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# 2 ALGEBRA

The solutions of the equation ${ax}^{2}+bx+c=0$ , where $a\ne 0$ , are given by: $x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$ .

## 2.2 VECTOR ALGEBRA

### 2.2.1 Definitions

Any quantity which is completely determined by its magnitude is called a scalar. Examples include: mass, density, and temperature. Any quantity which is completely determined by its magnitude and direction is called a vector. Examples include: velocity, acceleration, force. A vector quantity is usually represented by a boldfaced letter such as $V$ . Two vectors ${V}_{1}$ and ${V}_{2}$ are equal to one another if they have equal magnitudes and are acting in the same directions. A negative vector, written as $-V$ , is one which acts in the opposite direction to $V$ , but is of equal magnitude to it. The magnitude of $V$ is written $|V|$ or simply $v$ . The unit vector$\frac{V}{|V|}$ (when $|V|\ne 0\right)$ is that vector which has the same direction as $V$ , but has a magnitude of unity (sometimes represented as $\stackrel{\wedge }{V}$ ).

The vector sum of ${V}_{1}$ and ${V}_{2}$ is represented by ${V}_{1}+{V}_{2}$ . The vector sum of ${V}_{1}$ and $-{V}_{2}$ , or the difference of the vector ${V}_{2}$ from ${V}_{1}$ is represented by ${V}_{1}-{V}_{2}$ .

If $r$ is a scalar, then $rV=Vr$ and this represents a vector $r$ times the magnitude of $V$ , in the same direction as $V$ if $r$ is positive, and in the opposite direction if $r$ is negative. If $r$ and $s$ are scalars and ${V}_{1}$ , ${V}_{2}$ , ${V}_{3}$ are vectors then the following rules of scalars and vectors hold:
$\begin{array}{cc}\hfill {V}_{1}+{V}_{2}& ={V}_{2}+{V}_{1}\hfill \\ \multicolumn{1}{c}{\left(r+s\right){V}_{1}}& =r{V}_{1}+s{V}_{1}\hfill \\ \multicolumn{1}{c}{r\left({V}_{1}+{V}_{2}\right)}& =r{V}_{1}+r{V}_{2}\hfill \\ \multicolumn{1}{c}{{V}_{1}+\left({V}_{2}+{V}_{3}\right)}& =\left({V}_{1}+{V}_{2}\right)+{V}_{3}={V}_{1}+{V}_{2}+{V}_{3}\hfill \\ \multicolumn{1}{c}{}\end{array}$
The vector $0$ is a vector of zero length.

### 2.2.2 Vectors in Space

1. A plane is described by two distinct vectors ${V}_{1}$ and ${V}_{2}$ . Should these vectors not intersect each other, then one can be displaced parallel to itself until they do. Any other vector $V$ lying in this plane is given by
$V=r{V}_{1}+s{V}_{2}$
2. A position vector specifies the position in space of a point relative to a fixed origin. If ${V}_{1}$ and ${V}_{2}$ are the position vectors of the points $A$ and $B$ , relative to the origin $O$ , then any point $P$ on the line AB has a position vector $V$ given by
$V=r{V}_{1}+\left(1-r\right){V}_{2}$
The scalar '' $r$ '' can be taken as the metric representation of $P$ since $r=0$ implies $P=B$ and $r=1$ implies $P=A$ . If the point $P$ divides the line AB in the ratio $r:s$ then
$V=\left(\frac{r}{r+s}\right){V}_{1}+\left(\frac{s}{r+s}\right){V}_{2}$
3. The vectors ${V}_{1}$ , ${V}_{2},{V}_{3},\dots ,{V}_{n}$ are said to be linearly dependent if there exist scalars ${r}_{1},{r}_{2},{r}_{3},\dots ,{r}_{n}$ , not all zero, such that
${r}_{1}{V}_{1}+{r}_{2}{V}_{2}+\dots +{r}_{n}{V}_{n}=0$
4. A vector $V$ is linearly dependent upon the set of vectors $\left\{{V}_{1},{V}_{2},{V}_{3},\dots ,{V}_{n}\right\}$ if
$V={r}_{1}{V}_{1}+{r}_{2}{V}_{2}+{r}_{3}{V}_{3}+\dots +{r}_{n}{V}_{n}$
5. Three vectors are linearly dependent if and only if they are co-planar.
6. All points in space can be uniquely determined by linear dependence upon three base vectors i.e., three vectors any one of which is linearly independent of the other two. The simplest set of base vectors are the unit vectors along the coordinate axes. These are usually designated by $i$ , $j$ , and $k$ .
7. If $V$ is a vector in space, and $a$ , $b$ , and $c$ are the respective magnitudes of the projections of the vector along the axes then
$V=ai+bj+ck\mathrm{ }\text{and} \mathrm{ }|V|=\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}$
and the direction cosines of $V$ are
$cos\alpha =a/v,\mathrm{ }cos\beta =b/v,\mathrm{ }cos\gamma =c/v.$
8. The law of vector addition yields
${V}_{1}+{V}_{2}=\left({a}_{1}+{a}_{2}\right)i+\left({b}_{1}+{b}_{2}\right)j+\left({c}_{1}+{c}_{2}\right)k$
9. ### 2.2.3 The Scalar, Dot, or Inner Product of Two Vectors

