Fortsettelse kapittel 1

Skalar og vektorfelt

Skalarprodukt

Skalarproduktet (prikk-produkt, dot product) mellom to vektorer gir en skalar

$ \begin{align*} \vec{u} \cdot \vec{v} &= u_x v_x + u_y v_y + u_z v_z \\ \begin{pmatrix}{\red{u_x} \\ \green{u_y} \\ \blue{u_z}}\end{pmatrix} \cdot \begin{pmatrix}{\red{v_x} \\ \green{v_y} \\ \blue{v_z}}\end{pmatrix} &= \red{u_x v_x} + \green{u_y v_y} + \blue{u_z v_z} \\ \end{align*} $

Rigorøst skalarprodukt

$\vec{u} \cdot \vec{v} = (\red{u_x \mathbf{i}} + \green{u_y \mathbf{j}} +\blue{u_z \mathbf{k}}) \cdot (\red{v_x \mathbf{i}} + \green{v_y \mathbf{j}} +\blue{v_z \mathbf{k}})$
$ \require{cancel} \begin{split} &= \red{u_x v_x \mathbf{i} \cdot \mathbf{i}} + \red{u_x} \green{v_y} \cancel{\red{\mathbf{i}} \cdot \green{\mathbf{j}}} + \red{u_x} \blue{v_z} \cancel{\red{\mathbf{i}} \cdot \blue{\mathbf{k}}} \\ &+ \green{u_y} \red{v_x} \cancel{\green{\mathbf{j}} \cdot \red{\mathbf{i}}} + \green{u_y v_y \mathbf{j} \cdot \mathbf{j}} + \green{u_y} \blue{v_z} \cancel{\green{\mathbf{j}} \cdot \blue{\mathbf{k}}} \\ &+ \blue{u_z} \red{v_x} \cancel{\blue{\mathbf{k}} \cdot \red{\mathbf{i}}} + \blue{u_z} \green{v_y} \cancel{\blue{\mathbf{k}} \cdot \green{\mathbf{j}}} + \blue{u_z v_z \mathbf{k} \cdot \mathbf{k}} \\ &= \red{u_x v_x} + \green{u_y v_y} + \blue{u_z v_z} \end{split} $

Lengden av en vektor

Pythagoras

$ \begin{align*} |\vec{u}| &= \sqrt{\vec{u} \cdot \vec{u}}\\ &= \sqrt{u_x^2 + u_y^2 + u_z^2} \end{align*} $

Geometrisk tolkning $\mathbf{a} \cdot \mathbf{b}$

Lengden av a ganger lengden av b i retning av a

Alternativ geometrisk tolkning $\mathbf{a} \cdot \mathbf{b}$

Lengden av b ganger lengden av a i retning av b

Likningen for et plan
Planet til høyre har a som normalvektor.
VC oppgir at likningen for planet er
$\mathbf{r} \cdot \mathbf{a} = \text{konstant}$
Hvorfor?

$\mathbf{r} \cdot \mathbf{a}$ er lengden av a ganger lengden av r i retning av a (p i figuren). p er den samme for alle posisjonsvektorer i planet så derfor er $\mathbf{r} \cdot \mathbf{a}$ konstant.

Kryss-produkt

 $\vec{w} = \vec{u} \times \vec{v}$

$|\vec{w}|=|\vec{u}||\vec{v}|\sin \alpha$

Mer om kryss-produkt
Høyrehåndsregel $ \begin{align*} \red{\mathbf{i}} \times \green{\mathbf{j}} &= \blue{\mathbf{k}}\\ \blue{\mathbf{k}} \times \red{\mathbf{i}} &= \green{\mathbf{j}}\\ \green{\mathbf{j}} \times \blue{\mathbf{k}} &= \red{\mathbf{i}} \\ \red{{\mathbf{i}}} \times \red{\mathbf{i}} &= 0 \\ \green{{\mathbf{j}}} \times \green{\mathbf{j}} &= 0 \\ \blue{{\mathbf{k}}} \times \blue{\mathbf{k}} &= 0 \end{align*} $

Utledning av kryss-produkt

$ \vec{u} \times \vec{v} = (\red{u_x \mathbf{i}} + \green{u_y \mathbf{j}} +\blue{u_z \mathbf{k}}) \times (\red{v_x \mathbf{i}} + \green{v_y \mathbf{j}} +\blue{v_z \mathbf{k}}) $

$ \require{cancel} \begin{split} &= \red{u_x v_x \cancel{\mathbf{i} \times \mathbf{i}}} + \red{u_x} \green{v_y} {\red{\mathbf{i}} \times \green{\mathbf{j}}} + \red{u_x} \blue{v_z} \red{\mathbf{i}} \times \blue{\mathbf{k}} \\ &+ \green{u_y} \red{v_x} \green{\mathbf{j}} \times \red{\mathbf{i}} + \green{u_y v_y \cancel{\mathbf{j} \times \mathbf{j}}} + \green{u_y} \blue{v_z} \green{\mathbf{j}} \times \blue{\mathbf{k}} \\ &+ \blue{u_z} \red{v_x} \blue{\mathbf{k}} \times \red{\mathbf{i}} + \blue{u_z} \green{v_y} \blue{\mathbf{k}} \times \green{\mathbf{j}} + \blue{u_z v_z \cancel{\mathbf{k} \times \mathbf{k}}} \\ \end{split} $

Bruker at kryss-produktet er distribuert ved addisjon
$\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$

