Question Number 169885 by depressiveshrek last updated on 11/May/22
$$\mid\overset{\rightarrow} {{a}}\mid=\mathrm{13} \\ $$$$\mid\overset{\rightarrow} {{b}}\mid=\mathrm{19} \\ $$$$\mid\overset{\rightarrow} {{a}}+\overset{\rightarrow} {{b}}\mid=\mathrm{24} \\ $$$$\mid\overset{\rightarrow} {{a}}−\overset{\rightarrow} {{b}}\mid=? \\ $$
Answered by mr W last updated on 11/May/22
$$\mid\overset{\rightarrow} {{a}}−\overset{\rightarrow} {{b}}\mid={x} \\ $$$${x}^{\mathrm{2}} +\mathrm{24}^{\mathrm{2}} =\mathrm{2}×\left(\mathrm{13}^{\mathrm{2}} +\mathrm{19}^{\mathrm{2}} \right) \\ $$$$\Rightarrow{x}=\sqrt{\mathrm{2}×\left(\mathrm{13}^{\mathrm{2}} +\mathrm{19}^{\mathrm{2}} \right)−\mathrm{24}^{\mathrm{2}} }=\mathrm{22}\:\checkmark \\ $$
Commented by depressiveshrek last updated on 11/May/22
$${sorry}\:{where}\:{does}\:{the}\:\mathrm{2}\:{come}\:{from}? \\ $$
Commented by mr W last updated on 12/May/22
Commented by mr W last updated on 11/May/22
$$\overset{\rightarrow} {{c}}=\overset{\rightarrow} {{a}}+\overset{\rightarrow} {{b}} \\ $$$$\overset{\rightarrow} {{d}}=\overset{\rightarrow} {{a}}−\overset{\rightarrow} {{b}} \\ $$$${a}^{\mathrm{2}} =\left(\frac{{d}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{c}}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{2}×\left(\frac{{d}}{\mathrm{2}}\right)×\left(\frac{{c}}{\mathrm{2}}\right)×\mathrm{cos}\:\theta \\ $$$${b}^{\mathrm{2}} =\left(\frac{{d}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{c}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{2}×\left(\frac{{d}}{\mathrm{2}}\right)×\left(\frac{{c}}{\mathrm{2}}\right)×\mathrm{cos}\:\theta \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{2}×\left(\frac{{d}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{2}×\left(\frac{{c}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{c}^{\mathrm{2}} +{d}^{\mathrm{2}} =\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right) \\ $$