Unit Vectors

A unit vector is a vector with magnitude $1$. For any nonzero vector ${\bf{v}}$, we can use scalar multiplication to find a unit vector ${\bf{u}}$ that has the same direction as ${\bf{v}}$. To do this, we multiply the vector by the reciprocal of its magnitude:

Recall that when we defined scalar multiplication, we noted that $||k{\bf{v}}|| = |k| \cdot ||{\bf{v}}||$. For ${\bf{u}} = \frac{1}{||{\bf{v}}||}{\bf{v}}$, it follows that $||{\bf{u}}|| = \frac{1}{||{\bf{v}}||}(||{\bf{v}}||) = 1$. We say that ${\bf{u}}$ is the unit vector in the direction of ${\bf{v}}$ (Figure 18). The process of using scalar multiplication to find a unit vector with a given direction is called normalization.

Figure 1. The vector ${\bf{v}}$ and associated unit vector ${\bf{u}}=\frac{1}{||{\bf{v}}||}{\bf{v}}$. In this case, $||{\bf{v}}||>1$.

We have seen how convenient it can be to write a vector in component form. Sometimes, though, it is more convenient to write a vector as a sum of a horizontal vector and a vertical vector. To make this easier, let’s look at standard unit vectors. The standard unit vectors are the vectors ${\bf{i}} = \langle 1,0 \rangle$ and ${\bf{j}} = \langle 0,1 \rangle$ (Figure 2.18).

Figure 2. The standard unit vectors ${\bf{i}}$ and ${\bf{j}}$.

By applying the properties of vectors, it is possible to express any vector in terms of ${\bf{i}}$ and ${\bf{j}}$ in what we call a linear combination:

Thus, ${\bf{v}}$ is the sum of a horizontal vector with magnitude $x$, and a vertical vector with magnitude $y$, as in the following figure.

Figure 3. The vector ${\bf{v}}$ is the sum of $x{\bf{i}}$ and $y{\bf{j}}$.

Example: Using Standard Unit Vectors

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$||{\bf{w}}|| = \langle 3,-4 \rangle = 3{\bf{i}} - 4{\bf{j}}$

$\begin{array}{ccc}\hfill {\bf{u}} & =\hfill & \langle \cos{60°}, \sin{60°} \rangle \hfill \\ \hfill & =\hfill & \langle \frac{1}{2}, \frac{\sqrt{3}}{2} \rangle \hfill \\ \hfill & =\hfill & \frac{1}{2}{\bf{i}} - \frac{\sqrt{3}}{2}{\bf{j}}\end{array}$.

Figure 4. The terminal point of ${\bf{u}}$ lies on the unit circle $(\cos{\theta},\sin{\theta})$.

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Applications of Vectors

Because vectors have both direction and magnitude, they are valuable tools for solving problems involving such applications as motion and force. Recall the boat example and the quarterback example we described earlier. Here we look at two other examples in detail.

Example: Finding Resultant Force

Jane’s car is stuck in the mud. Lisa and Jed come along in a truck to help pull her out. They attach one end of a tow strap to the front of the car and the other end to the truck’s trailer hitch, and the truck starts to pull. Meanwhile, Jane and Jed get behind the car and push. The truck generates a horizontal force of $300$ lb on the car. Jane and Jed are pushing at a slight upward angle and generate a force of $150$ lb on the car. These forces can be represented by vectors, as shown in Figure 22. The angle between these vectors is $15°$. Find the resultant force (the vector sum) and give its magnitude to the nearest tenth of a pound and its direction angle from the positive $x$-axis.

Figure 5. Two forces acting on a car in different directions.

Show Solution

To find the effect of combining the two forces, add their representative vectors. First, express each vector in component form or in terms of the standard unit vectors. For this purpose, it is easiest if we align one of the vectors with the positive $x$-axis. The horizontal vector, then, has initial point $(0,0)$ and terminal point $(300,0)$. It can be expressed as $\langle 300, 0\rangle$ or $300{\bf{i}}$.The second vector has magnitude $150$ and makes an angle of $15°$ with the first, so we can express it as $\langle 150\cos({15°}), 150\sin({15°}) \rangle$, or $150\cos({15°}){\bf{i}} + 150\sin({15°}){\bf{j}}$. Then, the sum of the vectors, or resultant vector, is ${\bf{r}} = \langle 300,0 \rangle + \langle 150\cos({15°}), 150\sin({15°}) \rangle$ , and we have

$\begin{array}{ccc}\hfill {\bf{r}} & =\hfill \sqrt{(300 + 150\cos({15°}))^2 + (150\sin({15°}))^2} \hfill \\ \hfill & \approx 446.6 \end{array}$

The angle $θ$ made by ${\bf{r}}$ and the positive $x$-axis has $\tan{θ} = \frac{150\sin({15°})}{(300 + 150\cos({15°}))} \approx 0.09$, so $θ \approx \tan^{-1}{0.09} \approx 5°$, which means the resultant force ${\bf{r}}$ has an angle of $5°$ above the horizontal axis.

Example: Finding Resultant Velocity

An airplane flies due west at an airspeed of $425$ mph. The wind is blowing from the northeast at $40$ mph. What is the ground speed of the airplane? What is the bearing of the airplane?

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Let’s start by sketching the situation described (Figure 23).

Figure 5. Initially, the plane travels due west. The wind is from the northeast, so it is blowing to the southwest. The angle between the plane’s course and the wind is $45^{\circ}$ (Figure not drawn to scale.)

Set up a sketch so that the initial points of the vectors lie at the origin. Then, the plane’s velocity vector is ${\bf{p}} = -425{\bf{i}}$. The vector describing the wind makes an angle of $225°$ with the positive $x$-axis:

${\bf{w}} = \langle 40\cos({225°}), 40\sin({225°}) \rangle = \langle -\frac{40}{\sqrt{2}} - \frac{40}{\sqrt{2}} \rangle = - \frac{40}{\sqrt{2}}{\bf{i}} - \frac{40}{\sqrt{2}}{\bf{j}}$.

When the airspeed and the wind act together on the plane, we can add their vectors to find the resultant force:

${\bf{p}} + {\bf{w}} = -425{\bf{i}} + (- \frac{40}{\sqrt{2}}{\bf{i}} - \frac{40}{\sqrt{2}}{\bf{j}}) = (-425 - \frac{40}{\sqrt{2}}){\bf{i}} - \frac{40}{\sqrt{2}}{\bf{j}}$.

The magnitude of the resultant vector shows the effect of the wind on the ground speed of the airplane:

$||{\bf{p}} + {\bf{w}}|| = \sqrt{(-425 - \frac{40}{\sqrt{2}})^2 + (-\frac{40}{\sqrt{2}})^2} \approx 454.17$mph

As a result of the wind, the plane is traveling at approximately 454 mph relative to the ground.To determine the bearing of the airplane, we want to find the direction of the vector ${\bf{p}} + {\bf{w}}$:

$\begin{array}{ccc}\hfill \tan{θ} & =\hfill & \frac{-\frac{40}{\sqrt{2}}}{(-425 - \frac{40}{\sqrt{2}})} \approx 0.06 \hfill \\ \hfill θ & \approx \hfill &3.57° \end{array}$

The overall direction of the plane is $3.57°$ south of west.

Watch the following video to see the worked solution to the above Try IT.