2.1 Scalars and Vectors

OpenStax

2 Vectors

2.1 Scalars and Vectors

Learning Objectives

By the end of this section, you will be able to:

Describe the difference between vector and scalar quantities.
Identify the magnitude and direction of a vector.
Explain the effect of multiplying a vector quantity by a scalar.
Describe how one-dimensional vector quantities are added or subtracted.
Explain the geometric construction for the addition or subtraction of vectors in a plane.
Distinguish between a vector equation and a scalar equation.

Many familiar physical quantities can be specified completely by giving a single number and the appropriate unit. For example, “a class period lasts 50 min” or “the gas tank in my car holds 65 L” or “the distance between two posts is 100 m.” A physical quantity that can be specified completely in this manner is called a scalar quantity. Scalar is a synonym of “number.” Time, mass, distance, length, volume, temperature, and energy are examples of scalar quantities.

Scalar quantities that have the same physical units can be added or subtracted according to the usual rules of algebra for numbers. For example, a class ending 10 min earlier than 50 min lasts $50 min - 10 min = 40 min$ . Similarly, a 60-cal serving of corn followed by a 200-cal serving of donuts gives $60 cal + 200 cal = 260 cal$ of energy. When we multiply a scalar quantity by a number, we obtain the same scalar quantity but with a larger (or smaller) value. For example, if yesterday’s breakfast had 200 cal of energy and today’s breakfast has four times as much energy as it had yesterday, then today’s breakfast has $4 (200 cal) = 800 cal$ of energy. Two scalar quantities can also be multiplied or divided by each other to form a derived scalar quantity. For example, if a train covers a distance of 100 km in 1.0 h, its speed is 100.0 km/1.0 h = 27.8 m/s, where the speed is a derived scalar quantity obtained by dividing distance by time.

Many physical quantities, however, cannot be described completely by just a single number of physical units. For example, when the U.S. Coast Guard dispatches a ship or a helicopter for a rescue mission, the rescue team must know not only the distance to the distress signal, but also the direction from which the signal is coming so they can get to its origin as quickly as possible. Physical quantities specified completely by giving a number of units (magnitude) and a direction are called vector quantities. Examples of vector quantities include displacement, velocity, position, force, and torque. In the language of mathematics, physical vector quantities are represented by mathematical objects called vectors (Figure). We can add or subtract two vectors, and we can multiply a vector by a scalar or by another vector, but we cannot divide by a vector. The operation of division by a vector is not defined.

A photo of a dog. Below the photo is a horizontal arrow which starts below the dog’s tail and ends below the dog’s nose. The arrow is labeled Vector D, and its length is labeled as magnitude D. The start (tail) of the arrow is labeled “From rail of a vector origin” and its end (head) is labeled “To head of a vector end.” — **Figure 2.2** We draw a vector from the initial point or origin (called the “tail” of a vector) to the end or terminal point (called the “head” of a vector), marked by an arrowhead. Magnitude is the length of a vector and is always a positive scalar quantity. (credit: modification of work by Cate Sevilla)

Let’s examine vector algebra using a graphical method to be aware of basic terms and to develop a qualitative understanding. In practice, however, when it comes to solving physics problems, we use analytical methods, which we’ll see in the next section. Analytical methods are more simple computationally and more accurate than graphical methods. From now on, to distinguish between a vector and a scalar quantity, we adopt the common convention that a letter in bold type with an arrow above it denotes a vector, and a letter without an arrow denotes a scalar. For example, a distance of 2.0 km, which is a scalar quantity, is denoted by d = 2.0 km, whereas a displacement of 2.0 km in some direction, which is a vector quantity, is denoted by $\vec{d}$ .

Suppose you tell a friend on a camping trip that you have discovered a terrific fishing hole 6 km from your tent. It is unlikely your friend would be able to find the hole easily unless you also communicate the direction in which it can be found with respect to your campsite. You may say, for example, “Walk about 6 km northeast from my tent.” The key concept here is that you have to give not one but two pieces of information—namely, the distance or magnitude (6 km) and the direction (northeast).

