As we have seen numerous times in this book, Mathematics can be used to model and solve real-world problems. For many applications, real numbers suffice; that is, real numbers with the appropriate units attached can be used to answer questions like “How close is the nearest Sasquatch nest?”

There are other times though, when these kinds of quantities do not suffice. Perhaps it is important to know, for instance, how close the nearest Sasquatch nest is as well as the direction in which it lies. To answer questions like these which involve both a quantitative answer, or magnitude, along with a direction, we use the mathematical objects called vectors. ^[1] (The word ‘vector’ comes from the Latin vehere meaning ‘to convey’ or ‘to carry.’)

A vector is represented geometrically as a directed line segment where the magnitude of the vector is taken to be the length of the line segment and the direction is made clear with the use of an arrow at one endpoint of the segment. When referring to vectors in this text, we shall adopt ^[2] (Other textbook authors use bold vectors such as \(\textbf {v}\). We find that writing in bold font on the chalkboard is inconvenient at best, so we have chosen the ‘arrow’ notation.) the ‘arrow’ notation, so the symbol \(\overrightarrow {v}\) is read as ‘the vector \(v\)’. Below in blue is a typical vector \(\overrightarrow {v}\) with endpoints \(P\left (1, 2\right )\) and \(Q\left (4, 6\right )\).

The point \(P\) is called the initial point or tail of \(\overrightarrow {v}\) and the point \(Q\) is called the terminal point or head of \(\overrightarrow {v}\). Since we can reconstruct \(\overrightarrow {v}\) completely from \(P\) and \(Q\), we write \(\overrightarrow {v} = \overrightarrow {PQ}\), where the order of points \(P\) (initial point) and \(Q\) (terminal point) is important. (Think about this before moving on.)

While it is true that \(P\) and \(Q\) completely determine \(\overrightarrow {v}\), it is important to note that since vectors are defined in terms of their two characteristics, magnitude and direction, any directed line segment with the same length and direction as \(\overrightarrow {v}\) is considered to be the same vector as \(\overrightarrow {v}\), regardless of its initial point.

In the case of our vector \(\overrightarrow {v}\) above, any vector which moves three units to the right and four up ^[3] (If this idea of ‘over’ and ‘up’ seems familiar, it should. The slope of the line segment containing \(\overrightarrow {v}\) is \(\frac {4}{3}\).) from its initial point to arrive at its terminal point is considered the same vector as \(\overrightarrow {v}\). The notation we use to capture this idea is the component form of the vector, \(\overrightarrow {v} = \left <3,4\right >\), where the first number, \(3\), is called the \(x\)-component of \(\overrightarrow {v}\) and the second number, \(4\), is called the \(y\)-component of \(\overrightarrow {v}\) ^[4] (Note that there are a number of other notations for vectors. In fact, the Geogebra code to draw this vector uses the form \(\overrightarrow {v} = \left (\begin{array}{cc} 3 \\ 4 \end{array}\right ).\)) .

For example, if we wanted to reconstruct \(\overrightarrow {v} = \left <3,4\right >\) with initial point \(P'(-2,3)\), then we would find the terminal point of \(\overrightarrow {v}\) by adding \(3\) to the \(x\)-coordinate and adding \(4\) to the \(y\)-coordinate to obtain the terminal point \(Q'(1,7)\), as seen below if we set \(a=-2\) and \(b=3\).

The component form of a vector is what ties these very geometric objects back to Algebra and ultimately Trigonometry. We generalize our example in our definition below.

Using the language of components, we have that two vectors are equal if and only if their corresponding components are equal. That is, \(\left <v_1, v_2\right > = \left <v_1', v_2'\right >\) if and only if \(v_1 = v_1'\) and \(v_2 = v_2'\). (Again, think about this before reading on.)

