Q:

Find the unit vector in the direction of u = (-3,2).a. u = -3√13 / 13i + 2√13 / 13jb. u = -3√5 / 5i + 2√5 / 5jc. u = 3√5 / 5i - 2√5 / 5jd. u = 3√13 / 13i + 2√13 / 13j

Accepted Solution

A:
[tex]\bf \textit{unit vector for }(a,b)\implies \left( \cfrac{a}{\sqrt{a^2+b^2}}~~,~~\cfrac{b}{\sqrt{a^2+b^2}} \right)\\\\ -------------------------------\\\\ (-3,2)\qquad \stackrel{unit~vector}{\implies }\qquad \left( \cfrac{-3}{\sqrt{(-3)^2+2^2}}~~,~~\cfrac{2}{\sqrt{(-3)^2+2^2}} \right) \\\\\\ \left( -\cfrac{3}{\sqrt{13}}~~,~~ \cfrac{2}{\sqrt{13}}\right)\\\\ -------------------------------[/tex]

[tex]\bf \textit{and now let's \underline{rationalize} the denominator for each} \\\\\\ -\cfrac{3}{\sqrt{13}}\cdot \cfrac{\sqrt{13}}{\sqrt{13}}\implies -\cfrac{3\sqrt{13}}{13} \qquad \qquad \qquad \qquad \cfrac{2}{\sqrt{13}}\cdot \cfrac{\sqrt{13}}{\sqrt{13}}\implies \cfrac{2\sqrt{13}}{13} \\\\\\ \textit{and written in \underline{ai+bj form}}\qquad -\cfrac{3\sqrt{13}}{13}i~~~~+~~~~\cfrac{2\sqrt{13}}{13}j[/tex]