Difference between revisions of "Vector"

Revision as of 17:54, 29 August 2024

The word vector has many different definitions, depending on who is defining it and in what context. Physicists will often refer to a vector as "a quantity with a direction and magnitude." For Euclidean geometers, a vector is essentially a directed line segment. In many situations, a vector is best considered as an n-tuple of numbers (often real or complex). Most generally, but also most abstractly, a vector is any object which is an element of a given vector space.

A vector is usually graphically represented as an arrow. Vectors can be uniquely described in many ways. The two most common is (for 2-dimensional vectors) by describing it with its length (or magnitude) and the angle it makes with some fixed line (usually the x-axis) or by describing it as an arrow beginning at the origin and ending at the point $(x,y)$ . An $n$ -dimensional vector can be described in this coordinate form as an ordered $n$ -tuple of numbers within angle brackets or parentheses, $(x\,\,y\,\,z\,\,...)$ . The set of vectors over a field is called a vector space.

Description

Every vector $\overrightarrow{PQ}$ has a starting point $P\langle x_1, y_1\rangle$ and an endpoint $Q\langle x_2, y_2\rangle$ . Since the only thing that distinguishes one vector from another is its magnitude or length, and direction, vectors can be freely translated about a plane without changing. Hence, it is convenient to consider a vector as originating from the origin. This way, two vectors can be compared by only looking at their endpoints. This is why we only require $n$ values for an $n$ dimensional vector written in the form $(x\,\,y\,\,z\,\,...)$ . The magnitude of a vector, denoted $\|\vec{v}\|$ , is found simply by using the distance formula.

Addition of Vectors

For vectors $\vec{v}$ and $\vec{w}$ , with angle $\theta$ formed by them, $\|\vec{v}+\vec{w}\|^2=\|\vec{v}\|^2+\|\vec{w}\|^2+2\|\vec{v}\|\|\vec{w}\|\cos\theta$ .

$[asy] size(150); pen p=linewidth(1); MA("\theta",(5,-1),(2,3),(4,6),0.3,9,yellow); MC("\vec v",D((0,0)--(2,3),orange+p,Arrow),NW); D((2,3)--(3,4.5)); MC("\vec w",D((2,3)--(5,-1),green+p,Arrow),NE); MC(-10,"\vec{v}+\vec{w}",D((0,0)--(5,-1),red+p,Arrow),S); [/asy]$

Addition of vectors

From this it is simple to derive that for a real number $c$ , $c\vec{v}$ is the vector $\vec{v}$ with magnitude multiplied by $c$ . Negative $c$ corresponds to opposite directions.

Properties of Vectors

Since a vector space is defined over a field $K$ , it is logically inherent that vectors have the same properties as those elements in a field.

For any vectors $\vec{x}$ , $\vec{y}$ , $\vec{z}$ , and real numbers $a,b$ ,

$\vec{x}+\vec{y}=\vec{y}+\vec{x}$ (Commutative in +)
$(\vec{x}+\vec{y})+\vec{z}=\vec{x}+(\vec{y}+\vec{z})$ (Associative in +)
There exists the zero vector $\vec{0}$ such that $\vec{x}+\vec{0}=\vec{x}$ (Additive identity)
For each $\vec{x}$ , there is a vector $\vec{y}$ such that $\vec{x}+\vec{y}=\vec{0}$ (Additive inverse)
$1\vec{x}=\vec{x}$ (Unit scalar identity)
$(ab)\vec{x}=a(b\vec{x})$ (Associative in scalar)
$a(\vec{x}+\vec{y})=a\vec{x}+a\vec{y}$ (Distributive on vectors)
$(a+b)\vec{x}=a\vec{x}+b\vec{x}$ (Distributive on scalars)

Vector Operations

Dot (Scalar) Product

Consider two vectors $\bold{a}=\langle a_1,a_2,\ldots,a_n\rangle$ and $\bold{b}=\langle b_1, b_2,\ldots,b_n\rangle$ in $\mathbb{R}^n$ . The dot product is defined as $\bold{a}\cdot\bold{b}=\bold{b}\cdot\bold{a}=|\bold{a}| |\bold{b}|\cos\theta=a_1b_1+a_2b_2+\cdots+a_nb_n$ , where $\theta$ is the angle formed by the two vectors. This also yields the geometric interpretation of the dot product: from basic right triangle trigonometry, it follows that the dot product is equal to the length of the projection (i.e. the distance from the origin to the foot of the head of $\bold{a}$ to $\bold{b}$ ) of $\bold{a}$ onto $\bold{b}$ times the length of $\bold{b}$ . Note that the dot product is $0$ if and only if the two vectors are perpendicular.