We now set about defining operations on vectors. Suppose we are given two vectors \(\overrightarrow {v}\) and \(\overrightarrow {w}\). The sum, or resultant vector \(\overrightarrow {v} + \overrightarrow {w}\) is obtained as follows. First, plot \(\overrightarrow {v}\). Next, plot \(\overrightarrow {w}\) so that its initial point is the terminal point of \(\overrightarrow {v}\). To plot the vector \(\overrightarrow {v} + \overrightarrow {w}\) we begin at the initial point of \(\overrightarrow {v}\) and end at the terminal point of \(\overrightarrow {w}\). It is helpful to think of the vector \(\overrightarrow {v} + \overrightarrow {w}\) as the ‘net result’ of moving along \(\overrightarrow {v}\) then moving along \(\overrightarrow {w}\).

Our next example makes good use of resultant vectors and reviews bearings and the Law of Cosines. ^[5] (If necessary, review Sections bearings and TheLawofCosines.)

A plane leaves an airport with an airspeed ^[6] (That is, the speed of the plane relative to the air around it. If there were no wind, plane’s airspeed would be the same as its speed as observed from the ground. How does wind affect this? Keep reading!) of 175 miles per hour at a bearing of N\(40^{\circ }\)E. A 35 mile per hour wind is blowing at a bearing of S\(60^{\circ }\)E. Find the true speed of the plane, rounded to the nearest mile per hour, and the true bearing of the plane, rounded to the nearest degree.

Solution: For both the plane and the wind, we are given their speeds and their directions. Coupling speed (as a magnitude) with direction is the concept of velocity which we’ve seen a few times before. ^[7] (See Section circularmotion, for instance.)

We let \(\overrightarrow {v}\) denote the plane’s velocity and \(\overrightarrow {w}\) denote the wind’s velocity in the diagram below. The ‘true’ speed and bearing is found by analyzing the resultant vector, \(\overrightarrow {v} + \overrightarrow {w}\).

From the vector diagram, we get a triangle, the lengths of whose sides are the magnitude of \(\overrightarrow {v}\), which is 175, the magnitude of \(\overrightarrow {w}\), which is 35, and the magnitude of \(\overrightarrow {v} + \overrightarrow {w}\), which we’ll call \(c\).

From the given bearing information, we go through the usual geometry to determine that the angle between the sides of length 35 and 175 measures \(100^{\circ }\).

From the Law of Cosines, we determine \(c = \sqrt {31850 - 12250\cos (100^{\circ })} \approx 184\), which means the true speed of the plane is (approximately) \(184\) miles per hour.

To determine the true bearing of the plane, we need to determine the angle \(\alpha \). Using the Law of Cosines once more, ^[8] (Or, since our given angle, \(100^{\circ }\), is obtuse, we could use the Law of Sines without any ambiguity here.) we find \(\cos (\alpha ) = \frac {c^2+29400}{350c}\) so that \(\alpha \approx 11^{\circ }\).

Given the geometry of the situation, we add \(\alpha \) to the given \(40^{\circ }\) and find the true bearing of the plane to be (approximately) N\(51^{\circ }\)E. □

Our next step is to define addition of vectors component-wise to match the geometric action. ^[9] (Adding vectors ‘component-wise’ should seem hauntingly familiar. Compare this with how matrix addition was defined in section MatArithmetic. In more advanced courses. chief among them Linear Algebra, vectors are actaually defined as \(1\times n\) or \(n \times 1\) matrices, depending on the situation.)

In order for vector addition to enjoy the same kinds of properties as real number addition, it is necessary to extend our definition of vectors to include a ‘zero vector’, \(\overrightarrow {0} = \left <0, 0\right >\).

Geometrically, \(\overrightarrow {0}\) represents a point, which we can (very broadly) think of as a directed line segment with the same initial and terminal points. The reader may well object to the inclusion of \(\overrightarrow {0}\), since after all, vectors are supposed to have both a magnitude (length) and a direction.

While it seems clear that the magnitude of \(\overrightarrow {0}\) should be \(0\), it is not clear what its direction is. As we shall see, the direction of \(\overrightarrow {0}\) is in fact undefined, but this minor hiccup in the natural flow of things is worth the benefits we reap by including \(\overrightarrow {0}\) in our discussions. We have the following theorem.

The properties in Theorem vectoradditionprops are easily verified using the definition of vector addition, ^[10] (The interested reader is encouraged to compare Theorem vectoradditionprops and the ensuing discussion with Theorem matrixadditionprops in Section MatArithmetic.) and are a direct consequence of the definition of vector addition along with properties inherited from real number arithmetic.

