Definition, analytical expression and properties of scalar product

The scalar product between two vectors $\vec{u}$ and $\vec{v}$ , that is represented by $\vec{u} \cdot \vec{v}$ , is a real number that is obtained by multiplying the magnitude of $\vec{u}$ by the magnitude of $\vec{v}$ and by the cosine of the angle that is formed by $\vec{u}$ and $\vec{v}$ . $\vec{u} \cdot \vec{v} = | \vec{u} | | \vec{v} | \cos (\hat{u v})$

From the definition of the scalar product we have:

If $\vec{u} = \vec{0}$ or $\vec{v} = \vec{0}$ , then $\vec{u} \cdot \vec{v} = 0$ .
If $\vec{u}$ and $\vec{v}$ are perpendicular vectors and since $\cos (\hat{u v}) = \cos (90^{\circ}) = 0$ , we have $\vec{u} \cdot \vec{v} = 0$ .

Example

If $\vec{u} = (0, 2)$ , $\vec{v} = (3, 3)$ and $\hat{u v} = 45^{\circ}$ :

$\vec{u} \cdot \vec{v} = | \vec{u} | | \vec{v} | \cos (45^{\circ}) = 2 \cdot \sqrt{18} \frac{\sqrt{2}}{2} = \sqrt{36} = 6$

Example

If $| \vec{u} | = 3$ , $| \vec{v} | = 2$ and $\vec{u} \cdot \vec{v} = 0$ . What angle is formed by $\vec{u}$ and $\vec{v}$ ?

Since the formula of the scalar product is $\vec{u} \cdot \vec{v} = | \vec{u} | | \vec{v} | \cos (\hat{u v})$ , by replacing the information that we have, we will obtain: $\cos (\hat{u v}) = 0 \Rightarrow \hat{u v} = 90^{\circ}$

These two vectors are perpendicular.

Analytical expression of the scalar product:

Given $\vec{u} = (u_{1}, u_{2})$ and $\vec{v} = (v_{1}, v_{2})$ , its scalar product can be written as: $\vec{u} \cdot \vec{v} = u_{1} v_{1} + u_{2} v_{2}$

Example

If $\vec{u} = (3, 1)$ and $\vec{v} = (2, - 1)$ , then: $\vec{u} \cdot \vec{v} = 3 \cdot 2 + 1 \cdot (- 1) = 6 - 1 = 5$

Properties of the scalar product

The scalar product of a vector and itself is a positive real number: $\vec{u} \cdot \vec{u} ⩾ 0$ . If $\vec{u} \cdot \vec{u} = 0$ , then $\vec{u} = \vec{0}$ .
The scalar product is commutative: $\vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u}$ . Since the angle formed by $\vec{u}$ and $\vec{v}$ is $α$ and the angle formed by $\vec{v}$ and $\vec{u}$ is $- α$ , and we know that $\cos (- α) = c o s (α)$ .
The scalar product is pseudoassociative: $α (\vec{u} \cdot \vec{v}) = (α \vec{u}) \cdot \vec{v} = \vec{u} \cdot (α \vec{v})$ where $α$ is a real number.
The scalar product is a distributive with regard to the sum of vectors: $\vec{u} \cdot (\vec{v} + \vec{w}) = \vec{u} \cdot \vec{v} + \vec{u} \cdot \vec{w}$ .