Question

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

= 6i + 2sqrt3j

Answer 1

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