Question Number 36091 by Rio Mike last updated on 28/May/18
$$\mathrm{Given}\:\mathrm{the}\:\mathrm{position}\:\mathrm{vectors} \\ $$$${v}_{\mathrm{1}} =\:\mathrm{2}{i}\:−\:\mathrm{2}{j}\:{and}\:{v}_{\mathrm{2}} =\:\mathrm{2}{j}, \\ $$$$\mathrm{show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{unit}\:\mathrm{vector}\:\mathrm{in}\: \\ $$$$\mathrm{the}\:\mathrm{direction}\:\mathrm{of}\:\mathrm{the}\:\mathrm{vector}\: \\ $$$${v}_{\mathrm{1}} −\:{v}_{\mathrm{2}\:\:\:} \mathrm{is}\:\frac{\mathrm{1}}{\sqrt{\mathrm{5}}}\left(\mathrm{i}−\mathrm{2j}\right) \\ $$
Commented by prof Abdo imad last updated on 31/May/18
$${we}\:{have}\:{v}_{\mathrm{1}} −{v}_{\mathrm{2}} =\:\mathrm{2}{i}\:−\mathrm{2}{j}−\mathrm{2}{j}\:=\:\mathrm{2}{i}\:−\mathrm{4}{j}\:{so}\:{the}\: \\ $$$${unit}\:{vrctor}\:\:{in}\:{the}\:{direction}\:{of}\:{v}_{\mathrm{1}} \:−{v}_{\mathrm{2}} \:{is} \\ $$$${u}=\:\frac{\mathrm{1}}{\mid\mid{v}_{\mathrm{1}} −{v}_{\mathrm{2}} \mid\mid}\left({v}_{\mathrm{1}} −{v}_{\mathrm{2}} \right)\:=\frac{\mathrm{1}}{\sqrt{\mathrm{4}+\mathrm{16}}}\left(\mathrm{2}{i}−\mathrm{4}{j}\right) \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{5}}}\left(\mathrm{2}{i}−\mathrm{4}{j}\right)=\:\frac{\mathrm{1}}{\sqrt{\mathrm{5}}}\left({i}\:−\mathrm{2}{j}\right)\:. \\ $$