Question

Find the magnitude and direction (in degrees) of the vector. (Assume 0° ≤ < 360°.)

Answer 1

$6i~~ + ~~2\sqrt{3}j\implies < \stackrel{a}{6}~~,~~\stackrel{b}{2\sqrt{3}} > \\\\[-0.35em] ~\dotfill\\\\ \stackrel{magnitude}{\sqrt{a^2+b^2}}\implies \sqrt{6^2+(2\sqrt{3})^2}\implies \sqrt{36+(2^2\cdot 3)}\implies \sqrt{36+12}\implies \sqrt{48} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{direction}{tan^{-1}\left( \cfrac{b}{a}\right)}\implies tan^{-1}\left( \cfrac{2\sqrt{3}}{6} \right)\implies tan^{-1}\left( \cfrac{\sqrt{3}}{3} \right) \\\\\\ tan^{-1}\left( \cfrac{1}{\sqrt{3}} \right)\implies 30^o$