Question

Use the Law of Cosines to find the angle between the vectors. (Assume 0° ≤ ≤ 180°.)

Answer 1

We need to find the angle between the vectors:

`\begin{gathered} v=i+j \\ \\ w=3i-3j \end{gathered}`

If those vectors form the sides of a triangle, then the third side is formed by the vector:

`z=w-v=3i-3j-(i+j)=3i-i-3j-j=2i-4j`

Thus, using the Law of cosines, those vectors' magnitudes are related to the angle between w and v by the formula:

`\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=(|w|²+|v|²-|z|²)/(2|w||v|) \\ \\ \alpha=\arccos(|w|²+|v|²-|z|²)/(2\lvert w\rvert\lvert v\rvert) \end{gathered}`

We have:

`\begin{gathered} |w|=√(3²+(-3)²)=√(9+9)=√(2\cdot9)=3√(2) \\ \\ \lvert v\rvert=√(1²+1²)=√(1+1)=√(2) \\ \\ |z|=√(2²+(-4)²)=√(4+16)=√(20)=2√(5) \end{gathered}`

Therefore:

`\begin{gathered} \alpha=\arccos(18+2-20)/(2\cdot3√(2)\cdot√(2)) \\ \\ \alpha=\arccos(0)/(12) \\ \\ \alpha=\arccos0 \\ \\ \alpha=90\degree \end{gathered}`

Answer: 90º