Vector question. Let v=<-2,1> u=<3,-5> , find:1)v * u2)2)Find the angle formed between v and u

Answer:1. )[tex]\vec {v}.\vec{u}= -11[/tex]2.) The angle between v and u is 147.52°.Step-by-step explanation:Given:[tex]\vec {v}= ( -2 , 1 )\\\vec {u}= ( 3 , -5 )[/tex]To Find:1. [tex]\vec {v}.\vec{u}= ?[/tex]2.[tex]\theta = ?[/tex]Solution:[tex]\vec {v}.\vec{u}[/tex] is scalar product given as,[tex]\vec {v}.\vec{u}=|\vec {v}||\vec {u}|\cos \theta[/tex][tex]|\vec {v}|=|(-2, 1)|=\sqrt{(-2)^{2} +1^{2}}=\sqrt{5}\\|\vec {u}|=|(3, 5)|=\sqrt{(3)^{2} +(-5)^{2}}=\sqrt{34}[/tex][tex]\vec {v}.\vec{u}=(-2i +j).(3i-5j)\\[/tex]Here only i.i = j.j =1 and i.j = j.i = 0∴ [tex]\vec {v}.\vec{u}=(-2\times 3 +1\times -5)\\\vec {v}.\vec{u}=-6-5=-11 \\[/tex]Now, Substituting the above values we get[tex]-11=\sqrt{5}\times \sqrt{34}\cos \theta\\ \cos \theta=\frac{-11}{\sqrt{170}} \\ \cos \theta =-0.84366\\\therefore \theta =cos^{-1}(-0.84366)\\\therefore \theta =147.52\°[/tex]As it is negative mean [tex]\theta[/tex] is in Second Quadrant Because Cosine is negative in Second Quadrant.1. )[tex]\vec {v}.\vec{u}= -11[/tex]2.) The angle between v and u is 147.52°.