# Find the angle between two vectors

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

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

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

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The algebraic-geometry tag is inappropriate for this question. –  littleO Nov 1 '12 at 20:31

Yes, you are correctly using the standard way of computing the angle between two vectors, $\vec{u},\vec{v}\in\mathbb{R}^{2}$ through their inner product:
$$\theta=\arccos{\left(\frac{\vec{u}\cdot\vec{v}}{\|\vec{u}\|\|\vec{v}\|}\right)}$$
Which given your vectors $\vec{u}=[2,2]$ and $\vec{v}=[-1,2]$, gives us:
$$\theta=\arccos{\left(\frac{4-2}{\sqrt{2^{2}+2^{2}}\cdot\sqrt{(-1)^{2}+2^{2}}}\right)}\approx 71.57^{\circ}$$