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Engineering Statics: Open and Interactive

Section 2.5 Unit Vectors

A unit vector is a vector with a magnitude of one and no units. As such, a unit vector represents a pure direction. By convention, a unit vector is indicated by a hat over a vector symbol. This may sound like a new concept, but it’s a simple one, directly related to the unit circle, the Pythagorean Theorem, and the definitions of sine and cosine.

Subsection 2.5.1 Cartesian Unit Vectors

A unit vector can point in any direction, but because they occur so frequently the unit vectors in each of the three Cartesian coordinate directions are given their own symbols, which are:
  • \(\ihat\text{,}\) for the unit vector pointing in the \(x\) direction,
  • \(\jhat\text{,}\) for the unit vector pointing in the \(y\) direction, and
  • \(\khat\text{,}\) for the unit vector pointing in the \(z\) direction..


This interactive shows the graphical relations between the unit vector \(\hat{\vec{F}}\) and the unit circle. \(\ihat\) and \(\jhat\) are the standard unit vectors in the \(x\) and \(y\) directions.
Figure 2.5.1. Unit Vector Interactive
Applying the Pythagorean Theorem to the triangle gives the equation for a unit circle
\begin{equation*} \cos^2 \theta + \sin^2 \theta = 1^2 \end{equation*}
No matter what angle a unit vector makes with the \(x\) axis, \(\cos \theta\) and \(\sin \theta\) are its scalar components. This relation assumes that the angle \(\theta\) is measured from the \(x\) axis, if it is measured from the \(y\) axis the sine and cosine functions reverse, with \(\sin \theta\) defining the horizontal component and the \(\cos\ \theta\) defining the vertical component.
The \(x\) and \(y\) components of a point on the unit circle are also the scalar components of \(\hat{\vec{F}}\text{,}\) so
\begin{equation*} \left.\begin{aligned} F_x \amp = \cos \theta\\ F_y \amp = \sin \theta\\ \end{aligned}\right\} \implies \hat{\vec{F}} = \langle \cos \theta, \sin \theta \rangle\text{.} \end{equation*}

Subsection 2.5.2 Relation between Vectors and Unit Vectors

When a purely-directional unit vector is multiplied by a scalar value it is scaled by that amount. For example, when a unit vector pointing to the right is multiplied by \(\N{ 100}\) the result is a \(\N{100}\) vector pointing to the right. When a unit vector pointing up is multiplied by \(\N{ -50}\text{,}\) the negative magnitude flips the direction of the unit vector and the result is a \(\N{50}\) vector pointing down.
In general,
\begin{equation} \vec{F} = |\vec{F}|\ \hat{\vec{F}}\text{,}\tag{2.5.1} \end{equation}
where \(|\vec{F}|\) is the magnitude of vector \(\vec{F}\text{,}\) and \(\hat{\vec{F}}\) is the unit vector pointing in the direction of \(\vec{F}\text{.}\)
Solving equation (2.5.1) for \(\hat{\vec{F}}\) gives the approach to find the unit vector of known vector \(\vec{F}\text{.}\)
The process is straightforward— divide the vector by its magnitude.
\begin{equation} \hat{\vec{F}}=\frac{\vec{F}}{| \vec{F} | }\tag{2.5.2} \end{equation}
To emphasize that unit vectors are pure direction, recall that vectors consist of both a magnitude and direction, so when we divide a vector by its own magnitude we are just left with direction.
\begin{equation*} \textrm{unit vector} = \frac{\vec{F}}{| \vec{F} | } = \frac{\textrm{[vector]}}{\textrm{[magnitude]}}=\frac{\cancel{\textrm{[magnitude]}} \cdot\textrm{[direction]}}{\cancel{\textrm{[magnitude]}}} = \textrm{[direction]} \end{equation*}
This interactive shows vector \(\vec{F}\text{,}\) its associated unit vector \(\hat{\vec{F}}\text{,}\) and expressions for \(\vec{F}\) in terms of its unit vector \(\hat{\vec{F}}\text{.}\)


This interactive shows the graphical expression and component values of force \(\vec{F}\) and its unit vector \(\hat{\mathbf{F}}\text{.}\)
Figure 2.5.2. Unit Vectors

Example 2.5.3. Find unit vector of a force.

