Tweet
Login
Mathematics Crystal
You may switch between
tex
and
pdf
by changing the end of the URL.
Home
About Us
Materials
Site Map
Questions and Answers
Skills
Topic Notes
HSC
Integration
Others
Tangent
UBC
UNSW
Calculus Advanced
Challenges
Complex Numbers
Conics
Differentiation
Integration
Linear Algebra
Mathematical Induction
Motion
Others
Polynomial Functions
Probability
Sequences and Series
Trigonometry
/
Topics /
Linear Algebra /
Vector.tex
--Quick Links--
The Number Empire
Wolfram Mathematica online integrator
FooPlot
Calc Matthen
Walter Zorn
Quick Math
Lists of integrals
List of integrals of trigonometric functions
PDF
\documentclass[10pt]{article} \usepackage{amssymb,amsmath} \usepackage[hmargin=1cm,vmargin=0.5cm]{geometry} \begin{document} {\large Vectors} \begin{align*} \text{\bf Basics:}\quad& \text{In this context, an $n$-dimensional vector is represented as }\mathbf{A}=(a_1,a_2,\ldots,a_n).\\ &\mathbf{U=V}\quad\Leftrightarrow\quad u_r=v_r\quad\forall\:r\in\{\mathbb{N}:r\leq n\}.\qquad \mathbf{0}=(0,0,\ldots,0).\qquad \mathbf{-U}=(-u_1,-u_2,\ldots,-u_n).\\ &\mathbf{U+V}=(u_1+v_1,u_2+v_2,\ldots,u_n+v_n).\qquad \mathbf{U-V=U+(-V).}\qquad k\mathbf{U}=(ku_1,ku_2,\ldots,ku_n).\\ &\mathbf{U+(V+W)=(U+V)+W}.\quad \mathbf{U+V=V+U}.\quad \mathbf{U+0=0+U=U}.\quad \mathbf{U+(-U)=(-U)+U=0}.\\ &c(\mathbf{U+V})=c\mathbf{U}+c\mathbf{V}.\qquad (c+d)\mathbf{U}=c\mathbf{U}+d\mathbf{U}.\qquad (cd)\mathbf{U}=c(d\mathbf{U}).\qquad 1\mathbf{U}=\mathbf{U}.\\ \\ \text{\bf Presentation:}\quad& \text{In 3-dimension,}\quad\mathbf{i}=(1,0,0),\quad\mathbf{j}=(0,1,0),\quad\text{and}\quad\mathbf{k}=(0,0,1).\\ &(a,b,c)=(a,0,0)+(0,b,0)+(0,0,c)=a(1,0,0)+b(0,1,0)+c(0,0,1)=a\mathbf{i}+b\mathbf{j}+c\mathbf{k}.\\ \\ \text{\bf Length }&\text{\bf (magnitude)}:\quad|a\mathbf{i}+b\mathbf{j}+c\mathbf{k}|=\sqrt{a^2+b^2+c^2}.\\ &|c\mathbf{U}|=|c||\mathbf{U}|.\quad|\mathbf{U+V}|\leq|\mathbf{U}|+|\mathbf{V}|\quad\text{(The Triangle Inequality).}\\ \\ \text{\bf Dot product:}\quad& \mathbf{U\cdot V}=\sum_{r=1}^n u_r v_r\:.\\ &\mathbf{U\cdot U}=|\mathbf{U}|^2.\qquad \text{Proof:}\quad\mathbf{U\cdot U}=u_1^2+u_2^2+u_3^2=\left(\sqrt{u_1^2+u_2^2+u_3^2}\right)^2=|\mathbf{U}|^2.\\ &\because|\mathbf{U-V}|^2=|\mathbf{U}|^2+|\mathbf{V}|^2-2\mathbf{|U||V|}\cos\theta,\quad \text{where $\theta$ is the angle between $\mathbf{U}$ and $\mathbf{V}$,}\\ &\sum_{r=1}^n(u_r-v_r)^2=\sum_{r=1}^n u_r^2-2\sum_{r=1}^n u_r v_r+\sum_{r=1}^n v_r^2 =\sum_{r=1}^n u_r^2+\sum_{r=1}^n v_r^2-2\mathbf{|U||V|}\cos\theta,\quad -2\sum_{r=1}^n u_r v_r=-2\mathbf{|U||V|}\cos\theta,\\ &\therefore\mathbf{U\cdot V}=\mathbf{|U||V|}\cos\theta.