Scalars and Vectors

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).

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 [latex]{\stackrel{\to }{D}}_{AC}[/latex] when he reaches point C? We know that if he walks all the way to B, his displacement vector relative to A is [latex]{\stackrel{\to }{D}}_{AB}[/latex], which has magnitude [latex]{D}_{AB}=6\phantom{\rule{0.2em}{0ex}}\text{km}[/latex] 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 [latex]0.75{D}_{AB}=4.5\phantom{\rule{0.2em}{0ex}}\text{km}[/latex] away from the campsite at A. So, his displacement vector at the rest point C has magnitude [latex]{D}_{AC}=4.5\phantom{\rule{0.2em}{0ex}}\text{km}=0.75{D}_{AB}[/latex] and is parallel to the displacement vector [latex]{\stackrel{\to }{D}}_{AB}[/latex]. All of this can be stated succinctly in the form of the following vector equation:

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) [latex]\alpha =0.75[/latex]. The result, [latex]{\stackrel{\to }{D}}_{AC}[/latex], of such a multiplication is a new vector with a direction parallel to the direction of the original vector [latex]{\stackrel{\to }{D}}_{AB}[/latex].

In general, when a vector [latex]\stackrel{\to }{A}[/latex] is multiplied by a positive scalar [latex]\alpha[/latex], the result is a new vector [latex]\stackrel{\to }{B}[/latex] that is parallel to [latex]\stackrel{\to }{A}[/latex]:

The magnitude [latex]|\stackrel{\to }{B}|[/latex] of this new vector is obtained by multiplying the magnitude [latex]|\stackrel{\to }{A}|[/latex] of the original vector, as expressed by the scalar equation:

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 [latex]\alpha[/latex] is negative in the vector equation (Figure), then the magnitude [latex]|\stackrel{\to }{B}|[/latex] of the new vector is still given by (Figure), but the direction of the new vector [latex]\stackrel{\to }{B}[/latex] is antiparallel to the direction of [latex]\stackrel{\to }{A}[/latex]. These principles are illustrated in (Figure)(a) by two examples where the length of vector [latex]\stackrel{\to }{A}[/latex] is 1.5 units. When [latex]\alpha =2[/latex], the new vector [latex]\stackrel{\to }{B}=2\stackrel{\to }{A}[/latex] has length [latex]B=2A=3.0\phantom{\rule{0.2em}{0ex}}\text{units}[/latex] (twice as long as the original vector) and is parallel to the original vector. When [latex]\alpha =-2[/latex], the new vector [latex]\stackrel{\to }{C}=-2\stackrel{\to }{A}[/latex] has length [latex]C=|-2|A=3.0\phantom{\rule{0.2em}{0ex}}\text{units}[/latex] (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)(b)). What is his displacement vector [latex]{\stackrel{\to }{D}}_{AD}[/latex] when he finds the box at point D? What is his displacement vector [latex]{\stackrel{\to }{D}}_{DB}[/latex] from point D to the hole? We have already established that at rest point C his displacement vector is [latex]{\stackrel{\to }{D}}_{AC}=0.75{\stackrel{\to }{D}}_{AB}[/latex]. Starting at point C, he walks southwest (toward the campsite), which means his new displacement vector [latex]{\stackrel{\to }{D}}_{CD}[/latex] from point C to point D is antiparallel to [latex]{\stackrel{\to }{D}}_{AB}[/latex]. Its magnitude [latex]|{\stackrel{\to }{D}}_{CD}|[/latex] is [latex]{D}_{CD}=1.2\phantom{\rule{0.2em}{0ex}}\text{km}=0.2{D}_{AB}[/latex], so his second displacement vector is [latex]{\stackrel{\to }{D}}_{CD}=-0.2{\stackrel{\to }{D}}_{AB}[/latex]. His total displacement [latex]{\stackrel{\to }{D}}_{AD}[/latex] relative to the campsite is the vector sum of the two displacement vectors: vector [latex]{\stackrel{\to }{D}}_{AC}[/latex] (from the campsite to the rest point) and vector [latex]{\stackrel{\to }{D}}_{CD}[/latex] (from the rest point to the point where he finds his box):

