Review of Vector Mathematics

EML 4930:Dynamic systems I


Dot Product

The dot product of two vectors is a scalar, so the dot product is sometimes called the 'scalar product'. The dot product of two vectors $\overline {\mathbf{a}}$ and $\overline{\mathbf{b}}$ is denoted by $\overline{\mathbf{a}}.\overline{\mathbf{b}}$ (pronounced 'a dot b'). In dynamics the dot product is used to define the work and power, to reduce a vector to components, and to reduce vector equations to scalar equations.

Consider two vectors $\overline {\mathbf{a}}$ and $\overline{\mathbf{b}}$ with an angle $\theta$ between them as shown in Figure 1.

 
Figure: Vectors a and b
\begin{figure} \begin{center} \includegraphics [ height=2.2425in, width=2.7691in ]{vectorfig1.eps}\end{center} \end{figure}


assume that the Cartesian coordinate representation of $\overline {\mathbf{a}}$ and $\overline{\mathbf{b}}$ given by
\begin{displaymath} \overline{\mathbf{a}}=a_{x}\widehat{\mathbf{i}}+a_{y}\widehat{\mathbf{j}}+a_{z}\widehat{\mathbf{k}}\end{displaymath} (1)
\begin{displaymath} \overline{\mathbf{b}}=b_{x}\widehat{\mathbf{i}}+b_{y}\widehat{\mathbf{j}}+b_{z}\widehat{\mathbf{k}}\end{displaymath} (2)
The dot product is expressed as follows
\begin{align} \overline{\mathbf{a}}.\overline{\mathbf{b}} & =\vert\overline{\mat... ...ne{\mathbf{b}}\vert cos\theta\\ & =a_{x}b_{x}+a_{y}b_{y}+a_{z}b_{z}\end{align}
The following laws and special cases of the dot product are worth knowing well. It is assumed that $\alpha$ is a scalar.

Hence $\widehat{\mathbf{i}}.\widehat{\mathbf{j}}=\widehat{\mathbf{j}}.\widehat{\mathbf{k}}=\widehat{\mathbf{k}}.\widehat{\mathbf{i}}=0$

Hence $\widehat{\mathbf{i}}.\widehat{\mathbf{i}}=\widehat{\mathbf{j}}.\widehat{\mathbf{j}}=\widehat{\mathbf{k}}.\widehat{\mathbf{k}}=1$

Using dot product to find components

Suppose a vector $\overline{\mathbf{v}}$ has Cartesian coordinate representation
\begin{displaymath} \overline{\mathbf{v}}=v_{x}\widehat{\mathbf{i}}+v_{y}\widehat{\mathbf{j}}+v_{z}\widehat{\mathbf{k}}\end{displaymath} (3)
Then
\begin{align} v_{x} & =\overline{\mathbf{v}}.\widehat{\mathbf{i}}\\ v_{y} & =\... ...t{\mathbf{j}}\\ v_{z} & =\overline{\mathbf{v}}.\widehat{\mathbf{k}}\end{align}

Cross product

The cross product of two vectors $\overline {\mathbf{a}}$ and $\overline{\mathbf{b}}$ written as $\overline{\mathbf{a}}\times\overline{\mathbf{b}}$ and pronounced 'a cross b').Since the cross product of $\overline {\mathbf{a}}$ and $\overline{\mathbf{b}}$ is a vector, the cross product is also called the vector product to distinguish it from the scalar product ( the dot product). In dynamics the cross product is used to define the moment $\overline{\mathbf{M}}=\overline{\mathbf{r}}\times \overline{\mathbf{F}}$, and the angular momentum $\overline {\mathbf{H}}=\overline{\mathbf{r}}\times(m\overline{\mathbf{v}}).$ Also the cross product is part of the formulas for velocity and acceleration of points on spinning solids ( e.g. $\overline{\mathbf{v}}=\overline{\mathbf{\omega}}\times\overline{\mathbf{r}}).$

Consider two vectors $\overline {\mathbf{a}}$ and $\overline{\mathbf{b}}$ with an angle $\theta$ between them as shown in Figure 2. Again, assume that the cartesian coordinate representation of $\overline {\mathbf{a}}$ and $\overline{\mathbf{b}}$ are given by
\begin{displaymath} \overline{\mathbf{a}}=a_{x}\widehat{\mathbf{i}}+a_{y}\widehat{\mathbf{j}}+a_{z}\widehat{\mathbf{k}}\end{displaymath} (4)
\begin{displaymath} \overline{\mathbf{b}}=b_{x}\widehat{\mathbf{i}}+b_{y}\widehat{\mathbf{j}}+b_{z}\widehat{\mathbf{k}}\end{displaymath} (5)



 
Figure: Vectors a and b
\begin{figure} \begin{center} \includegraphics [ height=1.8144in, width=2.4517in ]{vecfig2.eps}\end{center} \end{figure}

The cross product may be expressed as follows
\begin{displaymath} \overline{\mathbf{a}}\times\overline{\mathbf{b}}=\vert\overl... ...rt\vert\overline{\mathbf{b}}\vert sin\theta\widehat{\mathbf{n}}\end{displaymath} (6)
where $\widehat{\mathbf{n}}$ is the unit vector perpendicular to the plane containing $\overline {\mathbf{a}}$ and $\overline{\mathbf{b}}$ . The cross product is also given by
\begin{align} \overline{\mathbf{a}}\times\overline{\mathbf{b}} & =\left\vert \b... ..._{z})\widehat{\mathbf{j}}+(a_{x}b_{y}-a_{y}b_{x})\widehat{\mathbf{k}}\end{align}
where $\vert$. $\vert$denotes the determinant.