Displacement is a general term used to describe a change in position, such as during a trip from the tent to the fishing hole. Displacement is an example of a vector quantity. If you walk from the tent (location A) to the hole (location B), as shown in Figure, the vector $\vec{D}$ , representing your displacement, is drawn as the arrow that originates at point A and ends at point B. The arrowhead marks the end of the vector. The direction of the displacement vector $\vec{D}$ is the direction of the arrow. The length of the arrow represents the magnitude D of vector $\vec{D}$ . Here, D = 6 km. Since the magnitude of a vector is its length, which is a positive number, the magnitude is also indicated by placing the absolute value notation around the symbol that denotes the vector; so, we can write equivalently that $D \equiv | \vec{D} |$ . To solve a vector problem graphically, we need to draw the vector $\vec{D}$ to scale. For example, if we assume 1 unit of distance (1 km) is represented in the drawing by a line segment of length u = 2 cm, then the total displacement in this example is represented by a vector of length $d = 6 u = 6 (2 cm) = 12 cm$ , as shown in Figure. Notice that here, to avoid confusion, we used $D = 6 km$ to denote the magnitude of the actual displacement and d = 12 cm to denote the length of its representation in the drawing.

An illustration of a lake, some distance northeast from a tent. North is up on the page, east to the right. The tent is labeled as location A, and the lake as location B. A straight arrow starts at A and ends at B. Three meandering paths, shown as dashed lines, also start at A and end at B. — **Figure 2.3** The displacement vector from point A (the initial position at the campsite) to point B (the final position at the fishing hole) is indicated by an arrow with origin at point A and end at point B. The displacement is the same for any of the actual paths (dashed curves) that may be taken between points A and B.

A ruler is shown, with distance measured in centimeters. A vector is shown as an arrow parallel to the ruler, extending from its end at 0 c m to 12 c m, and is labeled as vector D. — **Figure 2.4** A displacement $\vec{D}$ of magnitude 6 km is drawn to scale as a vector of length 12 cm when the length of 2 cm represents 1 unit of displacement (which in this case is 1 km).

Suppose your friend walks from the campsite at A to the fishing pond at B and then walks back: from the fishing pond at B to the campsite at A. The magnitude of the displacement vector ${\vec{D}}_{A B}$ from A to B is the same as the magnitude of the displacement vector ${\vec{D}}_{B A}$ from B to A (it equals 6 km in both cases), so we can write $D_{A B} = D_{B A}$ . However, vector ${\vec{D}}_{A B}$ is not equal to vector ${\vec{D}}_{B A}$ because these two vectors have different directions: ${\vec{D}}_{A B} \neq {\vec{D}}_{B A}$ . In Figure, vector ${\vec{D}}_{B A}$ would be represented by a vector with an origin at point B and an end at point A, indicating vector ${\vec{D}}_{B A}$ points to the southwest, which is exactly $180^{\circ}$ opposite to the direction of vector ${\vec{D}}_{A B}$ . We say that vector ${\vec{D}}_{B A}$ is antiparallel to vector ${\vec{D}}_{A B}$ and write ${\vec{D}}_{A B} = − {\vec{D}}_{B A}$ , where the minus sign indicates the antiparallel direction.

Two vectors that have identical directions are said to be parallel vectors—meaning, they are parallel to each other. Two parallel vectors $\vec{A}$ and $\vec{B}$ are equal, denoted by $\vec{A} = \vec{B}$ , if and only if they have equal magnitudes $| \vec{A} | = | \vec{B} |$ . Two vectors with directions perpendicular to each other are said to be orthogonal vectors. These relations between vectors are illustrated in Figure.

