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
- 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 $\vec{u}=\vec{0}$.
- 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 $\cos(-\alpha)=cos(\alpha)$.
- 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.
- The scalar product is a distributive with regard to the sum of vectors: $\vec{u}\cdot(\vec{v}+\vec{w})=\vec{u}\cdot\vec{v}+\vec{u}\cdot\vec{w}$.