The following laws and special cases of the cross product are worth knowing well. It is assumed that $\alpha$ is a scalar.

Hence $\widehat{\mathbf{i}}\times\widehat{\mathbf{i}}=\widehat{\mathbf{j}}\times\widehat{\mathbf{j}}=\widehat{\mathbf{k}}\times\widehat{\mathbf{k}}=0$

Hence $\vert$$\widehat{\mathbf{i}}\times\widehat{\mathbf{j}}\vert=\vert\widehat{\mathbf{j}}\t... ...hat{\mathbf{k}}\vert=\vert\widehat{\mathbf{k}}\times\widehat{\mathbf{i}}\vert=1$

$\widehat{\mathbf{i}}\times\widehat{\mathbf{k}}=-\widehat{\mathbf{j}},\quad\wide... ...f{i}},\quad\widehat{\mathbf{j}}\times\widehat{\mathbf{i}}=-\widehat{\mathbf{k}}$ (assuming a standard right handed coordinate system). See Figure 3 for convenient memorizing these formulas.

 
Figure: Mnemonic device to remember the cross products of the standard cartersian unit vectors
\begin{figure} \begin{center} \includegraphics [ height=3.269in, width=2.9577in ]{vecfig3.eps}\end{center} \end{figure}

The first distributive law above may be used ( in lieu of the determinant formula) to compute the cross products in terms of the cartesian components of the vectors. For example, suppose
\begin{displaymath} \overline{\mathbf{u}}=u_{x}\widehat{\mathbf{i}}+u_{y}\widehat{\mathbf{j}}\end{displaymath} (7)
\begin{displaymath} \overline{\mathbf{v}}=v_{x}\widehat{\mathbf{i}}+v_{y}\widehat{\mathbf{j}}\end{displaymath} (8)
Then
\begin{align} \overline{\mathbf{u}}\times\overline{\mathbf{v}} & =(u_{x}\widehat... ...idehat{\mathbf{k}}\\ & =(u_{x}v_{y}-u_{y}v_{x})\widehat{\mathbf{k}}\end{align}

Change of Basis

A vector $\overline{\mathbf{v}}$ in the (x,y) coordinate system with unit vectors $\widehat{\mathbf{i}}$ and $\widehat{\mathbf{j}}$ is given by
\begin{displaymath} \overline{\mathbf{v}}=v_{x}\widehat{\mathbf{i}}+v_{y}\widehat{\mathbf{j}}\end{displaymath} (9)
It is desired to express $\overline{\mathbf{v}}$ in the new coordinate system $(x^{\prime},y^{\prime})$ with unit vectors $\widehat{\mathbf{i}}^{\prime}$ and $\widehat{\mathbf{j}}^{\prime}$ as shown in Figure 4.

 
Figure: Coordinate system (x,y) and $(x^{\prime},y^{\prime})$
\begin{figure} \begin{center} \includegraphics [ height=1.7011in, width=2.6645in ]{vecfig4.eps}\end{center} \end{figure}

To represent $\overline{\mathbf{v}}$ in the new coordinate system, one must first express the unit vectors $\widehat{\mathbf{i}}$ and $\widehat{\mathbf{j}}$ in terms of the unit vectors $\widehat{\mathbf{i}}^{\prime}$ and $\widehat{\mathbf{j}}^{\prime}.$ To do this consider the diagram shown in Figure 5

 
Figure: Diagram of Unit Vectors
\begin{figure} \begin{center} \includegraphics [ height=1.8118in, width=2.6628in ]{vecfig5.eps}\end{center} \end{figure}

It is seen that
\begin{align} \widehat{\mathbf{i}} & =cos\theta\widehat{\mathbf{i}}^{\prime}+sin... ...a\widehat{\mathbf{i}}^{\prime}+cos\theta\widehat{\mathbf{j}}^{\prime}\end{align}
Using these expressions for $\widehat{\mathbf{i}}$ and $\widehat{\mathbf{j}}$ it follows that
\begin{align} \overline{\mathbf{v}} & =v_{x}\mathbf{(}cos\theta\widehat{\mathbf{... ...{\prime}+(v_{x}sin\theta+v_{y}cos\theta)\widehat{\mathbf{j}}^{\prime}\end{align}