Cross (Vector) Product

The cross product between two vectors $\bold{a}$ and $\bold{b}$ in $\mathbb{R}^3$ is defined as the vector whose length is equal to the area of the parallelogram spanned by $\bold{a}$ and $\bold{b}$ and whose direction is in accordance with the right-hand rule. Because of this, $|\bold{a}\times\bold{b}|=|\bold{a}| |\bold{b}|\sin\theta$ , where $\theta$ is the angle formed by the two vectors, and from the right-hand rule condition, $\bold{a}\times\bold{b}=-\bold{b}\times\bold{a}$ . Also, $\sin^2\theta+\cos^2\theta=1$ gives that $|\bold{a}|^2|\bold{b}|^2=|\bold{a}\cdot\bold{b}|^2+|\bold{a}\times\bold{b}|^2$ .

If $\bold{a}=\langle a_1,a_2,a_3\rangle$ and $\bold{b}=\langle b_1,b_2,b_3\rangle$ , then the cross product of $\bold{a}$ and $\bold{b}$ is given by

$\bold{a}\times\bold{b}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3\end{vmatrix}.$

where $\hat{i},\hat{j},\hat{k}$ are unit vectors along the coordinate axes, or equivalently, $\bold{a}\times\bold{b}=\langle a_2b_3-a_3b_2,a_3b_1-a_1b_3,a_1b_2-a_2b_1\rangle$ . Also, $\bold{a}\times\bold{a}=\bold{0}$ Through the use of bivectors (mentioned below), the cross product can be extended to any dimension.

Triple Scalar Product

The triple scalar product of three vectors $\bold{a,b,c}$ is defined as $(\bold{a}\times\bold{b})\cdot \bold{c}$ . Geometrically, the triple scalar product gives the signed volume of the parallelepiped determined by $\bold{a,b}$ and $\bold{c}$ . It follows that

$(\bold{a}\times\bold{b})\cdot \bold{c} = (\bold{c}\times\bold{a})\cdot \bold{b} = (\bold{b}\times\bold{c})\cdot \bold{a}.$

It can also be shown that

$(\bold{a}\times\bold{b})\cdot \bold{c} = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}.$

Triple Vector Product

The vector triple product of $\bold{a},\bold{b},\bold{c}$ is defined as the cross product of one vector, so that $\bold{a}\times(\bold{b}\times\bold{c})=\bold{b}(\bold{a}\cdot\bold{c})-\bold{c}(\bold{a}\cdot\bold{b})$ , which can be remembered by the mnemonic "BAC-CAB" (this relationship between the cross product and dot product is called the triple product expansion, or Lagrange's formula).

Bivectors

Sometimes when working with vectors (especially the cross-product) it can be hard to understand what is happening. For example, imagine describing a wheel on a car rotating with a vector. The resulting vector does not describe the rotation happening intuitively. This is where bivectors can help.

Just like vectors are oriented lines, bivectors are oriented areas. Consider two vectors $u$ and $v$ . The bivector formed by these vectors (denoted $u\wedge v$ ) represents the parallelogram formed by $u$ and $v$ . Thus, the magnitude $||u \wedge b||$ represents the area formed by this parallelogram.

Calculating the bivector requires only three properties. First, the bivector $u \wedge u = 0$ since there is no way to form a parallelogram. Second the bivector $u \wedge v = - v \wedge u$ . Finally, the wedge product ( $\wedge$ operation) is distributed.