Find the unit vector corresponding to a \(\N{100}\) force at 60° above the positive \(x\)-axis.
\begin{equation*} \hat{\vec{F}} = (1\; ; 60°) = \langle \cos 60°, \sin 60° \rangle \end{equation*}
In polar coordinates, the unit vector is a vector of magnitude 1, pointing in the same direction as the force, so, by inspection
\begin{align*} \vec{F} \amp = (\N{100}\, ; 60°)\\ \hat{\vec{F}} \amp = (1\, ; 60°) \end{align*}
In rectangular coordinates, first express \(\vec{F}\) in terms of its \(x\) and \(y\) components.
\begin{equation*} \left.\begin{aligned} F_x \amp = F \cos 60°\\ F_y \amp = F \sin 60°\\ \end{aligned}\right\} \implies \vec{F} = \langle F \cos 60°, F \sin 60° \rangle\text{.} \end{equation*}
\begin{equation*} \hat{\vec{F}} = \frac{\vec{F}}{F} = \frac{ \langle \N{100} \cos 60°, \N{100} \sin 60° \rangle} {\N{100}} = \langle \cos 60°, \sin 60° \rangle \end{equation*}

Subsection 2.5.3 Force Vectors from Position Vectors

Unit vectors are generally the best approach when working with forces and distances in three dimensions.
For example, when the locations of two points on the line of action of a force are known, the unit vector of the line of action can be found and used to determine the components of the force acting along that line. This can be accomplished as follows, where \(A\) and \(B\) are points on the line of action.
  1. Use the problem geometry to find \(\vec{AB}\text{,}\) the displacement vector from point \(A\) to point \(B\text{.}\)
    You can either subtract the coordinates of the starting point \(A\) from the coordinates of the destination point \(B\text{,}\)
    \begin{align*} A \amp = \left(A_x,A_y,A_z \right)\\ B \amp = \left(B_x,B_y,B_z \right)\\ \vec{AB} \amp = \left(B_x-A_x \right )\ihat+\left(B_y-A_y \right )\jhat+\left(B_z-A_z \right )\khat \textrm{, or} \end{align*}
    or, write the displacements directly by noting the change in the \(x\text{,}\) \(y\text{,}\) and \(z\) coordinates when moving from \(A\) to \(B\text{.}\)
    \begin{align*} AB_x \amp = \Delta x = B_x - A_x\\ AB_y \amp = \Delta y = AB_y = B_y - A_y\\ AB_z \amp = \Delta z = B_z - A_x\\ \vec{AB} \amp = AB_x\ \ihat+ AB_y\ \jhat+ AB_z\ \khat \end{align*}
    The result is the same with either method.
  2. Find the distance between point \(A\) and point \(B\) using the Pythagorean Theorem. This distance is also the magnitude of \(\vec{AB}\) or \(|\vec{AB}|\text{.}\)
    \begin{equation*} \left|\vec{AB}\right |=\sqrt{(AB_x)^2+(AB_y)^2+(AB_z)^2} \end{equation*}
  3. Find \(\widehat{\vec{AB}}\text{,}\) the unit vector from \(A\) to \(B\text{,}\) by dividing vector \(\vec{AB}\) by its magnitude. This is a unitless vector with a magnitude of 1 which points from \(A\) to \(B\text{.}\)
    \begin{equation*} \widehat{\vec{AB}}= \left \langle \frac{AB_x}{|\vec{AB}|},\frac{AB_y}{|\vec{AB}|},\frac{AB_z}{|\vec{AB}|} \right \rangle \end{equation*}
  4. Finally, multiply the magnitude of the force by the unit vector \(\widehat{\vec{AB}}\) to get force \(\vec{F}_{AB}.\)
    \begin{align*} \vec{F}_{AB} \amp = F_{AB} \; \widehat{\vec{AB}}\\ \amp = F_{AB} \left \langle \frac{AB_x}{|\vec{AB}|},\frac{AB_y}{|\vec{AB}|},\frac{AB_z}{|\vec{AB}|}\right \rangle \end{align*}
The interactive below can be used to visualize the displacement vector and its unit vector, and practice this procedure.


This interactive shows \(\vec{r}_{AB}\text{,}\) the displacement vector from \(\vec{A}\) to \(\vec{B}\) and the corresponding unit displacement vector \(\lambda_{AB}\text{.}\)
You may change \(\vec{A}\) and \(\vec{B}\) by moving the red and blue dots. Click the dot to switch between \(x\)-\(y\) mode and \(z\) mode. Coordinates of \(\vec{A}\) and \(\vec{B}\) can also be entered into the table directly.
Figure 2.5.4. Unit Vectors in Space

Example 2.5.5. Component in a Specified Direction.

