2.1 Vectors in the Plane

Vector Representation

A vector in a plane is represented by a directed line segment (an arrow). The endpoints of the segment are called the initial point and the terminal point of the vector. An arrow from the initial point to the terminal point indicates the direction of the vector. The length of the line segment represents its magnitude. We use the notation $\Vert \text{v}\Vert$ to denote the magnitude of the vector $\text{v}.$ A vector with an initial point and terminal point that are the same is called the zero vector, denoted $0.$ The zero vector is the only vector without a direction, and by convention can be considered to have any direction convenient to the problem at hand.

Vectors with the same magnitude and direction are called equivalent vectors. We treat equivalent vectors as equal, even if they have different initial points. Thus, if $\text{v}$ and $\text{w}$ are equivalent, we write

The arrows in Figure 2.2(b) are equivalent. Each arrow has the same length and direction. A closely related concept is the idea of parallel vectors. Two vectors are said to be parallel if they have the same or opposite directions. We explore this idea in more detail later in the chapter. A vector is defined by its magnitude and direction, regardless of where its initial point is located.

Figure 2.2 (a) A vector is represented by a directed line segment from its initial point to its terminal point. (b) Vectors ${\text{v}}_1$ through ${\text{v}}_5$ are equivalent.

The use of boldface, lowercase letters to name vectors is a common representation in print, but there are alternative notations. When writing the name of a vector by hand, for example, it is easier to sketch an arrow over the variable than to simulate boldface type: $\overrightarrow{v}.$ When a vector has initial point $P$ and terminal point $Q,$ the notation $\overrightarrow{PQ}$ is useful because it indicates the direction and location of the vector.

Example 2.1

Sketching Vectors

Sketch a vector in the plane from initial point $P(1,1)$ to terminal point $Q(8,5).$

Solution

See Figure 2.3. Because the vector goes from point $P$ to point $Q,$ we name it $\overrightarrow{PQ}.$

Figure 2.3 The vector with initial point $\left(1,1\right)$ and terminal point $\left(8,5\right)$ is named $\overrightarrow{PQ}.$

Combining Vectors

Vectors have many real-life applications, including situations involving force or velocity. For example, consider the forces acting on a boat crossing a river. The boat’s motor generates a force in one direction, and the current of the river generates a force in another direction. Both forces are vectors. We must take both the magnitude and direction of each force into account if we want to know where the boat will go.

A second example that involves vectors is a quarterback throwing a football. The quarterback does not throw the ball parallel to the ground; instead, he aims up into the air. The velocity of his throw can be represented by a vector. If we know how hard he throws the ball (magnitude—in this case, speed), and the angle (direction), we can tell how far the ball will travel down the field.

A real number is often called a scalar in mathematics and physics. Unlike vectors, scalars are generally considered to have a magnitude only, but no direction. Multiplying a vector by a scalar changes the vector’s magnitude. This is called scalar multiplication. Note that changing the magnitude of a vector does not indicate a change in its direction. For example, wind blowing from north to south might increase or decrease in speed while maintaining its direction from north to south.

As you might expect, if $k=\mathrm{-1},$ we denote the product $k\text{v}$ as

Note that $\text{−}\text{v}$ has the same magnitude as $\text{v},$ but has the opposite direction (Figure 2.4).

Figure 2.4 (a) The original vector v has length n units. (b) The length of $2\text{v}$ equals $2n$ units. (c) The length of $\text{v}\text{/}2$ is $n\text{/}2$ units. (d) The vectors $\text{v}$ and $\text{−}\text{v}$ have the same length but opposite directions.

Another operation we can perform on vectors is to add them together in vector addition, but because each vector may have its own direction, the process is different from adding two numbers. The most common graphical method for adding two vectors is to place the initial point of the second vector at the terminal point of the first, as in Figure 2.5(a). To see why this makes sense, suppose, for example, that both vectors represent displacement. If an object moves first from the initial point to the terminal point of vector $\text{v},$ then from the initial point to the terminal point of vector $\text{w},$ the overall displacement is the same as if the object had made just one movement from the initial point to the terminal point of the vector $v+w.$ For obvious reasons, this approach is called the triangle method. Notice that if we had switched the order, so that $\text{w}$ was our first vector and v was our second vector, we would have ended up in the same place. (Again, see Figure 2.5(a).) Thus, $v+w=w+v.$

A second method for adding vectors is called the parallelogram method. With this method, we place the two vectors so they have the same initial point, and then we draw a parallelogram with the vectors as two adjacent sides, as in Figure 2.5(b). The length of the diagonal of the parallelogram is the sum. Comparing Figure 2.5(b) and Figure 2.5(a), we can see that we get the same answer using either method. The vector $v+w$ is called the vector sum.