For the commutative property, we note that if \(\overrightarrow {v} = \left <v_1,v_2\right >\) and \(\overrightarrow {w} = \left <w_1,w_2\right >\) then

Geometrically, we can ‘see’ the commutative property by realizing that the sums \(\overrightarrow {v}+\overrightarrow {w}\) and \(\overrightarrow {w} + \overrightarrow {v}\) are the same directed diagonal determined by the parallelogram below.

The proofs of the associative and identity properties proceed similarly, and the reader is encouraged to verify them and provide accompanying diagrams.

The additive identity property is likewise verified algebraically using a calculation. If \(\overrightarrow {v} = \left <v_1,v_2\right >\) , then

From the commutative property of vector addition, we get that \(\overrightarrow {0} + \overrightarrow {v} = \overrightarrow {v}\) as well. Again, the reader is encouraged to visualize what this means geometrically. ^[11] (Recall, \(\overrightarrow {0}\) is represented geometrically as a point …)

Regarding additive inverses, we can verify by direct computation that if \(\overrightarrow {v} = \left < v_1, v_2 \right >\) and \(\overrightarrow {w} = \left < - v_1, -v_2 \right >\),

Once again, the commutative property of vector addition assures us that, likewise, \(\overrightarrow {w} + \overrightarrow {v} = \overrightarrow {0}\).

Moreover, additive inverses of vectors are unique. That is, given a vector \(\overrightarrow {v} = \left <v_1, v_2\right >\), there is precisely only one vector \(\overrightarrow {w}\) so that \(\overrightarrow {v} + \overrightarrow {w} = \overrightarrow {0}\).

To see this, suppose a vector \(\overrightarrow {w} = \left <w_1,w_2\right >\) satisfies \(\overrightarrow {v} + \overrightarrow {w} = \overrightarrow {0}\). By the definition of vector addition, we have \(\left <v_1 + w_1, v_2 + w_2\right > = \left <0,0\right >\). Hence, \(v_1 + w_1 = 0\) and \(v_2 + w_2 = 0\). We get \(w_1 = -v_1\) and \(w_2 = -v_2 \) so that \(\overrightarrow {w} = \left <-v_1 , -v_2 \right >\) as prescribed in Theorem vectoradditionprops.

Hence, every vector \(\overrightarrow {v}\) has one, and only one, additive inverse. In general, we denote the additive inverse of a vector \(\overrightarrow {v}\) with the (highly suggestive) notation \(- \overrightarrow {v}\).

Geometrically, the vectors \(\overrightarrow {v} = \left <v_1, v_2\right >\) and \(-\overrightarrow {v} = \left <-v_1, -v_2\right >\) have the same length, but opposite directions. As a result, when adding the vectors geometrically, the sum \(\overrightarrow {v} + (-\overrightarrow {v})\) results in starting at the initial point of \(\overrightarrow {v}\) and ending back at the initial point of \(\overrightarrow {v}\). That is, the net result of moving \(\overrightarrow {v}\) then \(-\overrightarrow {v}\) is not moving at all.

Using the additive inverse of a vector, we can define the difference of two vectors: \(\overrightarrow {v} - \overrightarrow {w} = \overrightarrow {v} + (-\overrightarrow {w})\). Looking at this at the level of components, we see if \(\overrightarrow {v} = \left <v_1,v_2\right >\) and \(\overrightarrow {w} = \left <w_1,w_2\right >\) then

In other words, like vector addition, vector subtraction works component-wise.

To interpret the vector \(\overrightarrow {v} - \overrightarrow {w}\) geometrically, we note

This means that the ‘net result’ of moving along \(\overrightarrow {w}\) then moving along \(\overrightarrow {v} - \overrightarrow {w}\) is just \(\overrightarrow {v}\) itself.

From the diagram below on the left, we see that \(\overrightarrow {v}-\overrightarrow {w}\) may be interpreted as the vector whose initial point is the terminal point of \(\overrightarrow {w}\) and whose terminal point is the terminal point of \(\overrightarrow {v}\).

