Question

if a = 15 80° and b = 12i + 16j what is the magnitude of a - b

Answer 1

$\vec{A}=(15,80^{\circ})$

     $=15cos80^{\circ}+15sin80^{\circ}$

     $=2.61i+14.77j$

$\vec{B}=12i-16j$

SO, Vector $({A-B}=(2.61i+14.77j)-(12i-16j)$

       $=-9.39i+30.77j$

Magnitude $|(A-B|=\sqrt{(-9.39)^2+(30.77)^2}=\sqrt{88.17+946.79}=\sqrt{1034.96}=32.17$