Figure 2.5 (a) When adding vectors by the triangle method, the initial point of $\text{w}$ is the terminal point of $\text{v}.$ (b) When adding vectors by the parallelogram method, the vectors $\text{v}$ and $\text{w}$ have the same initial point.

It is also appropriate here to discuss vector subtraction. We define $\text{v}-\text{w}$ as $v+\left(\text{−}\text{w}\right)=v+(\mathrm{-1})\text{w}.$ The vector $\text{v}-\text{w}$ is called the vector difference. Graphically, the vector $\text{v}-\text{w}$ is depicted by drawing a vector from the terminal point of $\text{w}$ to the terminal point of $\text{v}$ (Figure 2.6).

Figure 2.6 (a) The vector difference $\text{v}-\text{w}$ is depicted by drawing a vector from the terminal point of $\text{w}$ to the terminal point of $\text{v}.$ (b) The vector $\text{v}-\text{w}$ is equivalent to the vector $\text{v}+\left(\text{−}\text{w}\right).$

In Figure 2.5(a), the initial point of $\text{v}+\text{w}$ is the initial point of $\text{v}.$ The terminal point of $\text{v}+\text{w}$ is the terminal point of $\text{w}.$ These three vectors form the sides of a triangle. It follows that the length of any one side is less than the sum of the lengths of the remaining sides. So we have

This is known more generally as the triangle inequality. There is one case, however, when the resultant vector $\text{u}+\text{v}$ has the same magnitude as the sum of the magnitudes of $\text{u}$ and $\text{v}.$ This happens only when $\text{u}$ and $\text{v}$ have the same direction.

Vector Components

Working with vectors in a plane is easier when we are working in a coordinate system. When the initial points and terminal points of vectors are given in Cartesian coordinates, computations become straightforward.

Checkpoint 2.3

Which of the following vectors are equivalent?

We have seen how to plot a vector when we are given an initial point and a terminal point. However, because a vector can be placed anywhere in a plane, it may be easier to perform calculations with a vector when its initial point coincides with the origin. We call a vector with its initial point at the origin a standard-position vector. Because the initial point of any vector in standard position is known to be $\left(0,0\right),$ we can describe the vector by looking at the coordinates of its terminal point. Thus, if vector v has its initial point at the origin and its terminal point at $\left(x,y\right),$ we write the vector in component form as

When a vector is written in component form like this, the scalars x and y are called the components of $\text{v}.$

Recall that vectors are named with lowercase letters in bold type or by drawing an arrow over their name. We have also learned that we can name a vector by its component form, with the coordinates of its terminal point in angle brackets. However, when writing the component form of a vector, it is important to distinguish between $\langle x,y\rangle$ and $\left(x,y\right).$ The first ordered pair uses angle brackets to describe a vector, whereas the second uses parentheses to describe a point in a plane. The initial point of $\langle x,y\rangle$ is $\left(0,0\right);$ the terminal point of $\langle x,y\rangle$ is $\left(x,y\right).$

When we have a vector not already in standard position, we can determine its component form in one of two ways. We can use a geometric approach, in which we sketch the vector in the coordinate plane, and then sketch an equivalent standard-position vector. Alternatively, we can find it algebraically, using the coordinates of the initial point and the terminal point. To find it algebraically, we subtract the x-coordinate of the initial point from the x-coordinate of the terminal point to get the x component, and we subtract the y-coordinate of the initial point from the y-coordinate of the terminal point to get the y component.

To find the magnitude of a vector, we calculate the distance between its initial point and its terminal point. The magnitude of vector $\text{v}=\langle x,y\rangle$ is denoted $\Vert \text{v}\Vert ,$ or $\left|\text{v}\right|,$ and can be computed using the formula

Note that because this vector is written in component form, it is equivalent to a vector in standard position, with its initial point at the origin and terminal point $\left(x,y\right).$ Thus, it suffices to calculate the magnitude of the vector in standard position. Using the distance formula to calculate the distance between initial point $\left(0,0\right)$ and terminal point $\left(x,y\right),$ we have

Based on this formula, it is clear that for any vector $\text{v},$ $\Vert \text{v}\Vert \ge 0,$ and $\Vert \text{v}\Vert =0$ if and only if $v=0.$

The magnitude of a vector can also be derived using the Pythagorean theorem, as in the following figure.

Figure 2.13 If you use the components of a vector to define a right triangle, the magnitude of the vector is the length of the triangle’s hypotenuse.

We have defined scalar multiplication and vector addition geometrically. Expressing vectors in component form allows us to perform these same operations algebraically.

Now that we have established the basic rules of vector arithmetic, we can state the properties of vector operations. We will prove two of these properties. The others can be proved in a similar manner.

