### Key Questions

- What is a right-hand Cartesian coordinate system?
- What are direction cosine angles and why are they always less than 180°?
- How are spherical coordinates different than cylindrical coordinates?

- What is a right-hand Cartesian coordinate system?
- What are direction cosine angles and why are they always less than 180°?
- How are spherical coordinates different than cylindrical coordinates?

In this section we will discuss four methods to specify points and vectors in three-dimensional space.

The most commonly used method is an extension of two-dimensional rectangular coordinates to three-dimensions. Alternately, points and vectors in three dimensions can be specified in terms of direction cosines, or using spherical or cylindrical coordinate systems. These will be discussed in the following sections.

You will often need to convert from one representation to another. Good visualization skills are helpful here.

We can extend the two-dimensional Cartesian coordinate system into three dimensions easily by adding a \(z\) axis perpendicular to the two-dimensional Cartesian plane. The notation is similar to the notation used for two-dimensional vectors. Points and forces are expressed as ordered triples of rectangular coordinates following the same notation used previously.

\begin{align*}
P \amp = (x, y, z) \amp \vec{F} \amp = \langle F_x, F_y, F_z \rangle
\end{align*}

For nearly all three-dimensional problems, you will need the rectangular \(x\text{,}\) \(y\text{,}\) and \(z\) locations of points in space and components of vectors before proceeding with the computations. If you are given the components upfront, then you are set to move forward, but otherwise, you will need to transform one coordinate system into rectangular coordinates.

Does it matter which way the axes are oriented? Is it OK to make the \(x\) axis point left or the \(y\) axis point down?

In one sense, it doesn’t matter at all. The positive directions of the coordinate axes are arbitrary. On the other hand, it’s convenient for everyone if we can agree on a standard orientation. In mathematics and engineering, the default is a right-handed coordinate system, where the coordinate axes are oriented according to the right hand rule shown in the figure.

To apply the right-hand rule, orient your thumb and first two fingers at right angles to each other and align them with three coordinate axes. Starting with your thumb, name your the axes in alphabetical order \(x\)- \(y\)- \(z\text{.}\)

These are the labels for the three axes and your fingers point in their positive directions. If it is more convenient, you may name your thumb \(y\) or \(z\text{,}\) as long as you name the other two fingers in the same sequence \(y\)- \(z\)- \(x\) or \(z\)- \(x\)- \(y\text{.}\)

The direction of a vector in two-dimensional systems could be expressed clearly with a single angle measured from a reference axis, but adding an additional dimension means that one angle is no longer enough.

One way to define the direction of a three-dimensional vector is by using direction cosine angles, also commonly known as coordinate direction angles. The direction cosine angles are the angles between the positive \(x\text{,}\) \(y\text{,}\) and \(z\) axes to a given vector and are traditionally named \(\theta_x\text{,}\) \(\theta_y\text{,}\) and \(\theta_z\text{.}\) Three-dimensional vectors, components, and angles are often difficult to visualize because they do not commonly lie in the Cartesian planes.

We can relate the components of a vector to its direction cosine angles using the following equations.

\begin{align}
\cos {\theta_x} \amp = \frac{A_x}{\left |A \right |} \amp \cos \theta_y \amp = \frac {A_y}{\left |A \right |} \amp \cos \theta_z \amp = \frac {A_z}{\left |A \right |}\tag{2.4.1}
\end{align}

Note the component in the numerator of each direction cosine equation is positive or negative as defined by the coordinate system, and the vector magnitude in the denominator is always positive. From these equations, we can conclude that:

- Direction cosines are signed value between -1 and 1.
- Direction cosine angles must always be between \(\ang{0}\) and \(\ang{180}\) or\begin{equation*} \ang{0} \le \theta_n \le \ang{180}\text{.} \end{equation*}
- Any direction cosine angle greater than \(\ang{90}\) indicates a negative component along that respective axis. Spatially this is because all direction cosine angles are measured from the positive side of each axis. Mathematically this is because the cosine of any angle between 90 and 180 degrees is numerically negative.

A cable force pulls on an anchor ring centered on \(x\text{,}\) \(y\text{,}\) \(z\) axis origin.

