# 2.2: Scalars and Vectors (Part 1)

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- 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 \(\PageIndex{1}\)). 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.

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 \(\PageIndex{2}\), 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 ≡ |\(\vec{D}\)|. 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 = 6u = 6(2 cm) = 12 cm , as shown in Figure \(\PageIndex{3}\). 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.

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 \(\vec{D}_{AB}\) from A to B is the same as the magnitude of the displacement vector \(\vec{D}_{BA}\) from B to A (it equals 6 km in both cases), so we can write \(\vec{D}_{AB}\) = \(\vec{D}_{BA}\). However, vector \(\vec{D}_{AB}\) is not equal to vector \(\vec{D}_{BA}\) because these two vectors have different directions: \(\vec{D}_{AB}\) ≠ \(\vec{D}_{BA}\). In Figure 2.3, vector \(\vec{D}_{BA}\) would be represented by a vector with an origin at point B and an end at point A, indicating vector \(\vec{D}_{BA}\) points to the southwest, which is exactly 180° opposite to the direction of vector \(\vec{D}_{AB}\). We say that vector \(\vec{D}_{BA}\) is **antiparallel **to vector \(\vec{D}_{AB}\) and write \(\vec{D}_{AB}\) = \(-\vec{D}_{BA}\), 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 \(\PageIndex{4}\).

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.

**Alice**moves north at 6 knots and**Bob**moves west at 6 knots.**Alice**moves west at 6 knots and**Bob**moves west at 3 knots.**Alice**moves northeast at 6 knots and**Bob**moves south at 3 knots.**Alice**moves northeast at 6 knots and**Bob**moves southwest at 6 knots.**Alice**moves northeast at 2 knots and**Bob**moves closer to the shore northeast at 2 knots.

## 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 \(\PageIndex{5}\).

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 \(\PageIndex{5a}\)). What is his displacement vector \(\vec{D}_{AC}\) when he reaches point C? We know that if he walks all the way to B, his displacement vector relative to A is \(\vec{D}_{AB}\), which has magnitude D_{AB} = 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_{AB} = 4.5 km away from the campsite at A. So, his displacement vector at the rest point C has magnitude D_{AC} = 4.5 km = 0.75 D_{AB} and is parallel to the displacement vector \(\vec{D}_{AB}\). All of this can be stated succinctly in the form of the following **vector equation**:

\[\vec{D}_{AC} = 0.75\; \vec{D}_{AB} \ldotp \nonumber\]

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) \(\alpha\) = 0.75. The result, \(\vec{D}_{AC}\), of such a multiplication is a new vector with a direction parallel to the direction of the original vector \(\vec{D}_{AB}\). In general, when a vector \(\vec{D}_{A}\) is multiplied by a positive scalar \(\alpha\), the result is a new vector \(\vec{D}_{B}\) that is parallel to \(\vec{D}_{A}\):

\[\vec{B} = \alpha \vec{A} \label{2.1}\]

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**:

\[ B = | \alpha | A \ldotp \label{2.2}\]

In a scalar equation, both sides of the equation are numbers. Equation \ref{2.2} is a scalar equation because the magnitudes of vectors are scalar quantities (and positive numbers). If the scalar \(\alpha\) is **negative **in the vector equation Equation \ref{2.1}, then the magnitude |\(\vec{B}\)| of the new vector is still given by Equation \ref{2.2}, but the direction of the new vector \(\vec{B}\) is **antiparallel **to the direction of \(\vec{A}\). These principles are illustrated in Figure \(\PageIndex{6a}\) by two examples where the length of vector \(\vec{A}\) is 1.5 units. When \(\alpha\) = 2, the new vector \(\vec{B}\) = 2\(\vec{A}\) has length B = 2A = 3.0 units (twice as long as the original vector) and is parallel to the original vector. When \(\alpha\) = −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.