Proof of Distributive Property

Apply the distributive property for real numbers:

$$\begin{array}{cc}\hfill r\left(\text{u}+\text{v}\right)& =r·\langle x_1+x_2,y_1+y_2\rangle \hfill \\ & =\langle r\left(x_1+x_2\right),r\left(y_1+y_2\right)\rangle \hfill \\ & =\langle r{x}_1+r{x}_2,r{y}_1+r{y}_2\rangle \hfill \\ & =\langle r{x}_1,r{y}_1\rangle +\langle r{x}_2,r{y}_2\rangle \hfill \\ & =r\text{u}+r\text{v}.\hfill \end{array}$$

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Checkpoint 2.6

Prove the additive inverse property.

We have found the components of a vector given its initial and terminal points. In some cases, we may only have the magnitude and direction of a vector, not the points. For these vectors, we can identify the horizontal and vertical components using trigonometry (Figure 2.15).

Figure 2.15 The components of a vector form the legs of a right triangle, with the vector as the hypotenuse.

Consider the angle $\theta$ formed by the vector v and the positive x-axis. We can see from the triangle that the components of vector $\text{v}$ are $\langle \Vert \text{v}\Vert \text{cos}\theta ,\Vert \text{v}\Vert \text{sin}\theta \rangle .$ Therefore, given an angle and the magnitude of a vector, we can use the cosine and sine of the angle to find the components of the vector.

Example 2.6

Finding the Component Form of a Vector Using Trigonometry

Find the component form of a vector with magnitude 4 that forms an angle of $\mathrm{-45}\text{°}$ with the x-axis.

Solution

Let $x$ and $y$ represent the components of the vector (Figure 2.16). Then $x=4\text{cos}\left(\mathrm{-45}\text{°}\right)=2\sqrt{2}$ and $y=4\text{sin}\left(\mathrm{-45}\text{°}\right)=\mathrm{-2}\sqrt{2}.$ The component form of the vector is $\langle 2\sqrt{2},\mathrm{-2}\sqrt{2}\rangle .$

Figure 2.16 Use trigonometric ratios, $x=\Vert \text{v}\Vert \text{cos}\theta$ and $y=\Vert \text{v}\Vert \text{sin}\theta ,$ to identify the components of the vector.

Checkpoint 2.7

Find the component form of vector $\text{v}$ with magnitude $10$ that forms an angle of $120\text{°}$ with the positive x-axis.

Unit Vectors

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

Recall that when we defined scalar multiplication, we noted that $\Vert k\text{v}\Vert =\left|k\right|·\Vert \text{v}\Vert .$ For $\text{u}=\frac{1}{\Vert \text{v}\Vert }\text{v},$ it follows that $\Vert \text{u}\Vert =\frac{1}{\Vert \text{v}\Vert }\left(\Vert \text{v}\Vert \right)=1.$ We say that $\text{u}$ is the unit vector in the direction of$\text{v}$ (Figure 2.17). The process of using scalar multiplication to find a unit vector with a given direction is called normalization.

Figure 2.17 The vector $\text{v}$ and associated unit vector $\text{u}=\frac{1}{\Vert \text{v}\Vert }\text{v}.$ In this case, $\Vert \text{v}\Vert >1.$

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 $\text{i}=\langle 1,0\rangle$ and $\text{j}=\langle 0,1\rangle$ (Figure 2.18).

Figure 2.18 The standard unit vectors $\text{i}$ and $\text{j}.$

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

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

Figure 2.19 The vector $\text{v}$ is the sum of $x\text{i}$ and $y\text{j}.$

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 2.9

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 2.21. The angle between these vectors is $15\text{°}.$ 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 2.21 Two forces acting on a car in different directions.

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 $\left(0,0\right)$ and terminal point $\left(300,0\right).$ It can be expressed as $\langle 300,0\rangle$ or $300\text{i}.$

The second vector has magnitude $150$ and makes an angle of $15\text{°}$ with the first, so we can express it as $\langle 150\text{cos}\left(15\text{°}\right),150\text{sin}\left(15\text{°}\right)\rangle ,$ or $150\text{cos}\left(15\text{°}\right)\text{i}+150\text{sin}\left(15\text{°}\right)\text{j}.$ Then, the sum of the vectors, or resultant vector, is $\text{r}=\langle 300,0\rangle +\langle 150\text{cos}\left(15\text{°}\right),150\text{sin}\left(15\text{°}\right)\rangle ,$ and we have

$$\begin{array}{cc}\hfill \Vert \text{r}\Vert & =\sqrt{{\left(300+150\text{cos}\left(15\text{°}\right)\right)}^2+{\left(150\text{sin}\left(15\text{°}\right)\right)}^2}\hfill \\ & \approx \mathrm{446.6.}\hfill \end{array}$$

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