\qquad \theta=\cos^{-1}\left(\frac{\mathbf{U\cdot V}}{\mathbf{|U||V|}}\right).\qquad \mathbf{|U\cdot V|}\leq\mathbf{|U||V|}\quad\text{(The Cauchy-Schwarz Inequality).}\\ &\text{Orthogonal:}\quad\theta=\tfrac{\pi}{2}\:\Leftrightarrow\:\mathbf{U\cdot V}=0.\qquad \text{($\mathbf{i}, \mathbf{j}$ and $\mathbf{k}$ are orthogonal to each other.)}\\ &\text{Parallel:}\quad\theta=0\:\Leftrightarrow\:\mathbf{|U\cdot V|=|U||V|}.\\ &\text{Projection of $\mathbf{U}$ in the direction of $\mathbf{V}$:}\quad \text{proj}_\mathbf{v}\mathbf{U}=\mathbf{\frac{U\cdot V}{V\cdot V}{V}}.\qquad \mathbf{U}-\text{proj}_\mathbf{v}\mathbf{U}\text{ is orthogonal to }\mathbf{V}.\\ &\qquad\qquad\text{For a unit vector $\hat{\mathbf{v}}$, proj$_\mathbf{\hat{v}}\mathbf{U}=\mathbf{({U\cdot V}){V}}$}.\\ &\text{Consider }\mathbf{U\cdot i}\text{, the Direction Cosine of the $x$-axis is } \cos\alpha=\mathbf{\frac{U\cdot i}{|U||i|}}=\frac{a}{\mathbf{|U|}}.\\ &\text{Likewise, }\cos\beta=\mathbf{\frac{U\cdot j}{|U||j|}}=\frac{b}{\mathbf{|U|}}\text{ for the $y$-axis, and } \cos\gamma=\mathbf{\frac{U\cdot k}{|U||k|}}=\frac{c}{\mathbf{|U|}}\text{ for the $z$-axis.}\\ \\ \text{\bf Cross product:}\quad& \mathbf{U\times V}=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\u_1&u_2&u_3\\v_1&v_2&v_3\end{vmatrix}\quad \text{(expressed in 3-dimensional as an example.)}\\ &=\begin{vmatrix}u_2&u_3\\v_2&v_3\end{vmatrix}\mathbf{i} +\begin{vmatrix}u_3&u_1\\v_3&v_1\end{vmatrix}\mathbf{j} +\begin{vmatrix}u_1&u_2\\v_1&v_2\end{vmatrix}\mathbf{k} =(u_2 v_3-u_3 v_2)\mathbf{i}+(u_3 v_1-u_1 v_3)\mathbf{j}+(u_1 v_2-u_2 v_1)\mathbf{k}.\\ % &\mathbf{U\times V}\text{ is orthogonal to both $\mathbf{U}$ and $\mathbf{V}$.}\\ &\text{Proof:}\quad\mathbf{(U\times V)\cdot U}=(u_2 v_3-u_3 v_2)u_1+(u_3 v_1-u_1 v_3)u_2+(u_1 v_2-u_2 v_1)u_3=0.\quad \text{Likewise, }\mathbf{(U\times V)\cdot V}=0.\\ &\text{Note:}\quad\mathbf{i\times j}=(0-0,0-0,1-0)=\mathbf{k}.\quad \mathbf{U\times V}\text{ follows the xyz axis orientation, forming a right-hand system.}\\ \\ &\mathbf{U\times(V+W)=U\times V+U\times W}.\qquad \mathbf{(U+V)\times W=U\times W+V\times W}.\\ &(c\mathbf{U})\times\mathbf{V}=c\mathbf{U}\times\mathbf{V}=\mathbf{U}\times(c\mathbf{V}).\qquad \mathbf{0\times U=U\times 0=0}.\\ \\ \text{Watch-outs:}\quad& \mathbf{U\times U=0}.\quad\mathbf{U\times V=-V\times U}.\quad\mathbf{U\times(V\times W)\neq(U\times V)\times W}.