Examples: Scalar product

2.14.1 Examples: Scalar product

Let

\begin{array}{cc} \vec{a} = (\begin{matrix} a_{1} \\ ⋮ \\ a_{n} \end{matrix}) \in R & \vec{b} = (\begin{matrix} b_{1} \\ ⋮ \\ b_{n} \end{matrix}) \in R \end{array}

be two vectors with real components:

Definition 29 $\vec{a} \cdot \vec{b} = a_{1} b_{1} + \dots + a_{N} b_{N} = \sum_{j = 1}^{k} a_{j} b_{j}$ is called scalar product of $\vec{a}$ and $\vec{b}$
Note: $\vec{a}, \vec{b} \in R^{N}, \vec{a} \cdot \vec{b} \in R$ !!

Example:

\begin{array}{lccccccccl} \vec{a} & = & (\begin{matrix} 1 \\ 2 \\ 3 \\ 4 \end{matrix}), & \vec{b} & = & (\begin{matrix} 5 \\ 6 \\ 7 \\ 8 \end{matrix}) & \Rightarrow & \vec{a} \cdot \vec{b} & = & 5 + 12 + 21 + 32 \\ = & 70 \\ \vec{a} & = & (\begin{matrix} 1 \\ - 1 \\ 1 \end{matrix}), & \vec{b} & = & (\begin{matrix} 1 \\ 1 \\ 0 \end{matrix}) & \Rightarrow & \vec{a} \cdot \vec{b} & = & 1 - 1 + 0 = 0 \end{array}

Example from physics:
Work:

W = \vec{F} \cdot \vec{s} = | \vec{F} | | \vec{s} | \cdot \cos φ

\begin{array}{lcl} \cos ∠ (\vec{a}, \vec{b}) & = & \frac{\vec{a} \cdot \vec{b}}{| \vec{a} | \cdot | \vec{b} |} \\ \vec{a} \cdot \vec{b} & = & 0 if \vec{a} = \vec{0} or \vec{b} = \vec{0} or \vec{a} ⊥ \vec{b} \end{array}} in 3D

now definitions for $\vec{a}, \vec{b} \in R^{N}$

$\vec{a}, \vec{b} \neq \vec{0}$ . If $\vec{a} \cdot \vec{b} = 0$ then $\vec{a}$ is called orthogonal or perpendicular to $\vec{b}$ or $\vec{a} ⊥ \vec{b}$
$\frac{\vec{a} \cdot \vec{b}}{| \vec{a} | | \vec{b} |} \leq 1$ . We define $\cos φ = \frac{\vec{a} \cdot \vec{b}}{| \vec{a} | | \vec{b} |}$ where $φ$ is the angle between the two vectors $\vec{a}$ and $\vec{b}$ in $R$

Example:

$\vec{a} = (\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}) \vec{b} = (\begin{matrix} 2 \\ 2 \\ 0 \end{matrix}) \Rightarrow \cos φ = \frac{2 + 2 + 0}{\sqrt{3} \sqrt{8}} = \sqrt{\frac{2}{3}} \to φ = 0.615 = {35.26}^{\circ}$
$\begin{array}{rcl} \vec{a} & = & \begin{array}{crcc} (\begin{matrix} 1 \\ 1 \\ 1 \\ 1 \end{matrix}) & \vec{b} & = & (\begin{matrix} 2 \\ 2 \\ 0 \\ 0 \end{matrix}) \end{array} \Rightarrow \cos φ = \frac{2 + 2 + 0 + 0}{\sqrt{4} \sqrt{8}} = \frac{1}{\sqrt{2}} \Rightarrow φ = \frac{π}{4} = 45^{\circ} \\ \vec{a} & = & \begin{array}{crcc} (\begin{matrix} 1 \\ 1 \\ 1 \\ 1 \end{matrix}) & \vec{b} & = & (\begin{matrix} 1 \\ - 1 \\ - 1 \\ 1 \end{matrix}) \end{array} \Rightarrow \vec{a} \cdot \vec{b} = 1 - 1 - 1 + 1 = 0 \to \vec{a} ⊥ \vec{b} \end{array}$

With frame

Scalar product