It is also worth mentioning that in the parallelogram determined by the vectors \(\overrightarrow {v}\) and \(\overrightarrow {w}\) above on the right, the vector \(\overrightarrow {v}-\overrightarrow {w}\) is one of the diagonals – the other being \(\overrightarrow {v} + \overrightarrow {w}\).

Next, we discuss scalar multiplication – that is, taking a real number times a vector. We define scalar multiplication for vectors in the same way we defined it for matrices in Section MatArithmetic.

Scalar multiplication by \(k\) in vectors can be understood geometrically as scaling the vector (if \(k > 0\)) or scaling the vector and reversing its direction (if \(k < 0\)) as demonstrated below.

Note by definition scalarmultvector, \((-1)\overrightarrow {v} = (-1)\left <v_1,v_2\right > = \left <(-1)v_1, (-1)v_2\right > = \left <-v_1,-v_2\right > = -\overrightarrow {v}\), which is what we would expect. This and other properties of scalar multiplication are summarized in the theorem below.

The proof of Theorem vectorscalarmultprops, like the proof of Theorem vectoradditionprops, ultimately boils down to the definition of scalar multiplication and properties of real numbers.

For example, to prove the associative property, we let \(\overrightarrow {v} = \left <v_1,v_2\right >\). If \(k\) and \(r\) are scalars then

The reader is invited to think about what this property means geometrically. The remaining properties are proved similarly and are left as exercises.

Our next example demonstrates how Theorem vectorscalarmultprops allows us to do the same kind of algebraic manipulations with vectors as we do with variables – multiplication and division of vectors notwithstanding. If the pedantry seems familiar, it should. This is the same treatment we gave Example matrixaddscalarex in Section MatArithmetic. As in that example, we spell out the solution in excruciating detail to encourage the reader to think carefully about why each step is justified.

A vector whose initial point is \((0,0)\) is said to be in standard position. If \(\overrightarrow {v} = \left <v_1,v_2\right >\) is plotted in standard position, then its terminal point is necessarily \(\left (v_1,v_2\right )\). (Once more, think about this before reading on.)

Plotting a vector in standard position enables us to more easily quantify the concepts of magnitude and direction of the vector.

Recall the magnitude of vector \(\overrightarrow {v}\) is the length of the directed line segment representing \(\overrightarrow {v}\). When plotted in standard position, the length of this line segment is none other than the distance from the origin \((0,0)\) to the point \(\left (v_1,v_2\right )\). Hence, the magnitude of \(\overrightarrow {v}\), which we denote \(\| \overrightarrow {v} \|\), is given by \(\| \overrightarrow {v} \| = \sqrt {v_1^2 + v_2^2}\).

Turning to the notion of direction, we note that the point \(\left (v_1,v_2\right )\) is on the terminal side of the angle \(\theta \) depicted in the diagram above. From Theorem cosinesinecircle, we have \(v_1 = \| \overrightarrow {v} \| \cos (\theta )\) and \(v_2 = \| \overrightarrow {v} \| \sin (\theta )\). From the definition of scalar multiplication and vector equality, we get

This motivates the following definition.

A few remarks are in order. First, we note that if \(\overrightarrow {v} \neq 0\) then there are infinitely many angles \(\theta \) which satisfy Definition polarformvector. However, the fact that all of them must contain the same point \(\left (v_1,v_2\right )\) on their terminal sides means they are all coterminal.

Hence, if \(\theta \) and \(\theta '\) both satisfy the conditions of Definition polarformvector, then \(\cos (\theta ) = \cos (\theta ')\) and \(\sin (\theta ) = \sin (\theta ')\), and as such, \( \left < \cos (\theta ), \sin (\theta ) \right > = \left < \cos (\theta '), \sin (\theta ') \right >\) making \(\hat {v}\) is well-defined.

