Answer:
[tex]\theta=101.3\degree[/tex] to the nearest tenth
Step-by-step explanation:
The given vectors are;
u = – 3i + 3j and v = 3i + 2j
We use the dot product to find the angle between the two vectors.
[tex]u\bullet v=|u| |v|\cos \theta[/tex]
[tex](-3i+3j)\bullet (3i+2j)=|-3i+3j| |3i+3j|\cos \theta[/tex]
[tex](-3i+3j)\bullet (3i+2j)=\sqrt{(-3)^2+3^2} \sqrt{3^2+2^2}\cos \theta[/tex]
[tex]-9+6=3\sqrt{2} \sqrt{13}\cos \theta[/tex]
[tex]-3=3\sqrt{2} \sqrt{13}\cos \theta[/tex]
[tex]-1=\sqrt{26} \cos \theta[/tex]
[tex]-\frac{1}{\sqrt{26}}= \cos \theta[/tex]
[tex]\cos^{-1}(-\frac{1}{\sqrt{26}})= \theta[/tex]
[tex]\theta=101.3\degree[/tex]
