We need to find the angle between the vectors:
[tex]\begin{gathered} v=i+j \\ \\ w=3i-3j \end{gathered}[/tex]
If those vectors form the sides of a triangle, then the third side is formed by the vector:
[tex]z=w-v=3i-3j-(i+j)=3i-i-3j-j=2i-4j[/tex]
Thus, using the Law of cosines, those vectors' magnitudes are related to the angle between w and v by the formula:
[tex]\begin{gathered} |z|²=|w|²+|v|²-2|w||v|\cos\alpha \\ \\ 2|w||v|\cos\alpha=|w|²+|v|²-|z|² \\ \\ \cos\alpha=\frac{|w|²+|v|²-|z|²}{2|w||v|} \\ \\ \alpha=\arccos\frac{|w|²+|v|²-|z|²}{2\lvert w\rvert\lvert v\rvert} \end{gathered}[/tex]
We have:
[tex]\begin{gathered} |w|=\sqrt{3²+(-3)²}=\sqrt{9+9}=\sqrt{2\cdot9}=3\sqrt{2} \\ \\ \lvert v\rvert=\sqrt{1²+1²}=\sqrt{1+1}=\sqrt{2} \\ \\ |z|=\sqrt{2²+(-4)²}=\sqrt{4+16}=\sqrt{20}=2\sqrt{5} \end{gathered}[/tex]
Therefore:
[tex]\begin{gathered} \alpha=\arccos\frac{18+2-20}{2\cdot3\sqrt{2}\cdot\sqrt{2}} \\ \\ \alpha=\arccos\frac{0}{12} \\ \\ \alpha=\arccos0 \\ \\ \alpha=90\degree \end{gathered}[/tex]
Answer: 90º
