Question Number 207332 by mustafazaheen last updated on 12/May/24
$$\left(\overset{\rightarrow} {\mathrm{a}}×\overset{\rightarrow} {\mathrm{b}}\right)×\left(\overset{\rightarrow} {\mathrm{a}}\right)=? \\ $$$$\mathrm{how}\:\mathrm{is}\:\mathrm{the}\:\mathrm{solution} \\ $$
Commented by Ghisom last updated on 12/May/24
$$\left(\begin{pmatrix}{{a}_{\mathrm{1}} }\\{{a}_{\mathrm{2}} }\\{{a}_{\mathrm{3}} }\end{pmatrix}\:×\begin{pmatrix}{{b}_{\mathrm{1}} }\\{{b}_{\mathrm{2}} }\\{{b}_{\mathrm{3}} }\end{pmatrix}\:\right)×\begin{pmatrix}{{a}_{\mathrm{1}} }\\{{a}_{\mathrm{2}} }\\{{a}_{\mathrm{3}} }\end{pmatrix}\:= \\ $$$$=\begin{pmatrix}{{a}_{\mathrm{2}} {b}_{\mathrm{3}} −{a}_{\mathrm{3}} {b}_{\mathrm{2}} }\\{{a}_{\mathrm{3}} {b}_{\mathrm{1}} −{a}_{\mathrm{1}} {b}_{\mathrm{3}} }\\{{a}_{\mathrm{1}} {b}_{\mathrm{2}} −{a}_{\mathrm{2}} {b}_{\mathrm{1}} }\end{pmatrix}\:×\begin{pmatrix}{{a}_{\mathrm{1}} }\\{{a}_{\mathrm{2}} }\\{{a}_{\mathrm{3}} }\end{pmatrix}\:= \\ $$$$=\begin{pmatrix}{\left({a}_{\mathrm{2}} ^{\mathrm{2}} +{a}_{\mathrm{3}} ^{\mathrm{2}} \right){b}_{\mathrm{1}} −{a}_{\mathrm{1}} \left({a}_{\mathrm{2}} {b}_{\mathrm{2}} +{a}_{\mathrm{3}} {b}_{\mathrm{3}} \right)}\\{\left({a}_{\mathrm{1}} ^{\mathrm{2}} +{a}_{\mathrm{3}} ^{\mathrm{2}} \right){b}_{\mathrm{2}} −{a}_{\mathrm{2}} \left({a}_{\mathrm{1}} {b}_{\mathrm{1}} +{a}_{\mathrm{3}} {b}_{\mathrm{3}} \right)}\\{\left({a}_{\mathrm{1}} ^{\mathrm{2}} +{a}_{\mathrm{2}} ^{\mathrm{2}} \right){b}_{\mathrm{3}} −{a}_{\mathrm{3}} \left({a}_{\mathrm{1}} {b}_{\mathrm{1}} +{a}_{\mathrm{2}} {b}_{\mathrm{2}} \right)}\end{pmatrix} \\ $$
Answered by aleks041103 last updated on 12/May/24
$$\left(\overset{\rightarrow} {{a}}×\overset{\rightarrow} {{b}}\right)×\overset{\rightarrow} {{c}}=\left(\overset{\rightarrow} {{a}}.\overset{\rightarrow} {{c}}\right)\overset{\rightarrow} {{b}}−\left(\overset{\rightarrow} {{b}}.\overset{\rightarrow} {{c}}\right)\overset{\rightarrow} {{a}} \\ $$$$\Rightarrow\left(\overset{\rightarrow} {{a}}×\overset{\rightarrow} {{b}}\right)×\overset{\rightarrow} {{a}}=\left(\overset{\rightarrow} {{a}}.\overset{\rightarrow} {{a}}\right)\overset{\rightarrow} {{b}}−\left(\overset{\rightarrow} {{b}}.\overset{\rightarrow} {{a}}\right)\overset{\rightarrow} {{a}}= \\ $$$$=\mid\overset{\rightarrow} {{a}}\mid^{\mathrm{2}} \overset{\rightarrow} {{b}}−\left(\overset{\rightarrow} {{a}}.\overset{\rightarrow} {{b}}\right)\overset{\rightarrow} {{a}} \\ $$