You have to use the following formula to calculate the angle between TWO vectors:
[tex]\cos \left(\theta \right)\:=\frac{\vec{a\:}\cdot \vec{b\:}}{\left|\vec{a\:}\right|\cdot \left|\vec{b\:}\right|}[/tex]
In this case the vector u will be the vector a, and the vector v will be the vector b
To replace the formula, we have to know the dot product:
In this case a * b
Multiply each i and each j
[tex]\vec{a\:}\cdot\vec{b\:}=\text{ \lparen}ai*bi)+(aj*bj)[/tex][tex]\vec{a\:}\cdot\vec{b\:}=\text{ \lparen}3*-7)+(4*5)=-21+20=-1[/tex]
Now
[tex]|a|=\sqrt{ai^2+aj^2}=\sqrt{3^2+4^2}=\sqrt{25}=5[/tex][tex]|b|=\sqrt{bi^2+bj^2}=\sqrt{(-7)^2+5^2}=\sqrt{74}[/tex]
Now replace in
[tex]\cos(\theta)=\frac{\vec{a}\vec{b}}{\lvert\vec{a}\rvert\lvert\vec{b}\rvert}[/tex][tex]\cos \left(θ\right)=-\frac{1}{5\sqrt{74}}[/tex]
Clear Cos with ArcCos
[tex]θ=\arccos\left(\cos\left(θ\right)\right)=\arccos\left(-\frac{1}{5\sqrt{74}}\right)[/tex]
ANS:
[tex]θ=\text{ 91.33221985\degree}[/tex]