This product is represented as ${V}_{1}·{V}_{2}$ and is defined to be equal to ${v}_{1}{v}_{2}cos\theta$ , where ${v}_{1}=|{V}_{1}|,{v}_{2}=|{V}_{2}|$ , and $\theta$ is the angle from ${V}_{1}$ to ${V}_{2}$ . That is
${V}_{1}·{V}_{2}={v}_{1}{v}_{2}cos\theta ={a}_{1}{a}_{2}+{b}_{1}{b}_{2}+{c}_{1}{c}_{2}={V}_{2}·{V}_{1}$
Note the relations:
$\begin{array}{cc}\hfill \left({V}_{1}+{V}_{2}\right)·{V}_{3}& ={V}_{1}·{V}_{3}+{V}_{2}·{V}_{3}\hfill \\ \multicolumn{1}{c}{{V}_{1}·\left({V}_{2}+{V}_{3}\right)}& ={V}_{1}·{V}_{2}+{V}_{1}·{V}_{3}\hfill \\ \multicolumn{1}{c}{}\end{array}$
If ${V}_{1}$ is perpendicular to ${V}_{2}$ then ${V}_{1}·{V}_{2}=0$ , and if ${V}_{1}$ is parallel to ${V}_{2}$ then ${V}_{1}·{V}_{2}=|{V}_{1}|\phantom{\rule{0.2em}{0ex}}|{V}_{2}|$ . In particular:
$\begin{array}{cc}\hfill i·i& =j·j=k·k=1\hfill \\ \multicolumn{1}{c}{i·j}& =j·k=k·i=0\hfill \\ \multicolumn{1}{c}{}\end{array}$

### 2.2.4 The Vector or Cross Product of Two Vectors

This product is represented as ${V}_{1}×{V}_{2}$ and is defined as
${V}_{1}×{V}_{2}=|{V}_{1}||{V}_{2}|\phantom{\rule{0.2em}{0ex}}sin\theta \phantom{\rule{0.2em}{0ex}}1$
where $\theta$ is the angle from ${V}_{1}$ to ${V}_{2}$ and $1$ is a unit vector perpendicular to the plane of ${V}_{1}$ and ${V}_{2}$ and so directed that a right-handed screw driven in the direction of $1$ would carry ${V}_{1}$ into ${V}_{2}$ . Note that
$tan\theta =\frac{|{V}_{1}×{V}_{2}|}{{V}_{1}·{V}_{2}}$
The following rules apply to vector products:
$\begin{array}{cc}\hfill {V}_{1}×{V}_{2}& =-{V}_{2}×{V}_{1}\hfill \\ \multicolumn{1}{c}{{V}_{1}×\left({V}_{2}+{V}_{3}\right)}& ={V}_{1}×{V}_{2}+{V}_{1}×{V}_{3}\hfill \\ \multicolumn{1}{c}{\left({V}_{1}+{V}_{2}\right)×{V}_{3}}& ={V}_{1}×{V}_{3}+{V}_{2}×{V}_{3}\hfill \\ \multicolumn{1}{c}{{V}_{1}×\left({V}_{2}×{V}_{3}\right)}& ={V}_{2}\left({V}_{3}·{V}_{1}\right)-{V}_{3}\left({V}_{1}·{V}_{2}\right)\hfill \\ \multicolumn{1}{c}{i×i=0,\mathrm{ }j×j}& =0,\mathrm{ }k×k=0\hfill \\ \multicolumn{1}{c}{i×j=k,\mathrm{ }j×k}& =i,\mathrm{ }k×i=j\hfill \end{array}$
If ${V}_{1}={a}_{1}i+{b}_{1}j+{c}_{1}k$ , ${V}_{2}={a}_{2}i+{b}_{2}j+{c}_{2}k$ , and ${V}_{3}={a}_{3}i+{b}_{3}j+{c}_{3}k$ , then
${V}_{1}×{V}_{2}=|\begin{array}{lll}i\hfill & j\hfill & k\hfill \\ {a}_{1}\hfill & {b}_{1}\hfill & {c}_{1}\hfill \\ {a}_{2}\hfill & {b}_{2}\hfill & {c}_{2}\hfill \end{array}|=\left({b}_{1}{c}_{2}-{b}_{2}{c}_{1}\right)\mathrm{i}+\left({c}_{1}{a}_{2}-{c}_{2}{a}_{1}\right)\mathrm{j}+\left({a}_{1}{b}_{2}-{a}_{2}{b}_{1}\right)\mathrm{k}$
Note that, since ${V}_{1}×{V}_{2}=-{V}_{2}×{V}_{1}$ , the vector product is not commutative.

