Question Number 132650 by aupo14 last updated on 15/Feb/21
Commented by mr W last updated on 15/Feb/21
$${what}\:{do}\:{you}\:{mean}\:{with}\:\boldsymbol{{A\&B}}? \\ $$
Commented by aupo14 last updated on 15/Feb/21
Commented by mr W last updated on 15/Feb/21
$${what}\:{is}\:{that}? \\ $$
Commented by mr W last updated on 15/Feb/21
$${i}\:{know}\: \\ $$$$\boldsymbol{{A}}+\boldsymbol{{B}} \\ $$$$\boldsymbol{{A}}−\boldsymbol{{B}} \\ $$$$\boldsymbol{{A}}\bullet\boldsymbol{{B}} \\ $$$$\boldsymbol{{A}}×\boldsymbol{{B}} \\ $$$${but}\:{how}\:{is}\:\boldsymbol{{A\&B}}\:{defined}? \\ $$
Commented by aupo14 last updated on 15/Feb/21
Answered by mr W last updated on 15/Feb/21
Commented by mr W last updated on 16/Feb/21
$${a}=\frac{\boldsymbol{{A}}\centerdot\boldsymbol{{B}}}{\mid\boldsymbol{{A}}\mid} \\ $$$$\boldsymbol{{C}}=−\frac{{a}\boldsymbol{{A}}}{\mid\boldsymbol{{A}}\mid}+\boldsymbol{{B}}=−\frac{\boldsymbol{{A}}\centerdot\boldsymbol{{B}}}{\mid\boldsymbol{{A}}\mid^{\mathrm{2}} }\boldsymbol{{A}}+\boldsymbol{{B}} \\ $$$$=−\frac{\mathrm{1}×\mathrm{1}+\mathrm{1}×\left(−\mathrm{1}\right)+\mathrm{1}×\left(−\mathrm{1}\right)}{\mathrm{1}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }\left(\mathrm{1},\mathrm{1},\mathrm{1}\right)+\left(\mathrm{1},−\mathrm{1},−\mathrm{1}\right) \\ $$$$=\left(\frac{\mathrm{1}}{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{3}}\right)+\left(\mathrm{1},−\mathrm{1},−\mathrm{1}\right) \\ $$$$=\left(\frac{\mathrm{4}}{\mathrm{3}},−\frac{\mathrm{2}}{\mathrm{3}},−\frac{\mathrm{2}}{\mathrm{3}}\right) \\ $$$${generally} \\ $$$$\left(\mathrm{2}{p},\:−{p},\:−{p}\right)\:{is}\:{such}\:{a}\:{vector} \\ $$$${with}\:{p}\in{R}\:\wedge{p}\neq\mathrm{0} \\ $$