Applications of the scalar product

Magnitude of a vector.

The scalar product can be used to determine the length of a vector $$\vec{u}$$ since: $$$\vec{u}\cdot\vec{u}=|\vec{u}||\vec{u}|\cos(\widehat{uu})=|\vec{u}|^2$$$

from which: $$\vec{u}=\sqrt{\vec{u}\cdot\vec{u}}$$

So, we obtain, using the coordinates of the vector $$\vec{u}=(u_1,u_2)$$, $$$ \vec{u}=\sqrt{u_1^2+u_2^2}$$$

Angle between two vectors.

From the definition of the scalar product, $$\vec{u}\cdot\vec{u}=|\vec{u}||\vec{u}|\cos(\widehat{uu})$$ we can convert the cosine to another value: $$$\cos(\widehat{uv})=\dfrac{\vec{u}\cdot\vec{v}}{|\vec{u}||\vec{v}|}$$$

Applying the function arcosine to both sides of the equality we obtain (ang=angle): $$$ \text{ang}(\vec{u},\vec{v})=\arccos\Big(\dfrac{\vec{u}\cdot\vec{v}}{|\vec{u}||\vec{v}|}\Big)$$$

So, if we have two vectors $$\vec{u}=(u_1,u_2)$$ and $$\vec{v}=(v_1,v_2)$$ we have:

$$$ \text{ang}(\vec{u},\vec{v})=\text{ang}(\vec{v},\vec{u})=\arccos\Big(\dfrac{u_1 v_1+u_2 v_2}{\sqrt{u_1^2+u_2^2}\cdot\sqrt{v_1^2+v_2^2}}\Big)$$$