A rope pulls on an anchor ring centered at the origin with force \(\vec{F}=\lbf{\langle 20, -30, 60 \rangle}\text{.}\)

Find the magnitude of \(\vec{F}\) and the direction cosine angles, \(\theta_x\text{,}\) \(\theta_y\text{,}\) and \(\theta_z\) components.

\begin{gather*}
F=\lbf{70}\\
\theta_x=\ang{73.4}\\
\theta_y=\ang{115.38}\\
\theta_z=\ang{31.0}
\end{gather*}

Since the three components of \(F\) are perpendicular, we can apply the Pythagorean Theorem to find the magnitude of \(F\text{.}\)

\begin{align*}
F= |\vec{F}| \amp= \sqrt{{F_x}^2+{F_y}^2+{F_z}^2}\\
\amp = \lbf{\sqrt{20^2 + \left( -30 \right) ^2 + 60^2}}\\
\amp = \lbf{70}
\end{align*}

Direction cosine angles are equal to the inverse cosine of each Cartesian force component divided by the force magnitude.

\begin{align*}
\theta_x \amp = \cos^{-1} \left(\frac{F_x}{|\vec{F}|}\right) = \cos^{-1}\left( \frac{20}{70}\right) = \ang{73.4}\\
\theta_y \amp = \cos^{-1} \left(\frac{F_y}{|\vec{F}|}\right) = \cos^{-1}\left( \frac{-30}{70} \right) = \ang{115.38}\\
\theta_z \amp = \cos^{-1} \left(\frac{F_z}{|\vec{F}|}\right) = \cos^{-1}\left( \frac{60}{70} \right) = \ang{31.0}
\end{align*}

Direction cosine angles for \(\vec{F}\) measured from the postive axes.

Since the direction cosine angles are measured from the positive \(x\text{,}\) \(y\text{,}\) and \(z\) axes, the negative component of \(F_y\) means that \(\theta_y > \ang{90}\text{,}\) while both \(\theta_x\) and \(\theta_z\) are less than \(\ang{90}\) as their components are positive.

In spherical coordinates, points are specified with these three coordinates

- \(r\text{,}\) the radial distance from the origin to the tip of the vector,
- \(\theta\text{,}\) the azimuthal angle, measured counter-clockwise from the positive \(x\) axis to the projection of the vector onto the \(xy\) plane, and
- \(\phi\text{,}\) the polar angle from the \(z\) axis to the vector.

What are the differences between polar coordinates and terrestrial latitude/longitude locations?

In terrestrial measurements

- Coordinate \(r\) is not needed since all points are on the surface of the globe.
- Longitude is measured \(\ang{0}\) to \(\ang{180}\) East or West of the prime meridian, rather than \(\ang{0}\) to \(\ang{360}\) counter-clockwise from the \(x\) axis.
- Latitude is measured \(\ang{0}\) to \(\ang{90}\) North or South of the equator, whereas polar angle \(\phi\) is \(\ang{0}\) to \(\ang{180}\) measured from the “North Pole”.

When vectors are specified using cylindrical coordinates the magnitude of the vector is used instead of distance \(r\) from the origin to the point.

When the two given spherical angles are defined in the manner shown here, the rectangular components of the vector \(\vec{A} = (A\ ; \theta\ ; \phi) \) are found thus:

\begin{align}
A' \amp= A \sin \phi \tag{2.4.2}\\
A_z \amp= A \cos \phi\tag{2.4.3}\\
A_x \amp= A'\cos \theta = A \sin \phi \cos \theta\tag{2.4.4}\\
A_y \amp= A'\sin \theta = A \sin \phi \sin \theta\tag{2.4.5}
\end{align}

Reflect on the equations above. Can you think through the process of how they were derived? The generalized steps are as follows. First, draw an accurate sketch of the given information and define the right triangles related to both \(\theta\) and \(\phi\text{.}\) Then use trig identities on the right triangle involving the vector, the \(z\) axis and angle \(\phi\) to find \(A_z\text{,}\) and \(A'\text{,}\) the projection of \(\vec{A}\) onto the \(xy\) plane. Finally, use trig identities on the right triangle involving vector \(\vec{A}'\) and \(\theta\) to find the remaining components of \(\vec{A}\text{.}\)

A cable force pulls on an anchor ring centered on \(x\text{,}\) \(y\text{,}\) \(z\) axis origin.

A rope pulls on an anchor ring centered at the origin with force \(\vec{F}=\lbf{\langle 20, -30, 60 \rangle}\text{.}\)

Find the *spherical* coordinates of \(\vec{F}\text{.}\)

\begin{align*}
F \amp =\lbf{70}\\
\theta \amp =\ang{-56.31} \textrm{ or } \theta=\ang{303.69} \\
\phi \amp =\ang{31}
\end{align*}

To represent \(\vec{F}\) in spherical coordinates, we must find the radial distance \(r\text{,}\) the azimuthal angle \(\theta\text{,}\) and the polar angle \(\phi\text{.}\)

Coordinate \(r\) is simply the magnitude of force \(\vec{F}.\) Since the three components of \(\vec{F}\) are perpendicular, we can apply the Pythagorean Theorem to find it.

\begin{align*}
F = |\vec{F}| = r \amp= \sqrt{F_x^2+F_y^2+F_z^2}\\
\amp = \lbf{\sqrt{20^2 + \left( -30 \right) ^2 + 60^2}}\\
\amp = \lbf{70}
\end{align*}