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 \(\PageIndex{5b}\)). What is his displacement vector \(\vec{D}_{AD}\) when he finds the box at point D? What is his displacement vector \(\vec{D}_{DB}\) from point D to the hole? We have already established that at rest point C his displacement vector is \(\vec{D}_{AC}\) = 0.75 \(\vec{D}_{AB}\). Starting at point C, he walks southwest (toward the campsite), which means his new displacement vector \(\vec{D}_{CD}\) from point C to point D is antiparallel to \(\vec{D}_{AB}\). Its magnitude |\(\vec{D}_{CD}\)| is D_{CD} = 1.2 km = 0.2 D_{AB}, so his second displacement vector is \(\vec{D}_{CD}\) = −0.2 \(\vec{D}_{AB}\). His total displacement \(\vec{D}_{AD}\) relative to the campsite is the vector sum of the two displacement vectors: vector \(\vec{D}_{AC}\) (from the campsite to the rest point) and vector \(\vec{D}_{CD}\) (from the rest point to the point where he finds his box):

\[\vec{D}_{AD} = \vec{D}_{AC} + \vec{D}_{CD} \ldotp \label{2.3}\]

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 Equation \ref{2.3} are known, we can find the resultant \(\vec{D}_{AD}\) as follows:

\[\vec{D}_{AD} = \vec{D}_{AC} + \vec{D}_{CD} = 0.75\; \vec{D}_{AB} - 0.2\; \vec{D}_{AB} = (0.75 - 0.2) \vec{D}_{AB} = 0.55 \vec{D}_{AB} \ldotp \label{2.4}\]

When your friend finally reaches the pond at B, his displacement vector \(\vec{D}_{AB}\) from point A is the vector sum of his displacement vector \(\vec{D}_{AD}\) from point A to point D and his displacement vector \(\vec{D}_{DB}\) from point D to the fishing hole: \(\vec{D}_{AB}\) = \(\vec{D}_{AD}\) + \(\vec{D}_{DB}\) (see Figure \(\PageIndex{5c}\)). This means his displacement vector \(\vec{D}_{DB}\) is the difference of two vectors:

\[\vec{D}_{DB} = \vec{D}_{AB} − \vec{D}_{AD} = \vec{D}_{AB} + (− \vec{D}_{AD}) \ldotp \label{2.5}\]

Notice that a difference of two vectors is nothing more than a vector sum of two vectors because the second term in Equation \ref{2.5} is vector \(- \vec{D}_{AD}\) (which is antiparallel to \(\vec{D}_{AD}\)). When we substitute Equation \ref{2.4} into Equation \ref{2.5}, we obtain the second displacement vector:

\[\vec{D}_{DB} = \vec{D}_{AB} − \vec{D}_{AD} = \vec{D}_{AB} − 0.55\; \vec{D}_{AB} = (1.0 − 0.55)\; \vec{D}_{AB} = 0.45\; \vec{D}_{AB} \ldotp \label{2.6}\]

This result means your friend walked D_{DB} = 0.45 D_{AB} = 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 \(\PageIndex{6}\).

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 (\PageIndex{6b}\)). 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 (\PageIndex{6c}\)) 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**,

\[\vec{A} + \vec{B} = \vec{B} + \vec{A} \ldotp \label{2.7}\]

and **associative,**

\[ (\vec{A} + \vec{B}) + \vec{C} = \vec{A} + (\vec{B} + \vec{C}) \ldotp \label{2.8}\]

Moreover, multiplication by a scalar is **distributive**:

\[ \alpha_{1} \vec{A} + \alpha_{2} \vec{A} = (\alpha_{1} + \alpha_{2}) \vec{A} \ldotp \label{2.9}\]

We used the distributive property in Equation \ref{2.4} and Equation \ref{2.6}.

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 |\(\hat{u}\)| ≡ u = 1. The only role of a unit vector is to specify direction. For example, instead of saying vector \(\vec{D}_{AB}\) 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 \(\vec{D}_{AB}\) = (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

\[\vec{D}_{BA} = (−6.0\; km)\; \hat{u} \ldotp \nonumber\]