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 [latex]{\stackrel{\to }{D}}_{AD}[/latex] as follows:

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

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 [latex]\text{−}{\stackrel{\to }{D}}_{AD}[/latex] (which is antiparallel to [latex]{\stackrel{\to }{D}}_{AD}\right)[/latex]. When we substitute (Figure) into (Figure), we obtain the second displacement vector:

This result means your friend walked [latex]{D}_{DB}=0.45{D}_{AB}=0.45\left(6.0\phantom{\rule{0.2em}{0ex}}\text{km}\right)=2.7\phantom{\rule{0.2em}{0ex}}\text{km}[/latex] from the point where he finds his tackle box to the fishing hole.

When vectors [latex]\stackrel{\to }{A}[/latex] and [latex]\stackrel{\to }{B}[/latex] lie along a line (that is, in one dimension), such as in the camping example, their resultant [latex]\stackrel{\to }{R}=\stackrel{\to }{A}+\stackrel{\to }{B}[/latex] and their difference [latex]\stackrel{\to }{D}=\stackrel{\to }{A}-\stackrel{\to }{B}[/latex] 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 [latex]\stackrel{\to }{A}[/latex] and [latex]\stackrel{\to }{B}[/latex] 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 [latex]\stackrel{\to }{A}[/latex] is antiparallel to vector [latex]\stackrel{\to }{B}[/latex], 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 [latex]D=|A-B|[/latex] of the difference of their magnitudes. The direction of the difference vector [latex]\stackrel{\to }{D}[/latex] 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,

and associative,

Moreover, multiplication by a scalar is distributive:

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 [latex]\stackrel{^}{u}[/latex], has a magnitude of one and does not have any physical unit so that [latex]|\stackrel{^}{u}|\equiv u=1[/latex]. The only role of a unit vector is to specify direction. For example, instead of saying vector [latex]{\stackrel{\to }{D}}_{AB}[/latex] has a magnitude of 6.0 km and a direction of northeast, we can introduce a unit vector [latex]\stackrel{^}{u}[/latex] that points to the northeast and say succinctly that [latex]{\stackrel{\to }{D}}_{AB}=\left(6.0\phantom{\rule{0.2em}{0ex}}\text{km}\right)\stackrel{^}{u}[/latex]. Then the southwesterly direction is simply given by the unit vector [latex]\text{−}\stackrel{^}{u}[/latex]. In this way, the displacement of 6.0 km in the southwesterly direction is expressed by the vector

Algebra of Vectors in Two Dimensions

When vectors lie in a plane—that is, when they are in two dimensions—they can be multiplied by scalars, added to other vectors, or subtracted from other vectors in accordance with the general laws expressed by (Figure), (Figure), (Figure), and (Figure). However, the addition rule for two vectors in a plane becomes more complicated than the rule for vector addition in one dimension. We have to use the laws of geometry to construct resultant vectors, followed by trigonometry to find vector magnitudes and directions. This geometric approach is commonly used in navigation ((Figure)). In this section, we need to have at hand two rulers, a triangle, a protractor, a pencil, and an eraser for drawing vectors to scale by geometric constructions.

For a geometric construction of the sum of two vectors in a plane, we follow the parallelogram rule. Suppose two vectors [latex]\stackrel{\to }{A}[/latex] and [latex]\stackrel{\to }{B}[/latex] are at the arbitrary positions shown in (Figure). Translate either one of them in parallel to the beginning of the other vector, so that after the translation, both vectors have their origins at the same point. Now, at the end of vector [latex]\stackrel{\to }{A}[/latex] we draw a line parallel to vector [latex]\stackrel{\to }{B}[/latex] and at the end of vector [latex]\stackrel{\to }{B}[/latex] we draw a line parallel to vector [latex]\stackrel{\to }{A}[/latex] (the dashed lines in (Figure)). In this way, we obtain a parallelogram. From the origin of the two vectors we draw a diagonal that is the resultant [latex]\stackrel{\to }{R}[/latex] of the two vectors: [latex]\stackrel{\to }{R}=\stackrel{\to }{A}+\stackrel{\to }{B}[/latex] ((Figure)(a)). The other diagonal of this parallelogram is the vector difference of the two vectors [latex]\stackrel{\to }{D}=\stackrel{\to }{A}-\stackrel{\to }{B}[/latex], as shown in (Figure)(b). Notice that the end of the difference vector is placed at the end of vector [latex]\stackrel{\to }{A}[/latex].