Figure a: Two examples of vector A parallel to vector B. In one, A and B are on the same line, one after the other, but A is longer than B. In the other, A and B are parallel to each other with their tails aligned, but A is shorter than B. Figure b: An example of vector A antiparallel to vector B. Vector A points to the left and is longer than vector B, which points to the right. The angle between them is 180 degrees. Figure c: An example of vector A antiparallel to minus vector A: A points to the right and –A points to the left. Both are the same length. Figure d: Two examples of vector A equal to vector B: In one, A and B are on the same line, one after the other, and both are the same length. In the other, A and B are parallel to each other with their tails aligned, and both are the same length. Figure e: Two examples of vector A orthogonal to vector B: In one, A points down and B points to the right, meeting at a right angle, and both are the same length. In the other, points down and to the right and B points down and to the left, meeting A at a right angle. Both are the same length. — **Figure 2.5** Various relations between two vectors $\vec{A}$ and $\vec{B}$ . (a) $\vec{A} \neq \vec{B}$ because $A \neq B$ . (b) $\vec{A} \neq \vec{B}$ because they are not parallel and $A \neq B$ . (c) $\vec{A} \neq − \vec{A}$ because they have different directions (even though $| \vec{A} | = | - \vec{A} | = A)$ . (d) $\vec{A} = \vec{B}$ because they are parallel and have identical magnitudes A = B. (e) $\vec{A} \neq \vec{B}$ because they have different directions (are not parallel); here, their directions differ by $90^{\circ}$ —meaning, they are orthogonal.

Check Your Understanding

Two motorboats named Alice and Bob are moving on a lake. Given the information about their velocity vectors in each of the following situations, indicate whether their velocity vectors are equal or otherwise. (a) Alice moves north at 6 knots and Bob moves west at 6 knots. (b) Alice moves west at 6 knots and Bob moves west at 3 knots. (c) Alice moves northeast at 6 knots and Bob moves south at 3 knots. (d) Alice moves northeast at 6 knots and Bob moves southwest at 6 knots. (e) Alice moves northeast at 2 knots and Bob moves closer to the shore northeast at 2 knots.

Show Solution

a. not equal because they are orthogonal; b. not equal because they have different magnitudes; c. not equal because they have different magnitudes and directions; d. not equal because they are antiparallel; e. equal.

Algebra of Vectors in One Dimension

Vectors can be multiplied by scalars, added to other vectors, or subtracted from other vectors. We can illustrate these vector concepts using an example of the fishing trip seen in Figure.

Three illustrations of the same tent and lake northeast of the tent. North is up on the page. The tent location is point A, and the lake location is point B. A location between A and B, about 2/3 of the way from A to B, is labeled as point C. In figure a, the vector from A to B is shown as a blue arrow, starting at A and ending at B, and labeled vector D sub A B. The vector from A to C is shown as a red arrow, starting at A and ending at C and labeled vector D sub A C. Three meandering paths are shown as dashed lines that start at A and end at B. Figure b adds the following to the illustration of figure a: Point D is added about half way between point A and B. The vector from A to D is shown as a purple arrow, starting at A and ending at D and labeled vector D sub A D. The vector from D to B is shown as an orange arrow, starting at D and ending at B and labeled vector D sub D B. Figure c adds a green arrow from point C to point D and is labeled vector D sub C D. Vector D sub C D points in the direction opposite to that of the other vectors, toward the tent rather than toward the lake. — **Figure 2.6** Displacement vectors for a fishing trip. (a) Stopping to rest at point C while walking from camp (point A) to the pond (point B). (b) Going back for the dropped tackle box (point D). (c) Finishing up at the fishing pond.

Suppose your friend departs from point A (the campsite) and walks in the direction to point B (the fishing pond), but, along the way, stops to rest at some point C located three-quarters of the distance between A and B, beginning from point A (Figure(a)). What is his displacement vector ${\vec{D}}_{A C}$ when he reaches point C? We know that if he walks all the way to B, his displacement vector relative to A is ${\vec{D}}_{A B}$ , which has magnitude $D_{A B} = 6 km$ and a direction of northeast. If he walks only a 0.75 fraction of the total distance, maintaining the northeasterly direction, at point C he must be $0.75 D_{A B} = 4.5 km$ away from the campsite at A. So, his displacement vector at the rest point C has magnitude $D_{A C} = 4.5 km = 0.75 D_{A B}$ and is parallel to the displacement vector ${\vec{D}}_{A B}$ . All of this can be stated succinctly in the form of the following vector equation:

{\vec{D}}_{A C} = 0.75 {\vec{D}}_{A B} .