### 2.2.5 Scalar Triple Product

There is only one possible interpretation of the expression ${V}_{1}·{V}_{2}×{V}_{3}$ and that is ${V}_{1}·\left({V}_{2}×{V}_{3}\right)$ which is a scalar. This product is called the scalar triple product and is written as $\left[{V}_{1}{V}_{2}{V}_{3}\right]$ . Further
$\begin{array}{lll}\left[{V}_{1}{V}_{2}{V}_{3}\right]\hfill & =\hfill & {V}_{1}\cdot \left({V}_{2}×{V}_{3}\right)=\left({V}_{1}×{V}_{2}\right)\cdot {V}_{3}={V}_{2}\cdot \left({V}_{3}×{V}_{1}\right)\hfill \\ \hfill & =\hfill & |\begin{array}{lll}{a}_{1}\hfill & {b}_{1}\hfill & {c}_{1}\hfill \\ {a}_{2}\hfill & {b}_{2}\hfill & {c}_{2}\hfill \\ {a}_{3}\hfill & {b}_{3}\hfill & {c}_{3}\hfill \end{array}|\hfill \\ \hfill & =\hfill & |{V}_{1}||{V}_{2}||{V}_{3}|cos\varphi sin\theta ,\hfill \end{array}$
where $\theta$ is the angle between ${V}_{2}$ and ${V}_{3}$ and $\phi$ is the angle between ${V}_{1}$ and the normal to the plane of ${V}_{2}$ and ${V}_{3}$ . The determinant indicates that it can be considered as the volume of the parallelepiped whose three determining edges are ${V}_{1}$ , ${V}_{2}$ , and ${V}_{3}$ . Note that cyclic permutation of the subscripts does not change the value of the scalar triple product: $\left[{V}_{1}{V}_{2}{V}_{3}\right]=\left[{V}_{2}{V}_{3}{V}_{1}\right]=\left[{V}_{3}{V}_{1}{V}_{2}\right]$ but $\left[{V}_{1}{V}_{2}{V}_{3}\right]=-\left[{V}_{2}{V}_{1}{V}_{3}\right]$ and $\left[{V}_{1}{V}_{1}{V}_{2}\right]\equiv 0$ .

### 2.2.6 Vector Triple Product

The product ${V}_{1}×\left({V}_{2}×{V}_{3}\right)$ defines the vector triple product. The parentheses are vital to the definition.
$\begin{array}{lll}{V}_{1}×\left({V}_{2}×{V}_{3}\right)\hfill & =\hfill & \left({V}_{1}\cdot {V}_{3}\right){V}_{2}-\left({V}_{1}\cdot {V}_{2}\right){V}_{3}\hfill \\ \hfill & =\hfill & |\begin{array}{lll}i\hfill & j\hfill & k\hfill \\ {a}_{1}\hfill & {b}_{1}\hfill & {c}_{1}\hfill \\ |\begin{array}{ll}{b}_{2}\hfill & {c}_{2}\hfill \\ {b}_{3}\hfill & {c}_{3}\hfill \end{array}|\hfill & |\begin{array}{ll}{c}_{2}\hfill & {a}_{2}\hfill \\ {c}_{3}\hfill & {a}_{3}\hfill \end{array}|\hfill & |\begin{array}{ll}{a}_{2}\hfill & {b}_{2}\hfill \\ {a}_{3}\hfill & {b}_{3}\hfill \end{array}|\hfill \end{array}|\hfill \end{array}$
This is a vector, perpendicular to ${V}_{1}$ , lying in the plane of ${V}_{2}$ and ${V}_{3}$ . Similarly
$\begin{array}{lll}\left({V}_{1}×{V}_{2}\right)×{V}_{3}\hfill & =\hfill & |\begin{array}{lll}i\hfill & j\hfill & k\hfill \\ |\begin{array}{ll}{b}_{1}\hfill & {c}_{1}\hfill \\ {b}_{2}\hfill & {c}_{2}\hfill \end{array}|\hfill & |\begin{array}{ll}{c}_{1}\hfill & {a}_{1}\hfill \\ {c}_{2}\hfill & {a}_{2}\hfill \end{array}|\hfill & |\begin{array}{ll}{a}_{1}\hfill & {b}_{1}\hfill \\ {a}_{2}\hfill & {b}_{2}\hfill \end{array}|\hfill \\ {a}_{3}\hfill & {b}_{3}\hfill & {c}_{3}\hfill \end{array}|\hfill \\ {V}_{1}×\left({V}_{2}×{V}_{3}\right)\hfill & +\hfill & {V}_{2}×\left({V}_{3}×{V}_{1}\right)+{V}_{3}×\left({V}_{1}×{V}_{2}\right)\equiv 0\hfill \end{array}$
If ${V}_{1}×\left({V}_{2}×{V}_{3}\right)=\left({V}_{1}×{V}_{2}\right)×{V}_{3}$ then ${V}_{1},{V}_{2},{V}_{3}$ form an <i>orthogonal set</i>. Thus $\left\{i,j,k\right\}$ form an orthogonal set. Page 1 of 1  1/1 Entry Display
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