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The question is as follows:

Find the angle formed by u and v, given that u = [-1, 0, 1] and v = [0, 2, -2].

I thought about using a formula related to the Law of Cosines. $$\cos(C) = \frac{a^2 + b^2 - c^2}{ab} $$

The variable $a$ would be the magnitude of vector $u$ and the variable $b$ would be the magnitude of vector $v$. I got the values of 1 and $2\sqrt{2}$, respectively.

I remember that the third side of a triangle can be found by subtracting the two vectors, so I thought of doing $u - v$ (which would be denoted as side $c$). I did that and got $|u - v| = \sqrt{13}$.

After substituting the values to its corresponding places in the Law of Cosines formula, I got: $$\cos(C) = \frac{1^2 + (2\sqrt{2})^2 - (\sqrt{13})^2}{1(2\sqrt{2})}$$ The value that I get does not allow for me to find the $\arccos$ of it--I get an error message on my calculator.

I want to know what I am doing wrong, therefore, any help will be greatly appreciated!

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1 Answer

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by definition $$\cos\theta=<u,v>/(|u| |v|)$$

in this case you obtain $$\cos\theta=-1/\sqrt2$$

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