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 $$\vec{v}$$. $$$\vec{u}\cdot\vec{v}=|\vec{u}||\vec{v}|\cos(\widehat{uv})$$$

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 $$\cos(\widehat{uv})=\cos(90^\circ)=0$$, we have $$\vec{u}\cdot\vec{v}=0$$.

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

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

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 $$\vec{u}\cdot\vec{v}=|\vec{u}||\vec{v}|\cos(\widehat{uv})$$, by replacing the information that we have, we will obtain: $$$\cos(\widehat{uv})=0 \Rightarrow \widehat{uv}=90^\circ $$$

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: $$$\vec{u}\cdot\vec{v}=u_1 v_1+u_2 v_2$$$

If $$\vec{u}=(3,1)$$ and $$\vec{v}=(2,-1)$$, then: $$$\vec{u}\cdot\vec{v}=3\cdot2+1\cdot(-1)=6-1=5$$$

Properties of the scalar product

  1. The scalar product of a vector and itself is a positive real number: $$ \vec{u}\cdot\vec{u} \geqslant 0$$. If $$\vec{u}\cdot\vec{u}=0$$, then $$\vec{u}=\vec{0}$$.
  2. 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 $$\alpha$$ and the angle formed by $$\vec{v}$$ and $$\vec{u}$$ is $$-\alpha$$, and we know that $$\cos(-\alpha)=cos(\alpha)$$.
  3. The scalar product is pseudoassociative: $$\alpha(\vec{u}\cdot\vec{v})= (\alpha\vec{u})\cdot\vec{v}=\vec{u}\cdot(\alpha\vec{v})$$ where $$\alpha$$ is a real number.
  4. The scalar product is a distributive with regard to the sum of vectors: $$\vec{u}\cdot(\vec{v}+\vec{w})=\vec{u}\cdot\vec{v}+\vec{u}\cdot\vec{w}$$.