Review of Vector Mathematics
EML 3013C, 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 $\mbox{$\bar{{\bf a}}$}$ and $\mbox{$\bar{{\bf b}}$}$ is denoted by $\mbox{$\bar{{\bf a}}$}\cdot\mbox{$\bar{{\bf b}}$}$ (prounounced `a dot b'). In dynamics the dot product is used to define work and power, to reduce a vector to components, and to reduce vector equations to scalar equations.

Consider two vectors $\mbox{$\bar{{\bf a}}$}$ and $\mbox{$\bar{{\bf b}}$}$, with angle $\theta$ between them as shown in Figure 1.

 
\begin{figure} \begin{center} \includegraphics [height=1.8in, width=2.0in ] {vecfig1.eps} \end{center}\end{figure}
Figure 1: Vectors $\mbox{$\bar{{\bf a}}$}$ and $\mbox{$\bar{{\bf b}}$}$.

Assume that the cartesian coordinate representation of $\mbox{$\bar{{\bf a}}$}$ and $\mbox{$\bar{{\bf b}}$}$ are given by


The dot product may be expressed as follows.


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

Using dot product to find components

Suppose a vector $\mbox{$\bar{{\bf v}}$}$ has cartesian coordinate representation

\begin{displaymath} \mbox{$\bar{{\bf v}}$}= v_x\mbox{$\bf{\hat i}$}+ v_y\mbox{$\bf{\hat j}$}+ v_z\mbox{$\bf{\hat k}$}.\end{displaymath}

Then,




Cross product



The cross product of two vectors $\mbox{$\bar{{\bf a}}$}$ and $\mbox{$\bar{{\bf b}}$}$ is written $\mbox{$\bar{{\bf a}}$}\times\mbox{$\bar{{\bf b}}$}$ and pronounced `a cross b.' Since the cross product of $\mbox{$\bar{{\bf a}}$}$ and $\mbox{$\bar{{\bf 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 a moment, $\mbox{$\bar{\bf M}$}= \mbox{$\bar{{\bf r}}$}\times\mbox{$\bar{\bf F}$}$, and angular momentum, $\mbox{$\bar{\bf H}$}= \mbox{$\bar{{\bf r}}$}\times(m\mbox{$\bar{{\bf v}}$})$. Also, the cross product is part of the formulas for velocity and acceleration of points on spinning solids (e.g., $\mbox{$\bar{{\bf v}}$}= \mbox{$\bar{\bf \omega}$}\times\mbox{$\bar{{\bf r}}$})$.

Consider two vectors $\mbox{$\bar{{\bf a}}$}$ and $\mbox{$\bar{{\bf b}}$}$, with angle $\theta$ between them as shown in Figure 2. Again, assume that the cartesian coordinate representation of $\mbox{$\bar{{\bf a}}$}$ and $\mbox{$\bar{{\bf b}}$}$ are given by



 
Figure 2: Vectors $\mbox{$\bar{{\bf a}}$}$ and $\mbox{$\bar{{\bf b}}$}$.
\begin{figure} \begin{center} \includegraphics [height=1.8in, width=2.5in ] {vecfig2.eps} \end{center}\end{figure}

The cross product may be expressed as follows:


where $\mbox{$\bf{\hat n}$}$ is a unit vector perpendicular to the plane containing $\mbox{$\bar{{\bf a}}$}$ and $\mbox{$\bar{{\bf b}}$}$. (The direction of $\mbox{$\bf{\hat n}$}$ is determined by the right hand rule.) The cross product is also given by


where $\vert\cdot\vert$ denotes determinant.

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


 
Figure 3: Mnemonic device to remember the cross products of the standard cartesian unit vectors.
\begin{figure} \begin{center} \includegraphics [height=1.8in, width=2.5in ] {vecfig3.eps} \end{center}\end{figure}


The first distributive law above may be used (in lieu of the determinant formula) to compute cross products in terms of the cartesian components of the vectors. For example, suppose


Then,



Change of Basis

A vector $\mbox{$\bar{{\bf v}}$}$ in the (x,y) coordinate system with unit vectors $\mbox{$\bf{\hat i}$}$ and $\mbox{$\bf{\hat j}$}$ is given by

\begin{displaymath} \mbox{$\bar{{\bf v}}$}= v_x\mbox{$\bf{\hat i}$}+ v_y\mbox{$\bf{\hat j}$}.\end{displaymath}

It is desired to express $\mbox{$\bar{{\bf v}}$}$ in a new coordinate system $(x^\prime, y^\prime)$ with unit vectors $\mbox{$\bf{\hat i}$}^\prime$ and $\mbox{$\bf{\hat j}$}^\prime$ as shown in Figure 4.


 
Figure 4: Coordinate Systems (x,y) and $(x^\prime, y^\prime)$.
\begin{figure} \begin{center} \includegraphics [height=1.8in, width=2.5in ] {vecfig4.eps} \end{center}\end{figure}


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

 
Figure 5: Diagram of Unit Vectors.
\begin{figure} \begin{center} \includegraphics [height=1.8in, width=2.5in ] {vecfig5.eps} \end{center}\end{figure}

It is seen that

\begin{displaymath} \mbox{$\bf{\hat i}$}= \cos\theta\mbox{$\bf{\hat i}$}^\prime ... ...{$\bf{\hat i}$}^\prime + \cos\theta\mbox{$\bf{\hat j}$}^\prime.\end{displaymath}

Using these expressions for $\mbox{$\bf{\hat i}$}$ and $\mbox{$\bf{\hat j}$}$ it follows that


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