Find the angle between vector u=2i+sqrt(11)j and v=-3i-2j to the nearest degree

Answer: Last option 155°Step-by-step explanation:We have the components of the vectors u and v.Then, to find the angle between them, perform the following steps:1) Calculate the scalar product u * vIf [tex]u = 2i + \sqrt{11}j[/tex]  and [tex]v = -3i-2j[/tex]Then, the product scalar u * v is:[tex]u * v = (2)(-3) + (\sqrt{11})(-2)[/tex][tex]u * v = -6-2\sqrt{11}[/tex][tex]u * v = -12.633[/tex]2) Calculation of the magnitude of both vectors.[tex]| u | = \sqrt{(2) ^ 2 + (\sqrt{11})^2}\\\\| u | = \sqrt{4+11}\\\\| u | = \sqrt{15}[/tex][tex]| v | = \sqrt{(- 3) ^ 2 + (- 2) ^ 2}\\\\| v | = \sqrt{9 +4}\\\\| v | = \sqrt{13}[/tex]3) Now that you know the product point between the two vectors and the magnitude of each, then use the following formula to find an angle[tex]u * v = | u || v |cos(\alpha)[/tex][tex]-12.633 = \sqrt{15}\sqrt{13}*cos(\alpha)\\\\cos(\alpha) = \frac{-12.633}{\sqrt{15}\sqrt{13}}\\\\arcos(\frac{-12.633}{\sqrt{15}\sqrt{13}}) = \alpha\\\\\alpha =155\°[/tex]