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: $\vec{u} \cdot \vec{u} = | \vec{u} | | \vec{u} | \cos (\hat{u u}) = | \vec{u} |^{2}$

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

So, we obtain, using the coordinates of the vector $\vec{u} = (u_{1}, u_{2})$ , $\vec{u} = \sqrt{u_{1}^{2} + u_{2}^{2}}$

Example

For $\vec{u} = (3, 4)$ , we have that $| \vec{u} | = \sqrt{u_{1}^{2} + u_{2}^{2}} = \sqrt{3^{2} + 4^{2}} = \sqrt{25} = 5$

Angle between two vectors.

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

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

So, if we have two vectors $\vec{u} = (u_{1}, u_{2})$ and $\vec{v} = (v_{1}, v_{2})$ we have:

$ang (\vec{u}, \vec{v}) = ang (\vec{v}, \vec{u}) = \arccos (\frac{u_{1} v_{1} + u_{2} v_{2}}{\sqrt{u_{1}^{2} + u_{2}^{2}} \cdot \sqrt{v_{1}^{2} + v_{2}^{2}}})$

Example

Find the angle formed by $\vec{u} = (2, 3)$ and $\vec{v} = (- 1, 4)$ . In this case, applying the previous formula, we obtain: $ang (\vec{u}, \vec{v}) = \arccos (\frac{2 \cdot (- 1) + 3 \cdot 4}{\sqrt{2^{2} + 3^{2}} \cdot \sqrt{(- 1)^{2} + 4^{2}}}) = \arccos (0.67267) = 47^{\circ} 43^{'} 35^{″}$