\quad \text{(So $\mathbf{U\times V\times W}$ is undefined.)}\\ \end{align*} \begin{align*} \text{\bf Theorem:}\quad& \mathbf{|U\times V|^2=|U|^2|V|^2-(U\cdot V)^2}.\\ &\text{Proof:}\quad\text{LHS}=(u_2 v_3-u_3 v_2)^2+(u_3 v_1-u_1 v_3)^2+(u_1 v_2-u_2 v_1)^2\\ &=(u_2^2 v_3^2-2u_2 u_3 v_2 v_3+u_3^2 v_2^2)+(u_3^2 v_1^2-2u_3 u_1 v_3 v_1+u_1^2 v_3^2)+(u_1^2 v_2^2-2u_1 u_2 v_1 v_2+u_2^2 v_1^2)\\ &=(u_2^2 v_3^2+u_3^2 v_2^2+u_3^2 v_1^2+u_1^2 v_3^2+u_1^2 v_2^2+u_2^2 v_1^2)-2(u_2 u_3 v_2 v_3+u_3 u_1 v_3 v_1+u_1 u_2 v_1 v_2)\\ &\text{RHS}=(u_1^2+u_2^2+u_3^2)(v_1^2+v_2^2+v_3^2)-(u_1 v_1+u_2 v_2+u_3 v_3)^2\\ &=(u_1^2 v_1^2+u_1^2 v_2^2+u_1^2 v_3^2+u_2^2 v_1^2+u_2^2 v_2^2+u_2^2 v_3^2+u_3^2 v_1^2+u_3^2 v_2^2+u_3^2 v_3^2) -(u_1^2 v_1^2+u_2^2 v_2^2+u_3^2 v_3^2)\\ &\qquad-2(u_1 u_2 v_1 v_2+u_2 u_3 v_2 v_3+u_3 u_1 v_3 v_1)=\text{LHS}\qquad(QED)\\ \\ \text{\bf Theorem:}\quad& \mathbf{|U\times V|=|U||V|\sin\theta},\quad\text{where $\theta$ is the angle between $\mathbf{U}$ and $\mathbf{V}$.}\quad (\theta\in[0,\pi],\quad\text{so }\sin\theta\geq 0.)\\ &\text{Proof:}\quad\text{LHS}=\mathbf{\sqrt{|U\times V|^2}=\sqrt{|U|^2|V|^2-(U\cdot V)^2}=\sqrt{|U|^2|V|^2-(|U||V|\cos\theta)^2}}\\ &=\mathbf{\sqrt{|U|^2|V|^2(1-\cos^2\theta)}=|U||V|\sin\theta=}\text{RHS}.\\ \\ &\text{For the parallelogram formed by $\mathbf{U}$ and $\mathbf{V}$, its height on $\mathbf{U}$ is $\mathbf{|V|\sin\theta}$, so its area is $\mathbf{|U|\cdot|V|\sin\theta}$.}\\ &\boxed{\text{The area of the parallelogram formed by $\mathbf{U}$ and $\mathbf{V}$ is $\mathbf{|U\times V|}$.}}\\ \\ \text{\bf Triple Product:}\quad& \boxed{\text{The volume of the parallelepiped formed by $\mathbf{U, V}$ and $\mathbf{W}$ is $\mathbf{(U\times V)\cdot W=U\cdot(V\times W)}$.}}\\ &\text{$\mathbf{U,V}$ and $\mathbf{W}$ are in right-hand system ($\mathbf{U\times V}$ and $\mathbf{W}$ form an acute angle) to make a positive product.}\\ &\text{Proof:}\quad\text{Let }\mathbf{A=U\times V},\quad\text{then the area of the parallelogram formed by $\mathbf{U}$ and $\mathbf{V}$ is $\mathbf{|A|}$.}\\ &\text{The height of the parallelepiped, $H$, is the projection of $\mathbf{W}$ on $\mathbf{A}$. $H=\mathbf{\frac{A\cdot W}{|A|}}$.}\\ &\text{The volume of the parallelepiped is therefore }V=\mathbf{|A|}H=\mathbf{|A|\frac{A\cdot W}{|A|}=A\cdot W=(U\times V)\cdot W}.\\ &\text{Since $\mathbf{V,W}$ and $\mathbf{U}$ are in the same right-hand system, }V=\mathbf{(V\times W)\cdot U=U\cdot(V\times W)}.\\ &\therefore\quad V=\mathbf{(U\times V)\cdot W=U\cdot(V\times W)}.\\ \end{align*} \end{document}