If you earn $50 on Saturday and $60 on Sunday, your net income for the weekend is the sum of $50 and $60. With numbers, the word net implies addition. The same is true with vectors. For example, FIGURE 3.3 shows the displacement of a hiker who first hikes 4 miles to the east, then 3 miles to the north. The first leg of the hike is described by the displacement $\vec{A}=(4\text{mi,}\text{east})$. The second leg of the hike has displacement $\vec{B}=(3\text{mi,}\text{north})$. Vector $\vec{C}$ is the net displacement because it describes the net result of the hiker’s first having displacement $\vec{A}$, then displacement $\vec{B}$.

FIGURE 3.3 The net displacement $\vec{C}$ resulting from two displacements $\vec{A}$ and $\vec{B}$.

The net displacement $\vec{C}$ is an initial displacement $\vec{A}$ plus a second displacement $\vec{B}$, or

The sum of two vectors is called the resultant vector. It’s not hard to show that vector addition is commutative: $\vec{A}+\vec{B}=\vec{B}+\vec{A}$. That is, you can add vectors in any order you wish.

《 TACTICS BOX 1.1 showed the three-step procedure for adding two vectors, and it’s highly recommended that you turn back for a quick review. This tip-to-tail method for adding vectors, which is used to find $\vec{C}=\vec{A}+\vec{B}$ in Figure 3.3, is called graphical addition. Any two vectors of the same type—two velocity vectors or two force vectors—can be added in exactly the same way.

The graphical method for adding vectors is straightforward, but we need to do a little geometry to come up with a complete description of the resultant vector $\vec{C}$. Vector $\vec{C}$ of Figure 3.3 is defined by its magnitude C and by its direction. Because the three vectors $\vec{A}$, $\vec{B}$, and $\vec{C}$ form a right triangle, the magnitude, or length, of $\vec{C}$ is given by the Pythagorean theorem:

Notice that Equation 3.2 uses the magnitudes A and B of the vectors $\vec{A}$ and $\vec{B}$. The angle $\theta$, which is used in Figure 3.3 to describe the direction of $\vec{C}$, is easily found for a right triangle:

Altogether, the hiker’s net displacement is $\vec{C}=\vec{A}+\vec{B}=(5\text{mi,}37\text{°}\text{north}\text{of}\text{east})$.

EXAMPLE 3.1 Using graphical addition to find a displacement

A bird flies 100 m due east from a tree, then 50 m northwest (that is, $45\text{°}$ north of west). What is the bird’s net displacement?

VISUALIZE FIGURE 3.4 shows the two individual displacements, which we’ve called $\vec{A}$ and $\vec{B}$. The net displacement is the vector sum $\vec{C}=\vec{A}+\vec{B}$, which is found graphically.

FIGURE 3.4The bird’s net displacement is $\vec{C}=\vec{A}+\vec{B}$

FIGURE 3.4 Full Alternative Text

SOLVE The two displacements are $\vec{A}=(100\text{m,}\text{east})$ and $\vec{B}=$ $\stackrel{}{}\stackrel{}{(50\text{m, northwest).}}$ The net displacement $\vec{C}=\vec{A}+\vec{B}$ is found by drawing a vector from the initial to the final position. But describing $\vec{C}$ is a bit trickier than the example of the hiker because $\vec{A}$ and $\vec{B}$ are not at right angles. First, we can find the magnitude of $\vec{C}$ by using the law of cosines from trigonometry:

$$\begin{array}{ccc}\hfill C^2& \hfill =\hfill & A^2+B^2-2AB\text{cos}45\text{°}\hfill \\ \hfill & \hfill =\hfill & (100\text{m}{)}^2+{\left(50\text{m}\right)}^2-2\left(100\text{m}\right)\left(50\text{m}\right)\text{cos}45\text{°}\hfill \\ \hfill & \hfill =\hfill & 5430{\text{m}}^2\hfill \end{array}$$

Thus $C=\sqrt{5430{\text{m}}^2}=74\text{m}$. Then a second use of the law of cosines can determine angle $\varphi$ (the Greek letter phi):

$$\begin{array}{l}{B}^2=A^2+C^2-2AC\text{ cos }\varphi \hfill \\ \varphi ={\text{cos}}^{-1}\left[{\displaystyle \frac{A^2+C^2-B^2}{2AC}}\right]=29\text{°}\hfill \end{array}$$

The bird’s net displacement is

$$\vec{C}=(74\text{m,}29\text{°}\text{north}\text{of}\text{east})$$

It is often convenient to draw two vectors with their tails together, as shown in FIGURE 3.5a. To evaluate $\vec{D}+\vec{E}$, you could move vector $\vec{E}$ over to where its tail is on the tip of $\vec{D}$, then use the tip-to-tail rule of graphical addition. That gives vector $\vec{F}=\vec{D}+\vec{E}$ in Figure 3.5b. Alternatively, Figure 3.5c shows that the vector sum $\vec{D}+\vec{E}$ can be found as the diagonal of the parallelogram defined by $\vec{D}$ and $\vec{E}$. This method for vector addition is called the parallelogram rule of vector addition.

FIGURE 3.5 Two vectors can be added using the tip-to-tail rule or the parallelogram rule.

Vector addition is easily extended to more than two vectors. FIGURE 3.6 shows the path of a hiker moving from initial position 0 to position 1, then position 2, then position 3, and finally arriving at position 4. These four segments are described by displacement vectors $\vec{D}{}_1\text{,}\vec{D}{}_2\text{,}\vec{D}{}_3$, and $\vec{D}{}_4$. The hiker’s net displacement, an arrow from position 0 to position 4, is the vector $\vec{D}{}_{\text{net}}$. In this case,

The vector sum is found by using the tip-to-tail method three times in succession.

FIGURE 3.6 The net displacement after four individual displacements.