Question Number 133282 by rexford last updated on 20/Feb/21 Answered by mr W last updated on 25/Feb/21 $$\left(\boldsymbol{{b}}×\boldsymbol{{c}}\right)×\boldsymbol{{a}} \\ $$$$=\left[\left(\mathrm{1},\mathrm{2},−\mathrm{1}\right)×\left(\mathrm{1},\mathrm{1},−\mathrm{2}\right)\right]×\left(\mathrm{2},−\mathrm{1},\mathrm{1}\right) \\ $$$$=\left(−\mathrm{3},\mathrm{1},−\mathrm{1}\right)×\left(\mathrm{2},−\mathrm{1},\mathrm{1}\right) \\ $$$$=\left(\mathrm{0},\mathrm{1},\mathrm{1}\right) \\…
Question Number 133240 by rexford last updated on 20/Feb/21 Answered by Kunal12588 last updated on 20/Feb/21 $${r}\:\mathrm{cos}\:\gamma\:=\:\mathrm{2} \\ $$$$\mathrm{cos}\:\gamma\:=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{4}+\mathrm{1}+\mathrm{4}}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\gamma\:=\:\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{2}}{\mathrm{3}} \\ $$ Answered…
Question Number 133206 by rexford last updated on 20/Feb/21 Answered by EDWIN88 last updated on 20/Feb/21 $$\mathrm{i}+\mathrm{j}+\mathrm{3k}=\begin{pmatrix}{\mathrm{1}}\\{\mathrm{1}}\\{\mathrm{3}}\end{pmatrix}\:;\:\mathrm{3i}−\mathrm{3j}+\mathrm{k}\:=\:\begin{pmatrix}{\:\:\:\mathrm{3}}\\{−\mathrm{3}}\\{\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$$−\mathrm{4i}+\mathrm{5j}\:=\:\begin{pmatrix}{−\mathrm{4}}\\{\:\:\:\mathrm{5}}\\{\:\:\:\mathrm{0}}\end{pmatrix} \\ $$$$\Leftrightarrow\:\begin{pmatrix}{\mathrm{1}}\\{\mathrm{1}}\\{\mathrm{3}}\end{pmatrix}\:\mathrm{x}\:+\begin{pmatrix}{\:\:\:\mathrm{3}}\\{−\mathrm{3}}\\{\:\:\:\mathrm{1}}\end{pmatrix}\:\mathrm{y}\:+\begin{pmatrix}{−\mathrm{4}}\\{\:\:\:\mathrm{5}}\\{\:\:\:\mathrm{0}}\end{pmatrix}\:\mathrm{z}\:=\:\lambda\:\begin{pmatrix}{\mathrm{x}}\\{\mathrm{y}}\\{\mathrm{z}}\end{pmatrix} \\ $$$$\Leftrightarrow\:\begin{pmatrix}{\mathrm{x}+\mathrm{3y}−\mathrm{4z}}\\{\mathrm{x}−\mathrm{3y}+\mathrm{5z}}\\{\mathrm{3x}+\mathrm{y}+\mathrm{0z}}\end{pmatrix}\:=\:\lambda\begin{pmatrix}{\mathrm{x}}\\{\mathrm{y}}\\{\mathrm{z}}\end{pmatrix} \\ $$$$\Leftrightarrow\:\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\mathrm{3}\:\:\:\:\:\:−\mathrm{4}}\\{\mathrm{1}\:\:\:−\mathrm{3}\:\:\:\:\:\:\:\:\mathrm{5}}\\{\mathrm{3}\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{0}}\end{pmatrix}\:\begin{pmatrix}{\mathrm{x}}\\{\mathrm{y}}\\{\mathrm{z}}\end{pmatrix}\:=\:\lambda\:\begin{pmatrix}{\mathrm{x}}\\{\mathrm{y}}\\{\mathrm{z}}\end{pmatrix}…
Question Number 2045 by prakash jain last updated on 31/Oct/15 $${I}\mathrm{s}\:{the}\:{following}\:{series}\:{absolutely}\:{convergent}? \\ $$$${S}_{\mathrm{1}} =\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)} \\ $$$${Is}\:{the}\:{following}\:{series}\:{absolutely}\:{convergent}? \\ $$$${S}_{\mathrm{2}} =\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\left(\frac{\mathrm{1}}{{n}}−\:\frac{\mathrm{1}}{{n}+\mathrm{1}}\right) \\ $$…
Question Number 1865 by Denbang last updated on 18/Oct/15 $${proof}\:{that}\:\sqrt{\mathrm{2}}\:{is}\:{an}\:{irrational}\:{number} \\ $$$$ \\ $$ Answered by 123456 last updated on 18/Oct/15 $$\mathrm{suppuse}\:\mathrm{by}\:\mathrm{absurf}\:\mathrm{that}\:\sqrt{\mathrm{2}}\in\mathbb{Q},\:\mathrm{then} \\ $$$$\exists\left({p},{q}\right)\in\mathbb{Z},{q}\neq\mathrm{0}\:\mathrm{such}\:\mathrm{that}\:\sqrt{\mathrm{2}}=\frac{{p}}{{q}},\left({p},{q}\right)=\mathrm{1} \\…
Question Number 132904 by bramlexs22 last updated on 17/Feb/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{between}\:\mathrm{the}\:\mathrm{line}\: \\ $$$$\frac{\mathrm{x}−\mathrm{2}}{\mathrm{3}}=\:\frac{\mathrm{y}+\mathrm{1}}{−\mathrm{1}}=\:\frac{\mathrm{z}−\mathrm{3}}{\mathrm{2}}\:\mathrm{and}\:\mathrm{the}\: \\ $$$$\mathrm{plane}\:\mathrm{3x}+\mathrm{4y}+\mathrm{z}+\mathrm{5}\:=\:\mathrm{0}\: \\ $$ Answered by mr W last updated on 17/Feb/21 $${l}:\:\left(\mathrm{3},−\mathrm{1},\:\mathrm{2}\right)…
Question Number 1537 by 2closedStringsMeet last updated on 17/Aug/15 $${Show}\:{by}\:{use}\:{of}\:{the}\:{characteristics}\:{of} \\ $$$${a}\:{vector}−{space},\:{that}\:{the}\:{set}\:\overset{\rightarrow} {{x}}\in\mathbb{R}_{+} \\ $$$$\mathrm{with}\:\mathrm{the}\:\mathrm{following}\:\mathrm{operations}\:\oplus\:{and}\: \\ $$$${builds}\:{a}\:{vector}−{space}. \\ $$$$\ast{Use}\:{vector}−{space}\:{axioms} \\ $$$$\ast\overset{\rightarrow} {{x}}\oplus\overset{\rightarrow} {{y}}\:\:\:\:{be}\:\:\:\overset{\rightarrow} {{x}}\centerdot\overset{\rightarrow} {{y}}…
Question Number 1479 by yyy last updated on 12/Aug/15 $$ \\ $$$$ \\ $$ Answered by 123456 last updated on 12/Aug/15 $$\xi:\mathbb{R}×\mathrm{X}\rightarrow\mathbb{R} \\ $$$$\xi_{\mathrm{X}} \left({s}\right)=\underset{{x}\in\mathrm{X}}…
Question Number 1016 by 123456 last updated on 14/May/15 $$\left[\left(\overset{\rightarrow} {{x}}×\overset{\rightarrow} {{a}}\right)×\overset{\rightarrow} {{b}}\right]×\overset{\rightarrow} {{x}}=\overset{\rightarrow} {{c}} \\ $$ Answered by rpatle69@gmail.com last updated on 14/May/15 $$\left[−\overset{\rightarrow}…