Azimuthal angle \(\theta\) measures the angle between the \(x\) axis and the projection of \(\vec{F}\) onto the \(xy\) plane, \(F_{zy}\text{.}\)

Using a right triangle with sides \(F_x\text{,}\) \(F_y\text{,}\) and \(F_{xy} \text{,}\) we can find \(\theta\) using the inverse tangent of the ratio of the opposite to adjacent sides.

\begin{equation*}
\theta = \tan^{-1} \frac{F_y}{F_x} = \tan^{-1}\left( \frac{-30}{20}\right) = \ang{-56.31}
\end{equation*}

This angle is negative because it is measured *clockwise* from the positive \(x\) axis, opposite the standard CCW direction.

The polar angle \(\phi\) is measured down from the \(+z\) axis to the vector \(\vec{F}\text{.}\) We can find it using a right triangle with sides \(F\text{,}\) \(F_z\text{,}\) and \(F_{xy}\text{.}\) Note that \(\phi\) is the same as the direction cosine angle \(\theta_z\text{.}\)

\begin{equation*}
\phi = \theta_z =\cos^{-1} \frac{F_z}{|\vec{F}|} = \cos^{-1}\left( \frac{60}{70} \right) = \ang{31.0}
\end{equation*}

Also notice that the azimuthal angle \(\theta\) is smaller than the direction cosine angle \(\theta_x\text{,}\) since \(\theta\) is in the \(xy\) plane, but \(\theta_x\) is a 3D angle from the \(x\) axis to the vector \(\vec{F}.\)

The cylindrical coordinate system is seldom used in statics, however, it is useful in certain geometries. Cylindrical coordinates extend two-dimensional polar coordinates by adding a \(z\) coordinate indicating the distance above or below the \(xy\) plane.

Points are specified with these three cylindrical coordinates.

- \(r\text{,}\) the radius of the cylinder. This is the distance from the origin to the projection of the tip of the vector onto the \(xy\) plane,
- \(\theta\text{,}\) the azimuthal angle, measured counter-clockwise from the positive \(x\) axis to the projection of the vector onto the \(xy\) plane
- \(z\text{,}\) the vertical height of the vector tip.

Unfortunately, not all problems give the angles \(\theta\) and \(\phi\) as defined here; so you will need to find them from the given angles in other situations.

You can use the interactive diagram in this section to practice visualizing and finding the components of a vector in all of these coordinate systems. You should be able to find the \(x\text{,}\) \(y\text{,}\) and \(z\) coordinates given direction angles, spherical or cylindrical coordinates, and vice-versa.

A cable force pulls on an anchor ring centered on \(x\text{,}\) \(y\text{,}\) \(z\) axis origin.

A rope pulls on an anchor ring centered at the origin with force \(\vec{F}=\lbf{\langle 20, -30, 60 \rangle}\text{.}\)

Find the *cylindrical* coordinates of \(\vec{F}\text{.}\)

\begin{align*}
F \amp =\lbf{36.06}\\
\theta \amp =\ang{-56.31} \textrm{ or } \theta=\ang{303.69} \\
z \amp =\lbf{60}
\end{align*}

To represent \(\vec{F}\) in cylindrical coordinates, we must find the radial distance, \(r\text{,}\) the azimuthal angle, \(\theta\text{,}\) and the axial coordinate, \(z\text{.}\)

In cylindrical coordinates, \(r\) is the radius of the cylinder rather than the radius of the enclosing sphere. \(r\) is the projection of \(\vec{F}\) onto the \(xy\) plane, \(F_{xy}\text{,}\) and can be found by applying the Pythagorean Theorem to the \(x\) and \(y\) components of \(\vec{F}\text{.}\)

\begin{align*}
r = F_{xy} \amp = \sqrt{F_x^2+F_y^2}\\
\amp = \lbf{\sqrt{20^2 + \left( -30 \right) ^2}}\\
\amp = \lbf{36.06}
\end{align*}

The azimuthal angle \(\theta\) is the same in both cylindrical and spherical coordinates. It measures the angle between the \(x\) axis and the projection of \(\vec{F}\) onto the \(xy\) plane. \(\theta\) can be found using a right triangle in the \(xy\) plane with sides \(F_x\) and \(F_y\text{.}\)

\begin{equation*}
\theta = \tan^{-1} \frac{F_y}{F_x} = \tan^{-1}\left( \frac{-30}{20}\right) = \ang{-56.31}
\end{equation*}

Finally, the \(z\) component is the vertical component of the force, \(F_z\text{,}\) which was given.

\begin{equation*}
F_z = \lbf{60.0}
\end{equation*}