For 2D vectors $u$ and $v$ , the resulting bivector is \begin{align} u \wedge v &= (u_1 e_1 + u_2 e_2)\wedge (v_1 e_1 + v_2 e_2) \\ &= u_1 e_1\wedge (v_1 e_1 + v_2 e_2) + u_2 e_2\wedge (v_1 e_1 + v_2 e_2) \\ &= u_1 e_1\wedge v_1 e_1 + u_1 e_1\wedge v_2 e_2 + u_2 e_2\wedge v_1 e_1 + u_2 e_2\wedge v_2 e_2 \\ &= u_1 v_2 e_1\wedge e_2 - u_2 v_1 e_1\wedge e_2 \\ &= (u_1 v_2 - u_2 v_1)e_1\wedge e_2 \\ \end{align} where $e_1\wedge e_2$ represents the bivector for a unit square. This result is the same as calculating the determinate! Thus bivectors can help to calculate determinates.

Now consider 3D vectors $u$ and $v$ . The resulting bivector becomes \begin{align} u \wedge v &= (u_1 e_1 + u_2 e_2 + u_3 e_3)\wedge (v_1 e_1 + v_2 e_2 + u_3 e_3) \\ &= u_1 e_1\wedge (v_1 e_1 + v_2 e_2 + v_3 e_3) + u_2 e_2\wedge (v_1 e_1 + v_2 e_2 + v_3 e_3) +u_3 e_3\wedge (v_1 e_1 + v_2 e_2 + v_3 e_3) \\ &= u_1 e_1\wedge v_1 e_1 + u_1 e_1\wedge v_2 e_2 + u_1 e_1\wedge v_3 e_3 + \\ & \: \: \: u_2 e_2\wedge v_1 e_1 + u_2 e_2\wedge v_2 e_2 + u_2 e_2\wedge v_3 e_3 + \\ & \: \: \: u_3 e_3\wedge v_1 e_1 + u_3 e_3\wedge v_2 e_2 + u_3 e_3\wedge v_3 e_3 \\ &= u_1 e_1\wedge v_2 e_2 + u_1 e_1\wedge v_3 e_3 + u_2 e_2\wedge v_1 e_1 + u_2 e_2\wedge v_3 e_3 + u_3 e_3\wedge v_1 e_1 + u_3 e_3\wedge v_2 e_2 \\ &= u_1v_2 e_1\wedge e_2 - u_1 v_3 e_3\wedge e_1 - u_2 v_1 e_1\wedge e_2 + u_2 v_3 e_2\wedge e_3 + u_3v_1 e_3\wedge e_1 - u_3v_2 e_2\wedge e_3 \\ &=(u_2 v_3 - u_3v_2) e_2\wedge e_3 +(u_3 v_1 e_3 - u_1 v_3) e_3\wedge e_1 + (u_1v_2 - u_2 v_1) e_1\wedge e_2 \\ \end{align} which is the same as the cross product $u\times v$ ! Thus, instead of using determinates to calculate the cross-product as above, it is possible to use bivectors instead.

Exercise

Through the use of the bivector, it is possible to extend the definition of the cross-product to higher dimensions. Try to calculate the 4D cross-product using the bivector properties mentioned above.

Discussion

This is a thread about what vectors are.

@@ Line 44: / Line 44: @@
 If <math>\bold{a}=\langle a_1,a_2,a_3\rangle</math> and <math>\bold{b}=\langle b_1,b_2,b_3\rangle</math>, then the cross product of <math>\bold{a}</math> and <math>\bold{b}</math> is given by
 <center><math>\bold{a}\times\bold{b}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3\end{vmatrix}.</math></center>
-where <math>\hat{i},\hat{j},\hat{k}</math> are [[unit vector]]s along the coordinate axes, or equivalently, <math>\bold{a}\times\bold{b}=\langle a_2b_3-a_3b_2,a_3b_1-a_1b_3,a_1b_2-a_2b_1\rangle</math>. Also, <math>\bold{a}\times\bold{a}=\bold{0}</math>
+where <math>\hat{i},\hat{j},\hat{k}</math> are [[unit vector]]s along the coordinate axes, or equivalently, <math>\bold{a}\times\bold{b}=\langle a_2b_3-a_3b_2,a_3b_1-a_1b_3,a_1b_2-a_2b_1\rangle</math>. Also, <math>\bold{a}\times\bold{a}=\bold{0}</math> Through the use of ''bivectors'' (mentioned below), the cross product can be extended to any dimension.
 ===Triple Scalar Product===

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