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I would like to know if my way of solution is good.

I have to find the angle between the vector $[2,2]$ and $[-1,2]$

$$\theta = \arccos{\left( \frac{[2,2]\cdot[-1,2]}{\sqrt{8} \cdot \sqrt{5}}\right)} = \arccos{\left( \frac{2}{\sqrt{8}\cdot\sqrt{5}}\right)} = \arccos{\left(\frac{2\sqrt{5}}{5\sqrt{8}}\right)} \approx 71.57^{\circ}$$

thanks in advance

  • 0
    The algebraic-geometry tag is inappropriate for this question.2012-11-01

1 Answers 1