In a vector equation, both sides of the equation are vectors. The previous equation is an example of a vector multiplied by a positive scalar (number) $α = 0.75$ . The result, ${\vec{D}}_{A C}$ , of such a multiplication is a new vector with a direction parallel to the direction of the original vector ${\vec{D}}_{A B}$ .

In general, when a vector $\vec{A}$ is multiplied by a positive scalar $α$ , the result is a new vector $\vec{B}$ that is parallel to $\vec{A}$ :

\vec{B} = α \vec{A} .

The magnitude $| \vec{B} |$ of this new vector is obtained by multiplying the magnitude $| \vec{A} |$ of the original vector, as expressed by the scalar equation:

B = | α | A .

In a scalar equation, both sides of the equation are numbers. Figure is a scalar equation because the magnitudes of vectors are scalar quantities (and positive numbers). If the scalar $α$ is negative in the vector equation Figure, then the magnitude $| \vec{B} |$ of the new vector is still given by Figure, but the direction of the new vector $\vec{B}$ is antiparallel to the direction of $\vec{A}$ . These principles are illustrated in Figure(a) by two examples where the length of vector $\vec{A}$ is 1.5 units. When $α = 2$ , the new vector $\vec{B} = 2 \vec{A}$ has length $B = 2 A = 3.0 units$ (twice as long as the original vector) and is parallel to the original vector. When $α = - 2$ , the new vector $\vec{C} = - 2 \vec{A}$ has length $C = | - 2 | A = 3.0 units$ (twice as long as the original vector) and is antiparallel to the original vector.

Figure a shows vector A pointing to the right. It has magnitude A=1.5. Vector B=2 time vector A points to the right and has magnitude B = 2 A = 3.0. Vector C = -2 times vector A and has magnitude B = 2.0. Figure b shows vector A points to the right and has magnitude A=1.5. Vector B is shown below vector A, with their tails aligned. Vector B points to the right and has magnitude 2.0. In another view, Vector A is shown with vector B starting at the head of A and extending further to the right. Below them is a vector, labeled as vector R = vector A plus vector B, pointing to the right whose tail is aligned with the tail of vector A and whose head is aligned with the head of vector B. The magnitude of vector R is equal to magnitude A plus magnitude B = 3.5. Figure c shows vector A points to the right and has magnitude A=1.5. Vector B is shown below vector A, with their tails aligned. Vector minus B points to the right and has magnitude 3.2. In another view, Vector A is shown with vector minus B pointing to the left and with its head meeting the head of vector A. Below them is a vector, labeled as vector D = vector A minus vector B, shorter than B and pointing to the left whose head is aligned with the head of vector B. The magnitude of vector D is equal to magnitude of quantity A minus B = 1.7. — **Figure 2.7** Algebra of vectors in one dimension. (a) Multiplication by a scalar. (b) Addition of two vectors $(\vec{R}$ is called the resultant of vectors $\vec{A}$ and $\vec{B})$ . (c) Subtraction of two vectors $(\vec{D}$ is the difference of vectors $\vec{A}$ and $\vec{B})$ .

Now suppose your fishing buddy departs from point A (the campsite), walking in the direction to point B (the fishing hole), but he realizes he lost his tackle box when he stopped to rest at point C (located three-quarters of the distance between A and B, beginning from point A). So, he turns back and retraces his steps in the direction toward the campsite and finds the box lying on the path at some point D only 1.2 km away from point C (see Figure(b)). What is his displacement vector ${\vec{D}}_{A D}$ when he finds the box at point D? What is his displacement vector ${\vec{D}}_{D B}$ from point D to the hole? We have already established that at rest point C his displacement vector is ${\vec{D}}_{A C} = 0.75 {\vec{D}}_{A B}$ . Starting at point C, he walks southwest (toward the campsite), which means his new displacement vector ${\vec{D}}_{C D}$ from point C to point D is antiparallel to ${\vec{D}}_{A B}$ . Its magnitude $| {\vec{D}}_{C D} |$ is $D_{C D} = 1.2 km = 0.2 D_{A B}$ , so his second displacement vector is ${\vec{D}}_{C D} = - 0.2 {\vec{D}}_{A B}$ . His total displacement ${\vec{D}}_{A D}$ relative to the campsite is the vector sum of the two displacement vectors: vector ${\vec{D}}_{A C}$ (from the campsite to the rest point) and vector ${\vec{D}}_{C D}$ (from the rest point to the point where he finds his box):

{\vec{D}}_{A D} = {\vec{D}}_{A C} + {\vec{D}}_{C D} .

