Question Number 19976 by Tinkutara last updated on 19/Aug/17
$$\mathrm{Three}\:\mathrm{vectors}\:\overset{\rightarrow} {{A}},\:\overset{\rightarrow} {{B}}\:\mathrm{and}\:\overset{\rightarrow} {{C}}\:\mathrm{add}\:\mathrm{up}\:\mathrm{to} \\ $$$$\mathrm{zero}.\:\mathrm{Find}\:\mathrm{which}\:\mathrm{is}\:\mathrm{false}. \\ $$$$\left({a}\right)\:\left(\overset{\rightarrow} {{A}}×\overset{\rightarrow} {{B}}\right)×\overset{\rightarrow} {{C}}\:\mathrm{is}\:\mathrm{not}\:\mathrm{zero}\:\mathrm{unless}\:\overset{\rightarrow} {{B}},\:\overset{\rightarrow} {{C}} \\ $$$$\mathrm{are}\:\mathrm{parallel} \\ $$$$\left({b}\right)\:\left(\overset{\rightarrow} {{A}}×\overset{\rightarrow} {{B}}\right)\centerdot\overset{\rightarrow} {{C}}\:\mathrm{is}\:\mathrm{not}\:\mathrm{zero}\:\mathrm{unless}\:\overset{\rightarrow} {{B}},\:\overset{\rightarrow} {{C}} \\ $$$$\mathrm{are}\:\mathrm{parallel} \\ $$$$\left({c}\right)\:\mathrm{If}\:\overset{\rightarrow} {{A}},\:\overset{\rightarrow} {{B}},\:\overset{\rightarrow} {{C}}\:\mathrm{define}\:\mathrm{a}\:\mathrm{plane},\:\left(\overset{\rightarrow} {{A}}×\overset{\rightarrow} {{B}}×\overset{\rightarrow} {{C}}\right) \\ $$$$\mathrm{is}\:\mathrm{in}\:\mathrm{that}\:\mathrm{plane} \\ $$$$\left({d}\right)\:\left(\overset{\rightarrow} {{A}}×\overset{\rightarrow} {{B}}\right).\overset{\rightarrow} {{C}}\:=\:\mid\overset{\rightarrow} {{A}}\mid\mid\overset{\rightarrow} {{B}}\mid\mid\overset{\rightarrow} {{C}}\mid\:\rightarrow\:{C}^{\mathrm{2}} \:=\:{A}^{\mathrm{2}} \:+\:{B}^{\mathrm{2}} \\ $$
Answered by ajfour last updated on 19/Aug/17
$$\left({a}\right),\left({b}\right),\left({d}\right)\:{are}\:{false}\:.\:{for}\:\left(\overset{\rightarrow} {{A}}×\overset{\rightarrow} {{B}}\right).\overset{\rightarrow} {{C}}\:\:\:\:{to}\:{be} \\ $$$${zero}\:\overset{\rightarrow} {{C}}\:{just}\:{need}\:{to}\:{be}\:{in}\:{the}\:{plane} \\ $$$${of}\:\overset{\rightarrow} {{A}}\:{and}\:\overset{\rightarrow} {{B}}\:.\:{For}\:\left(\overset{\rightarrow} {{A}}×\overset{\rightarrow} {{B}}\right)×\overset{\rightarrow} {{C}}\:\:{to}\:{be} \\ $$$${zero}\:\overset{\rightarrow} {{C}}\:\:{need}\:{to}\:{be}\:{parallel}\:{to}\:\overset{\rightarrow} {{A}}×\overset{\rightarrow} {{B}}\:. \\ $$$${If}\:\left(\overset{\rightarrow} {{A}}×\overset{\rightarrow} {{B}}\right).\overset{\rightarrow} {{C}}=\mid\overset{\rightarrow} {{A}}\mid\mid\overset{\rightarrow} {{B}}\mid\mid\overset{\rightarrow} {{C}}\mid\:\:\:,\:\overset{\rightarrow} {{A}},\:\overset{\rightarrow} {{B}},\:\overset{\rightarrow} {{C}} \\ $$$${just}\:{need}\:{be}\:{mutually}\:{perpendicular}. \\ $$
Commented by ajfour last updated on 19/Aug/17
$$\left(\hat {{i}}×\hat {{j}}\right)×\hat {{j}}\:=\:−\hat {{i}}\:.\:\left({a}\right)\:{is}\:{false}\:. \\ $$
Commented by Tinkutara last updated on 20/Aug/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$