Question Number 23442 by chernoaguero@gmail.com last updated on 30/Oct/17
$$\mathrm{Find}\:\mathrm{a}\:\mathrm{unit}\:\mathrm{vector}\:\mathrm{which}\:\mathrm{is}\:\mathrm{perpendicula} \\ $$$$\mathrm{to}\:\mathrm{a}\:\mathrm{vector}\:\mathrm{A}\left(\mathrm{3coma5coma1}\right) \\ $$$$ \\ $$$$\mathrm{Sorry}\:\mathrm{for}\:\mathrm{writing}\:\mathrm{coma}\:\mathrm{cuz}\:\mathrm{i}\:\mathrm{dnt}\:\mathrm{see}\:\mathrm{a}\:\mathrm{key}\:\mathrm{for}\:\mathrm{it} \\ $$
Commented by $@ty@m last updated on 31/Oct/17
Commented by chernoaguero@gmail.com last updated on 31/Oct/17
$$\mathrm{oopz}\:\mathrm{u}\:\mathrm{are}\:\mathrm{right}\:\mathrm{i}\:\mathrm{was}\:\mathrm{seachin}\:\mathrm{for}\:\mathrm{it} \\ $$$$\mathrm{since}\:\mathrm{but}\:\mathrm{could}\:\mathrm{not}\:\mathrm{see}\:\mathrm{iti} \\ $$
Answered by $@ty@m last updated on 31/Oct/17
$${Given}, \\ $$$$\overset{\rightarrow} {{OA}}=\mathrm{3}\hat {{i}}+\mathrm{5}\hat {{j}}+\hat {{k}} \\ $$$${Let}\:\overset{\rightarrow} {{OB}}={x}\hat {{i}}+{y}\hat {{j}}+{z}\hat {{k}}\:\:{such}\:{that} \\ $$$$\overset{\rightarrow} {{OA}}\bot\overset{\rightarrow} {{OB}}\:\:\&\:\mid{OB}\mid=\mathrm{1}\:\:….\left(\mathrm{1}\right) \\ $$$$\:\Rightarrow\:\overset{\rightarrow} {{OA}}.\overset{\rightarrow} {{OB}}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{3}{x}+\mathrm{5}{y}+{z}=\mathrm{0}\:….\left(\mathrm{2}\right) \\ $$$${There}\:{would}\:{be}\:{infinite}\:{unit} \\ $$$${vectors}\:\:\overset{\rightarrow} {{OB}}={x}\hat {{i}}+{y}\hat {{j}}+{z}\hat {{k}}\:\: \\ $$$${satisfying}\:{conditions}\:\left(\mathrm{1}\right)\:{and}\:\left(\mathrm{2}\right). \\ $$$${For}\:{example}, \\ $$$${Let}\:{x}=\mathrm{0} \\ $$$${then}\:{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{1}\:\&\:\mathrm{5}{y}+{z}=\mathrm{0} \\ $$$$\Rightarrow{y}=\pm\frac{\sqrt{\mathrm{26}}}{\mathrm{26}}\:\&\:{z}=\mp\frac{\mathrm{5}\sqrt{\mathrm{26}}}{\mathrm{26}} \\ $$$$\therefore\:{one}\:{of}\:{the}\:{possible}\:{unit}\:{vectors} \\ $$$${perpendicular}\:{to}\:\overset{\rightarrow} {{OA}}\:{is} \\ $$$$\frac{\sqrt{\mathrm{26}}}{\mathrm{26}}\hat {{j}}−\frac{\mathrm{5}\sqrt{\mathrm{26}}}{\mathrm{26}}\hat {{k}} \\ $$
Commented by chernoaguero@gmail.com last updated on 01/Nov/17
$$\mathrm{Thank}\:\mathrm{u}\:\mathrm{so}\:\mathrm{much} \\ $$