The vector sum of two (or more) vectors is called the resultant vector or, for short, the resultant. When the vectors on the right-hand-side of Figure are known, we can find the resultant ${\vec{D}}_{A D}$ as follows:

{\vec{D}}_{A D} = {\vec{D}}_{A C} + {\vec{D}}_{C D} = 0.75 {\vec{D}}_{A B} - 0.2 {\vec{D}}_{A B} = (0.75 - 0.2) {\vec{D}}_{A B} = 0.55 {\vec{D}}_{A B} .

When your friend finally reaches the pond at B, his displacement vector ${\vec{D}}_{A B}$ from point A is the vector sum of his displacement vector ${\vec{D}}_{A D}$ from point A to point D and his displacement vector ${\vec{D}}_{D B}$ from point D to the fishing hole: ${\vec{D}}_{A B} = {\vec{D}}_{A D} + {\vec{D}}_{D B}$ (see Figure(c)). This means his displacement vector ${\vec{D}}_{D B}$ is the difference of two vectors:

{\vec{D}}_{D B} = {\vec{D}}_{A B} - {\vec{D}}_{A D} = {\vec{D}}_{A B} + (− {\vec{D}}_{A D}) .

Notice that a difference of two vectors is nothing more than a vector sum of two vectors because the second term in Figure is vector $− {\vec{D}}_{A D}$ (which is antiparallel to ${\vec{D}}_{A D})$ . When we substitute Figure into Figure, we obtain the second displacement vector:

{\vec{D}}_{D B} = {\vec{D}}_{A B} - {\vec{D}}_{A D} = {\vec{D}}_{A B} - 0.55 {\vec{D}}_{A B} = (1.0 - 0.55) {\vec{D}}_{A B} = 0.45 {\vec{D}}_{A B} .

This result means your friend walked $D_{D B} = 0.45 D_{A B} = 0.45 (6.0 km) = 2.7 km$ from the point where he finds his tackle box to the fishing hole.

When vectors $\vec{A}$ and $\vec{B}$ lie along a line (that is, in one dimension), such as in the camping example, their resultant $\vec{R} = \vec{A} + \vec{B}$ and their difference $\vec{D} = \vec{A} - \vec{B}$ both lie along the same direction. We can illustrate the addition or subtraction of vectors by drawing the corresponding vectors to scale in one dimension, as shown in Figure.

To illustrate the resultant when $\vec{A}$ and $\vec{B}$ are two parallel vectors, we draw them along one line by placing the origin of one vector at the end of the other vector in head-to-tail fashion (see Figure(b)). The magnitude of this resultant is the sum of their magnitudes: R = A + B. The direction of the resultant is parallel to both vectors. When vector $\vec{A}$ is antiparallel to vector $\vec{B}$ , we draw them along one line in either head-to-head fashion (Figure(c)) or tail-to-tail fashion. The magnitude of the vector difference, then, is the absolute value $D = | A - B |$ of the difference of their magnitudes. The direction of the difference vector $\vec{D}$ is parallel to the direction of the longer vector.

In general, in one dimension—as well as in higher dimensions, such as in a plane or in space—we can add any number of vectors and we can do so in any order because the addition of vectors is commutative,

\vec{A} + \vec{B} = \vec{B} + \vec{A},

and associative,

$(\vec{A} + \vec{B}) + \vec{C} = \vec{A} + (\vec{B} + \vec{C}) .$

Moreover, multiplication by a scalar is distributive:

α_{1} \vec{A} + α_{2} \vec{A} = (α_{1} + α_{2}) \vec{A} .