It follows from the parallelogram rule that neither the magnitude of the resultant vector nor the magnitude of the difference vector can be expressed as a simple sum or difference of magnitudes A and B, because the length of a diagonal cannot be expressed as a simple sum of side lengths. When using a geometric construction to find magnitudes [latex]|\stackrel{\to }{R}|[/latex] and [latex]|\stackrel{\to }{D}|[/latex], we have to use trigonometry laws for triangles, which may lead to complicated algebra. There are two ways to circumvent this algebraic complexity. One way is to use the method of components, which we examine in the next section. The other way is to draw the vectors to scale, as is done in navigation, and read approximate vector lengths and angles (directions) from the graphs. In this section we examine the second approach.

If we need to add three or more vectors, we repeat the parallelogram rule for the pairs of vectors until we find the resultant of all of the resultants. For three vectors, for example, we first find the resultant of vector 1 and vector 2, and then we find the resultant of this resultant and vector 3. The order in which we select the pairs of vectors does not matter because the operation of vector addition is commutative and associative (see (Figure) and (Figure)). Before we state a general rule that follows from repetitive applications of the parallelogram rule, let’s look at the following example.

Suppose you plan a vacation trip in Florida. Departing from Tallahassee, the state capital, you plan to visit your uncle Joe in Jacksonville, see your cousin Vinny in Daytona Beach, stop for a little fun in Orlando, see a circus performance in Tampa, and visit the University of Florida in Gainesville. Your route may be represented by five displacement vectors [latex]\stackrel{\to }{A},[/latex] [latex]\stackrel{\to }{B}[/latex], [latex]\stackrel{\to }{C}[/latex], [latex]\stackrel{\to }{D}[/latex], and [latex]\stackrel{\to }{E}[/latex], which are indicated by the red vectors in (Figure). What is your total displacement when you reach Gainesville? The total displacement is the vector sum of all five displacement vectors, which may be found by using the parallelogram rule four times. Alternatively, recall that the displacement vector has its beginning at the initial position (Tallahassee) and its end at the final position (Gainesville), so the total displacement vector can be drawn directly as an arrow connecting Tallahassee with Gainesville (see the green vector in (Figure)). When we use the parallelogram rule four times, the resultant [latex]\stackrel{\to }{R}[/latex] we obtain is exactly this green vector connecting Tallahassee with Gainesville: [latex]\stackrel{\to }{R}=\stackrel{\to }{A}+\stackrel{\to }{B}+\stackrel{\to }{C}+\stackrel{\to }{D}+\stackrel{\to }{E}[/latex].

Drawing the resultant vector of many vectors can be generalized by using the following tail-to-head geometric construction. Suppose we want to draw the resultant vector [latex]\stackrel{\to }{R}[/latex] of four vectors [latex]\stackrel{\to }{A}[/latex], [latex]\stackrel{\to }{B}[/latex], [latex]\stackrel{\to }{C}[/latex], and [latex]\stackrel{\to }{D}[/latex] ((Figure)(a)). We select any one of the vectors as the first vector and make a parallel translation of a second vector to a position where the origin (“tail”) of the second vector coincides with the end (“head”) of the first vector. Then, we select a third vector and make a parallel translation of the third vector to a position where the origin of the third vector coincides with the end of the second vector. We repeat this procedure until all the vectors are in a head-to-tail arrangement like the one shown in (Figure). We draw the resultant vector [latex]\stackrel{\to }{R}[/latex] by connecting the origin (“tail”) of the first vector with the end (“head”) of the last vector. The end of the resultant vector is at the end of the last vector. Because the addition of vectors is associative and commutative, we obtain the same resultant vector regardless of which vector we choose to be first, second, third, or fourth in this construction.