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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.

If a unit vector is multiplied by a scalar value, the vector is scaled by that amount, so for instance if a unit vector pointing to the right is multiplied by \(\N{ 100}\) the result is a \(\N{100}\) vector pointing to the right, if a unit vector point up is multiplied by \(\N{ -50}\) the result is a \(\N{50}\) vector pointing down.

Subsection 2.5.1 Cartesian Unit Vectors

Because they occur frequently, the unit vectors in each of the three Cartesian coordinate directions are given their own symbol, which are:

  • The unit vector pointing in the \(x\) direction is \(\ihat\text{,}\)
  • The unit vector pointing in the \(y\) direction is \(\jhat\text{,}\) and
  • The unit vector pointing in the \(z\) direction is \(\khat\text{.}\)

Unit vectors may sound like a new concept, but you have been using unit vector components since you learned your trigonometric identities of sine and cosine.

Lets use the equation (2.3.10) above:

\begin{equation*} \vec{F}=F \left \langle \cos \theta, \sin \theta \right \rangle \text{.} \end{equation*}

Now focus on the vector components inside the angled brackets in the equation above. Draw a right triangle with the perpendicular sides horizontal and vertical and a hypotenuse of one. Insert the angle \(\theta\) between the horizontal side and the hypotenuse. What equations do you find for the horizontal and vertical sides? You should discover that \(\cos \theta\) and \(\sin \theta\) are the horizontal and vertical components of the unit vector along \(\vec{F}\text{.}\)

To further prove this fact, let us look at unit vectors in the context of the Pythagorean Theorem. The Pythagorean Theorem tells us that the sum of the squares of the length of both orthogonal legs of a right triangle must equal the square of the length of the hypotenuse. So, if \(\cos \theta\) and \(\sin \theta\) are the orthogonal components of the unit vector \(\hat{\vec{F}}\) and the length of the unit vector is one, then putting these values into the Pythagorean Theorem gives us:

\begin{equation*} \cos^2 \theta + \sin^2 \theta = 1 \end{equation*}

Reflect for a few seconds if the equation above looks familiar. This is the equation for a circle centered at the origin with a radius of one unit. Hence, no matter what the angle \(\theta\) from a horizontal axis to a line is, the terms \(\cos \theta\) and \(\sin \theta\) make up the components of the unit vector along the line. Note that if the angle \(\theta\) is measured from a vertical axis, then the sine and cosine flip, with \(\sin \theta\) defining the horizontal component and the \(\cos\ \theta\) defining the vertical component of the unit vector.

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

Figure 2.5.1. Unit Vectors

Given the properties of unit vectors, there are a couple of conceptual checks you can make after computing unit vector components which can save you subsequent errors.

  1. All 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 just one unit in magnitude.
  2. 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 2.5.2 Unit Vectors and Direction Cosines

Looking closely at the right side of equation (2.4.1), notice that each equation consists of a component divided by the total vector magnitude, which you may recall are the same equations used for 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{2.5.1} \end{equation}

Subsection 2.5.3 Unit Vectors Using Components

Unit vectors are not limited to the \(x\text{,}\) \(y\text{,}\) and \(z\) coordinate directions; you can find a unit vector in any direction. To find the unit vector \(\widehat{\vec{F}}\) in the direction of an arbitrary vector \(\vec{F}\text{.}\)

  1. Define the vector in component form
    \begin{equation*} \vec{F} = \left \langle F_x, F_y \right \rangle\text{,} \end{equation*}
  2. Compute the magnitude of the vector using the Pythagorean Theorem
    \begin{equation*} F = | \vec{F} | = \sqrt {{F_x}^2 + {F_y}^2}\text{,} \end{equation*}
  3. Divide the components by the overall length
    \begin{equation*} \widehat{\vec{F}}=\frac{\vec{F}}{| \vec{F} | } = \frac{\left \langle F_x, F_y \right \rangle}{F}\text{.} \end{equation*}

To again emphasize that unit vectors are pure direction, we can use the definitions of vectors and scalars to show that dividing a vector by its magnitude just leaves the direction.

\begin{equation*} \text{unit vector} = \frac{\text{vector}}{\text{magnitude}}=\frac{\cancel{\text{magnitude}} \cdot\text{direction}}{\cancel{\text{magnitude}}} = \text{direction} \end{equation*}