# Problem: Find the angle between each of the following pairs of vectors A =Axi + Ayj and B =Bxi + ByjAx2= 3.20, Ay2 = 6.00; Bx2 = 11.8, By2 = 6.80.

###### FREE Expert Solution

Vector dot product:

$\overline{)\stackrel{\mathbf{⇀}}{\mathbf{A}}{\mathbf{·}}\stackrel{\mathbf{⇀}}{\mathbf{B}}{\mathbf{=}}{\mathbf{|}}{\mathbf{A}}{\mathbf{|}}{\mathbf{|}}{\mathbf{B}}{\mathbf{|}}{\mathbf{c}}{\mathbf{o}}{\mathbf{s}}{\mathbf{\theta }}}$

Vector magnitude:

$\overline{)\mathbf{|}\stackrel{\mathbf{⇀}}{\mathbit{A}}\mathbf{|}{\mathbf{=}}\sqrt{{{\mathbit{A}}_{\mathbit{x}}}^{\mathbf{2}}\mathbf{+}{{\mathbit{A}}_{\mathbit{y}}}^{\mathbf{2}}}}$

$\begin{array}{rcl}{\mathbf{\theta }}_{{\mathbf{A}}_{\mathbf{2}}{\mathbf{B}}_{\mathbf{2}}}& \mathbf{=}& {\mathbf{cos}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\frac{{\stackrel{\mathbf{⇀}}{\mathbf{A}}}_{\mathbf{2}}\mathbf{·}{\stackrel{\mathbf{⇀}}{\mathbf{B}}}_{\mathbf{2}}}{\mathbf{|}{\stackrel{\mathbf{⇀}}{\mathbf{A}}}_{\mathbf{2}}\mathbf{|}\mathbf{|}{\stackrel{\mathbf{⇀}}{\mathbf{B}}}_{\mathbf{2}}\mathbf{|}}\mathbf{\right)}\end{array}$ ###### Problem Details

Find the angle between each of the following pairs of vectors =Ax+ Ayj and B =Bx+ Byj

Ax2= 3.20, Ay2 = 6.00; Bx2 = 11.8, By2 = 6.80.