\:×\:\left(\mathrm{B}\:×\:\mathrm{C}\right)\:\:\:\mathrm{dt} \\ $$$$\mathrm{where},\:\:\:\:\:\:\:\:\:\mathrm{A}\:\:=\:\:\mathrm{ti}\:−\:\mathrm{3j}\:+\:\mathrm{2tk},\:\:\:\:\:\:\:\mathrm{B}\:\:=\:\:\mathrm{i}\:−\:\mathrm{2j}\:+\:\mathrm{2k}, \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{C}\:\:=\:\:\mathrm{3i}\:+\:\mathrm{tj}\:−\:\mathrm{k} \\ $$ Commented by tanmay.chaudhury50@gmail.com last updated…
Question Number 55948 by Tawa1 last updated on 06/Mar/19 Answered by kaivan.ahmadi last updated on 07/Mar/19 $${a}.\:\:{A}×{B}=\begin{vmatrix}{{i}\:\:\:\:\:{j}\:\:\:\:\:\:{k}}\\{\mathrm{3}\:\:\:−\mathrm{1}\:\:\:\mathrm{2}}\\{\mathrm{2}\:\:\:\mathrm{1}\:\:\:\:−\mathrm{1}}\end{vmatrix}\begin{vmatrix}{{i}\:\:\:\:{j}\:\:\:\:\:\:\:{k}}\\{\mathrm{3}\:\:\:−\mathrm{1}\:\:\:\mathrm{2}}\\{\mathrm{2}\:\:\:\:\:\mathrm{1}\:\:\:−\mathrm{1}}\end{vmatrix}= \\ $$$$\left({i}+\mathrm{4}{j}+\mathrm{3}{k}\right)−\left(−\mathrm{2}{k}−\mathrm{3}{j}+\mathrm{2}{i}\right)=−\mathrm{3}{i}+\mathrm{7}{j}+\mathrm{5}{k} \\ $$$$\left({A}×{B}\right)×{C}=\begin{vmatrix}{{i}\:\:\:\:\:\:\:\:\:{j}\:\:\:\:\:{k}}\\{−\mathrm{3}\:\:\:\mathrm{7}\:\:\:\:\:\mathrm{5}}\\{\mathrm{1}\:\:\:\:\:\:−\mathrm{2}\:\:\:\mathrm{2}}\end{vmatrix}\begin{vmatrix}{{i}\:\:\:\:\:\:\:\:\:\:{j}\:\:\:\:\:{k}}\\{−\mathrm{3}\:\:\:\:\:\mathrm{7}\:\:\:\:\mathrm{5}}\\{\mathrm{1}\:\:\:\:\:−\mathrm{2}\:\:\:\:\mathrm{2}}\end{vmatrix}= \\ $$$$\left(\mathrm{14}{i}+\mathrm{5}{j}+\mathrm{6}{k}\right)−\left(\mathrm{7}{k}−\mathrm{6}{j}−\mathrm{10}{i}\right)= \\ $$$$\mathrm{24}{i}+\mathrm{11}{j}−{k}…
Question Number 55860 by Easyman32 last updated on 05/Mar/19 Answered by tanmay.chaudhury50@gmail.com last updated on 05/Mar/19 $${jones}\:{probablity}\:{of}\:{winning}={a} \\ $$$${john}\:{probablity}\:{of}\:{winning}=\mathrm{3}{a} \\ $$$${david}….=\mathrm{2}×\mathrm{3}{a}=\mathrm{6}{a} \\ $$$${a}+\mathrm{3}{a}+\mathrm{6}{a}=\mathrm{1} \\ $$$${a}=\frac{\mathrm{1}}{\mathrm{10}}…
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Question Number 120801 by TITA last updated on 02/Nov/20 Commented by liberty last updated on 02/Nov/20 $$\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