Determine the components of a \(\kN{5}\) force \(\vec{F}\) acting at point \(A\text{,}\) in the direction of a line from \(A\) to \(B\text{.}\) Given: \(A =\m{ \left ( 2,3,-2.1 \right )}\) and \(B = \m{\left ( -2.5, 1.5, 2.2 \right )}\)
We will take the solution one step at a time.
Draw a good diagram.
The interactive in Figure 2.5.4 may be useful for this problem.
Find the displacement vector from \(A\) to \(B\text{.}\)
\begin{align*} \vec{AB} \amp=\m{\langle-4.5,-1.5,4.3\rangle} \end{align*}
\begin{align*} \vec{AB} \amp =\left(B_x-A_x \right )\ihat+\left(B_y-A_y \right )\jhat+\left(B_z-A_z \right )\khat\\ \amp =\m{\left [ \left(-2.5-2 \right )\ihat+\left(1.5-3 \right )\jhat+\left(2.2-(-2.1) \right )\khat \right ]}\\ \amp =\m{ \left(-4.5\ihat-1.5\jhat+4.3 \khat \right )}\\ \amp=\m{\langle-4.5,-1.5,4.3\rangle} \end{align*}
Find the magnitude of the displacement vector.
\begin{align*} \left|\vec{AB}\right |\amp = \m{6.402} \end{align*}
\begin{align*} \left|\vec{AB}\right |\amp =\sqrt{(\Delta_x)^2+(\Delta_y)^2+(\Delta_z)^2}\\ \amp =\sqrt{\m{(-4.5)^2+(-1.5)^2+4.3^2 }^2}\\ \amp =\sqrt{40.99 \m{}^2 }\\ \amp = \m{6.402} \end{align*}
Find the unit vector pointing from \(A\) to \(B\text{.}\)
\begin{align*} \widehat{\vec{AB}} \amp =\left\langle -0.7,-0.23,0.67\right \rangle \end{align*}
\begin{align*} \widehat{\vec{AB}}\amp= \left \langle \frac{\Delta_x}{|\vec{AB}|},\frac{\Delta_y}{|\vec{AB}|},\frac{\Delta_z}{|\vec{AB}|} \right \rangle \\ \amp =\left \langle \frac{-4.5}{6.402},\frac{-1.5}{6.402},\frac{4.3}{6.402}\right \rangle\\ \widehat{\vec{AB}} \amp =\left\langle -0.7,-0.23,0.67\right \rangle \end{align*}
Find the force vector.
\begin{gather*} \vec{F}_{AB}= \kN{ \left \langle -3.51,-1.17,3.36 \right \rangle } \end{gather*}
\begin{align*} \vec{F}_{AB} \amp = F_{AB} \; \widehat{\vec{AB}}\\ \amp =\kN{5} \left \langle -0.7,-0.23,0.67\right \rangle\\ \amp = \kN{ \left \langle -3.51,-1.17,3.36 \right \rangle } \end{align*}
Given the properties of unit vectors, there are some conceptual checks you can make after computing unit vector components which can prevent subsequent errors.
  • The signs of unit vector components need to match the signs of the original position vector. A unit vector has the same line of action and sense as the position vector but is scaled down to one unit in magnitude.
  • Components of a unit vector must be between -1 and 1. If the magnitude of a unit vector is one, then it is impossible for it to have rectangular components larger than one.

Subsection 2.5.4 Unit Vectors and Direction Cosines

If you look closely at the right side of equation (2.4.1), you will see that each equation consists of a component divided by the total vector magnitude. These are the same equations just used to find unit vectors. Thus, the cosine of each direction cosine angle collectively also computes the components of the unit vector; hence we can write an equation for \(\hat{\vec{A}}\text{,}\)i.e., the unit vector along \(\vec{A}\text{.}\)
\begin{equation*} \hat{\vec{A}}=\cos \theta_x\ \ihat +\cos \theta_y\ \jhat + \cos \theta_z\ \khat \end{equation*}
Combining the Pythagorean Theorem with our knowledge of unit vectors and direction cosine angles gives this result: if you know two of the three direction cosine angles you can manipulate the following equation to find the third.
\begin{equation} \cos^2 \theta_x + \cos^2 \theta_y +\cos^2 \theta_z = 1\tag{2.5.3} \end{equation}