Projections

As we have seen, addition combines two vectors to create a resultant vector. But what if we are given a vector and we need to find its component parts? We use vector projections to perform the opposite process; they can break down a vector into its components. The magnitude of a vector projection is a scalar projection. For example, if a child is pulling the handle of a wagon at a [latex]55^\circ[/latex] angle, we can use projections to determine how much of the force on the handle is actually moving the wagon forward (Figure 1). We return to this example and learn how to solve it after we see how to calculate projections.

Figure 2. The projection of [latex]{\bf{v}}[/latex] onto [latex]{\bf{u}}[/latex] shows the component of vector [latex]{\bf{v}}[/latex] in the direction of [latex]{\bf{u}}[/latex].

Sometimes it is useful to decompose vectors—that is, to break a vector apart into a sum. This process is called the resolution of a vector into components. Projections allow us to identify two orthogonal vectors having a desired sum. For example, let [latex]\mathbf{v} = \langle 6,-4 \rangle[/latex] and let [latex]\mathbf{u} = \langle 3,1 \rangle[/latex]. We want to decompose the vector [latex]\mathbf{v}[/latex] into orthogonal components such that one of the component vectors has the same direction as [latex]\mathbf{u}[/latex].

We first find the component that has the same direction as [latex]\mathbf{u}[/latex] by projecting [latex]\mathbf{v}[/latex] onto [latex]\mathbf{u}[/latex]. Let [latex]\mathbf{p} = \text{proj}_\text{u}\mathbf{v}[/latex]. Then, we have

Now consider the vector [latex]\mathbf{q} = \mathbf{v} - \mathbf{p}[/latex]. We have

Clearly, by the way we defined [latex]\mathbf{q}[/latex], we have [latex]\mathbf{v} = \mathbf{q} + \mathbf{p}[/latex], and

Therefore, [latex]\mathbf{q}[/latex] and [latex]\mathbf{p}[/latex] are orthogonal.

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

Example: scalar projection of velocity

A container ship leaves port traveling [latex]15^\circ[/latex] north of east. Its engine generates a speed of [latex]20[/latex] knots along that path (see the following figure). In addition, the ocean current moves the ship northeast at a speed of 2 knots. Considering both the engine and the current, how fast is the ship moving in the direction [latex]15^\circ[/latex] north of east? Round the answer to two decimal places.

Figure 3. Find how fast the ship is moving in the direction of [latex]15^{\circ}[/latex] north of east.

Show Solution

Let [latex]\mathbf{v}[/latex] be the velocity vector generated by the engine, and let [latex]\mathbf{w}[/latex] be the velocity vector of the current. We already know [latex]||\mathbf{v}|| = 20[/latex] along the desired route. We just need to add in the scalar projection of [latex]\mathbf{w}[/latex] onto [latex]\mathbf{v}[/latex]. We get

[latex]\begin{align*} \text{comp}_\text{v}\mathbf{w} &= \frac{\mathbf{v} \cdot \mathbf{w}}{||\mathbf{v}||}\\ &= \frac{||\mathbf{v}|| \, ||\mathbf{w}||\cos{(30^\circ)}}{||\mathbf{v}||}\\ &=||\mathbf{w}||\cos{(30^\circ)} \\ &= 2\frac{\sqrt{3}}{2} = \sqrt{3} \approx 1.73 \text{ knots.}\\\end{align*}[/latex]

The ship is moving at 21.73 knots in the direction [latex]15^\circ[/latex] north of east.

try it

Repeat the previous example, but assume the ocean current is moving southeast instead of northeast, as shown in the following figure.

Figure 4. Find how fast the ship is moving in the direction [latex]45^{\circ}[/latex] south of east.

Show Solution

[latex]21[/latex] knots.

Work

Now that we understand dot products, we can see how to apply them to real-life situations. The most common application of the dot product of two vectors is in the calculation of work.

From physics, we know that work is done when an object is moved by a force. When the force is constant and applied in the same direction the object moves, then we define the work done as the product of the force and the distance the object travels: [latex]W=Fd[/latex]. We saw several examples of this type in earlier chapters. Now imagine the direction of the force is different from the direction of motion, as with the example of a child pulling a wagon. To find the work done, we need to multiply the component of the force that acts in the direction of the motion by the magnitude of the displacement. The dot product allows us to do just that. If we represent an applied force by a vector F and the displacement of an object by a vector s, then the work done by the force is the dot product of F and s.

Let’s revisit the problem of the child’s wagon introduced earlier. Suppose a child is pulling a wagon with a force having a magnitude of [latex]8[/latex] lb on the handle at an angle of [latex]55^\circ[/latex]. If the child pulls the wagon [latex]50[/latex] ft, find the work done by the force (Figure 5).

Figure 5. The horizontal component of the force is the projection of [latex]{\bf{F}}[/latex] onto the positive [latex]x[/latex]-axis.

We have

In U.S. standard units, we measure the magnitude of force [latex]||\mathbf{F}||[/latex] in pounds. The magnitude of the displacement vector [latex]||\overrightarrow{PQ}||[/latex] tells us how far the object moved, and it is measured in feet. The customary unit of measure for work, then, is the foot-pound. One foot-pound is the amount of work required to move an object weighing [latex]1[/latex] lb a distance of [latex]1[/latex] ft straight up. In the metric system, the unit of measure for force is the newton (N), and the unit of measure of magnitude for work is a newton-meter (N·m), or a joule (J).

try it

A constant force of [latex]30[/latex] lb is applied at an angle of [latex]60^\circ[/latex] to pull a handcart [latex]10[/latex] ft across the ground (Figure 6). What is the work done by this force?

Figure 6. Find the force [latex]{\bf{F}}[/latex].

Show Solution

[latex]150[/latex] ft-lb