$ \require{cancel} \begin{split} &= \red{u_x v_x \cancel{\mathbf{i} \times \mathbf{i}}} + \red{u_x} \green{v_y} {\red{\mathbf{i}} \times \green{\mathbf{j}}} + \red{u_x} \blue{v_z} \red{\mathbf{i}} \times \blue{\mathbf{k}} \\ &+ \green{u_y} \red{v_x} \green{\mathbf{j}} \times \red{\mathbf{i}} + \green{u_y v_y \cancel{\mathbf{j} \times \mathbf{j}}} + \green{u_y} \blue{v_z} \boxed{\green{\mathbf{j}} \times \blue{\mathbf{k}}} \\ &+ \blue{u_z} \red{v_x} \blue{\mathbf{k}} \times \red{\mathbf{i}} + \blue{u_z} \green{v_y} \boxed{\blue{\mathbf{k}} \times \green{\mathbf{j}}} + \blue{u_z v_z \cancel{\mathbf{k} \times \mathbf{k}}} \\ \end{split} $

$ \green{\mathbf{j}} \times \blue{\mathbf{k}} = \red{\mathbf{i}}, \quad \blue{\mathbf{k}} \times \green{\mathbf{j}} = -\red{\mathbf{i}} $

$ \rightarrow (\green{u_y} \blue{v_z} - \blue{u_z} \green{v_y}) \, \red{\mathbf{i}} $
$ \require{cancel} \begin{split} &= \red{u_x v_x \cancel{\mathbf{i} \times \mathbf{i}}} + \red{u_x} \green{v_y} {\red{\mathbf{i}} \times \green{\mathbf{j}}} + \red{u_x} \blue{v_z} \boxed{\red{\mathbf{i}} \times \blue{\mathbf{k}}} \\ &+ \green{u_y} \red{v_x} \green{\mathbf{j}} \times \red{\mathbf{i}} + \green{u_y v_y \cancel{\mathbf{j} \times \mathbf{j}}} + \green{u_y} \blue{v_z} {\green{\mathbf{j}} \times \blue{\mathbf{k}}} \\ &+ \blue{u_z} \red{v_x} \boxed{\blue{\mathbf{k}} \times \red{\mathbf{i}}} + \blue{u_z} \green{v_y} {\blue{\mathbf{k}} \times \green{\mathbf{j}}} + \blue{u_z v_z \cancel{\mathbf{k} \times \mathbf{k}}} \\ \end{split} $

$ \red{\mathbf{i}} \times \blue{\mathbf{k}} = \green{\mathbf{j}}, \quad \blue{\mathbf{k}} \times \red{\mathbf{i}} = -\green{\mathbf{j}} $

$ \rightarrow (\red{u_x} \blue{v_z} - \blue{u_z} \red{v_x}) \, \green{\mathbf{j}} $
$ \begin{split} \vec{u} \times \vec{v} &= (\green{u_y} \blue{v_z} - \blue{u_z} \green{v_y})\, \red{\mathbf{i}} \\ &+ (\blue{u_z} \red{v_x} - \red{u_x}\blue{v_z} )\, \green{\mathbf{j}} \\ &+ (\red{u_x}\green{v_y} - \green{u_y \red{v_x}}) \, \blue{\mathbf{k}} \end{split} $

Huskeregel

Ta determinanten av følgende matrise

$ \begin{align*} \vec{u} \times \vec{v} &= \begin{vmatrix} \red{\mathbf{i}} & \green{\mathbf{j}} & \blue{\mathbf{k}} \\ \red{u_x} & \green{u_y} & \blue{u_z} \\ \red{v_x} & \green{v_y} & \blue{v_z} \end{vmatrix} \\ &= (\green{u_y} \blue{v_z} - \blue{u_z} \green{v_y})\, \red{\mathbf{i}} \\ &+ (\blue{u_z} \red{v_x} - \red{u_x} \blue{v_z} )\, \green{\mathbf{j}} \\ &+ (\red{u_x} \green{v_y} - \green{u_y} \red{v_x})\, \blue{\mathbf{k}} \end{align*} $

Normalvektor til plan

To vilkårlige vektorer i $\mathbb{R}^3$ spenner ut et plan

Sammendrag av forrige slide

  • To vektorer $\vec{u}$ og $\vec{v}$ i $\mathbb{R}^3$ spenner ut et plan
  • Kryssproduktet gir en ny vektor $\vec{a}$ med retning normal til planet
    $\vec{a} = \vec{u} \times \vec{v}$
  • En enhetsvektor $\vec{n}_a$ med lengde 1 og samme retning som $\vec{a}$ skapes med
    $\vec{n}_a = \frac{\vec{a}}{|\vec{a}|}$

Derivasjon av vektorer

Også komponentvis

$ \begin{split} \frac{d \vec{u}}{dt} &= \frac{d}{dt}\left(\red{u_x \mathbf{i}} + \green{u_y \mathbf{j}} +\blue{u_z \mathbf{k}} \right) \\ &= \frac{d \red{u_x}}{dt} \red{\mathbf{i}} + \frac{d \green{u_y}}{dt} \green{\mathbf{j}} + \frac{d \blue{u_z}}{dt} \blue{\mathbf{k}} \end{split} $

Siden enhetsvektorene er uavhengige av tid!

Liten repetisjon av derivasjon

 $f'(x)=\frac{d f(x)}{dx} \underset{\Delta x \to 0}{\approx} \frac{f(x+\Delta x) - f(x)}{\Delta x}$

Derivasjon av posisjonsvektor

 $\vec{v} = \frac{d \vec{r}(t)}{dt} \underset{\Delta t \to 0}{\approx} \frac{\vec{r}(t+\Delta t) - \vec{r}(t)}{\Delta t}$

Øvingsoppgaver i Geogebra

https://geogebra.org/groups

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