We used the distributive property in Figure and Figure.

When adding many vectors in one dimension, it is convenient to use the concept of a unit vector. A unit vector, which is denoted by a letter symbol with a hat, such as $\hat{u}$ , has a magnitude of one and does not have any physical unit so that $| \hat{u} | \equiv u = 1$ . The only role of a unit vector is to specify direction. For example, instead of saying vector ${\vec{D}}_{A B}$ has a magnitude of 6.0 km and a direction of northeast, we can introduce a unit vector $\hat{u}$ that points to the northeast and say succinctly that ${\vec{D}}_{A B} = (6.0 km) \hat{u}$ . Then the southwesterly direction is simply given by the unit vector $− \hat{u}$ . In this way, the displacement of 6.0 km in the southwesterly direction is expressed by the vector

{\vec{D}}_{B A} = (- 6.0 km) \hat{u} .

Example

A Ladybug Walker

A long measuring stick rests against a wall in a physics laboratory with its 200-cm end at the floor. A ladybug lands on the 100-cm mark and crawls randomly along the stick. It first walks 15 cm toward the floor, then it walks 56 cm toward the wall, then it walks 3 cm toward the floor again. Then, after a brief stop, it continues for 25 cm toward the floor and then, again, it crawls up 19 cm toward the wall before coming to a complete rest (Figure). Find the vector of its total displacement and its final resting position on the stick.

Strategy

If we choose the direction along the stick toward the floor as the direction of unit vector $\hat{u}$ , then the direction toward the floor is $+ \hat{u}$ and the direction toward the wall is $− \hat{u}$ . The ladybug makes a total of five displacements:

\begin{matrix} {\vec{D}}_{1} = (15 cm) (+ \hat{u}), \\ {\vec{D}}_{2} = (56 cm) (− \hat{u}), \\ {\vec{D}}_{3} = (3 cm) (+ \hat{u}), \\ {\vec{D}}_{4} = (25 cm) (+ \hat{u}), and \\ {\vec{D}}_{5} = (19 cm) (− \hat{u}) . \end{matrix}

The total displacement $\vec{D}$ is the resultant of all its displacement vectors.

Five illustrations of a ladybug on a ruler leaning against a wall. The +u hat direction is toward the floor parallel to the ruler, and the – u hat direction is up along the ruler. In the first illustration, the ladybug is located near the middle of the ruler and vector D sub 1 points down the ruler. In the second illustration, the ladybug is located lower, where the head of vector D sub 1 is in the first illustration, and vector D sub 2 points up the ruler. In the third illustration, the ladybug is located higher, where the head of vector D sub 2 is in the second illustration, and vector D sub 3 points down the ruler. In the fourth illustration, the ladybug is located lower, where the head of vector D sub 3 is in the third illustration, and vector D sub 4 points down the ruler. In the fifth illustration, the ladybug is located lower, where the head of vector D sub 4 is in the fourth illustration, and vector D sub 5 points up the ruler. — **Figure 2.8** Five displacements of the ladybug. Note that in this schematic drawing, magnitudes of displacements are not drawn to scale. (credit: modification of work by “Persian Poet Gal”/Wikimedia Commons)

Solution

Show Answer

The resultant of all the displacement vectors is $\begin{array}{cc} \vec{D} & = {\vec{D}}_{1} + {\vec{D}}_{2} + {\vec{D}}_{3} + {\vec{D}}_{4} + {\vec{D}}_{5} \\ = (15 cm) (+ \hat{u}) + (56 cm) (− \hat{u}) + (3 cm) (+ \hat{u}) + (25 cm) (+ \hat{u}) + (19 cm) (− \hat{u}) \\ = (15 - 56 + 3 + 25 - 19) cm \hat{u} \\ = - 32 cm \hat{u} . \end{array}$ In this calculation, we use the distributive law given by (Figure). The result reads that the total displacement vector points away from the 100-cm mark (initial landing site) toward the end of the meter stick that touches the wall. The end that touches the wall is marked 0 cm, so the final position of the ladybug is at the (100 – 32)cm = 68-cm mark.