For \(\overrightarrow {0} = \left < 0, 0 \right >\), note that \(\| \overrightarrow {0} \| = \sqrt {0^2 + 0^2} = 0\). Hence, \(\| \overrightarrow {0} \| \left < \cos (\theta ), \sin (\theta ) \right > = 0 \left < \cos (\theta ), \sin (\theta ) \right > = <0,0>\) for every angle \(\theta \). In other words, every angle \(\theta \) satisfies the equation \(\overrightarrow {v} = \left < \| \overrightarrow {v} \| \cos (\theta ), \| \overrightarrow {v} \| \sin (\theta ) \right >\) in Definition polarformvector, so for this reason, \(\hat {0}\) is undefined.

The following theorem summarizes the important facts about the magnitude and direction of a vector.

The proof of the first property in Theorem magdirprops is a direct consequence of the definition of \(\| \overrightarrow {v} \|\). Given \(\overrightarrow {v} = \left < v_1 ,v_2\right >\), then \(\| \overrightarrow {v} \| = \sqrt {v_1^2 + v_2^2}\) which is by definition greater than or equal to \(0\). Moreover, \(\sqrt {v_1^2 + v_2^2} = 0\) if and only of \(v_1^2 + v_2^2 = 0\) if and only if \(v_1 = v_2 = 0\). Hence, \(\| \overrightarrow {v} \| = 0\) if and only if \(\overrightarrow {v} = \left <0,0\right > = \overrightarrow {0}\), as required.

The second property is a result of the definition of magnitude and scalar multiplication along with a property of radicals. If \(\overrightarrow {v} = \left < v_1 ,v_2\right >\) and \(k\) is a scalar then

The equation \(\overrightarrow {v} = \| \overrightarrow {v} \| \hat {v}\) in Theorem magdirprops is a consequence of the definitions of \(\| \overrightarrow {v} \|\) and \(\hat {v}\) and was worked out in the discussion just prior to Definition polarformvector on page 10. In words, the equation \(\overrightarrow {v} = \| \overrightarrow {v} \| \hat {v}\) says that any given vector is the product of its magnitude and its direction – an important concept to keep in mind when studying and using vectors.

The formula for \(\hat {v}\) stated in Theorem magdirprops is a consequence of solving \(\overrightarrow {v} = \| \overrightarrow {v} \| \hat {v}\) for \(\hat {v}\) by multiplying ^[12] (Of course, to go from \(\overrightarrow {v} = \| \overrightarrow {v} \| \hat {v}\) to \(\hat {v} = \left ( \frac {1}{\|\overrightarrow {v}\|}\right ) \overrightarrow {v}\), we are essentially ‘dividing both sides’ of the equation by the scalar \(\| \overrightarrow {v} \|\). The authors encourage the reader, however, to work out the details carefully to gain an appreciation of the properties in play.) both sides of the equation by \(\frac {1}{\| \overrightarrow {v} \|}\) and using the properties of Theorem vectorscalarmultprops. We leave these details to the reader. We are overdue for an example.

Note that the second and third solutions to number preludetounitvector in Example polarformvecex above work for any nonzero vector, \(\overrightarrow {w}\). We will have more to say about this shortly.

The process exemplified by number resolvecomponents in Example polarformvecex above by which we take information about the magnitude and direction of a vector and find the component form of a vector is called resolving a vector into its components. As an application of this process, we revisit Example vectorbearingex below.

In part preludetounitvector of Example polarformvecex, we saw that the length of the direction vector, \(\hat {w}\), \(\| \hat {w} \| = 1\). Vectors of length \(1\) play such an important role that they are given a special name.

Note that if \(\overrightarrow {v}\) is a unit vector, then necessarily, \(\overrightarrow {v} = \| \overrightarrow {v} \| \hat {v} = 1 \cdot \hat {v} = \hat {v}\). Conversely, in the solution of part preludetounitvector of Example polarformvecex, two different arguments show for any nonzero vector \(\overrightarrow {v}\), \(\| \hat {v} \| = 1\), so \(\hat {v}\) is a unit vector.

In other words, unit vectors are direction vectors and vice-versa. Indeed, the vector \(\hat {v}\) which we have defined as ‘the direction of \(\overrightarrow {v}\)’ is often described as ‘the unit vector in the direction of \(\overrightarrow {v}\).’

In practice, if \(\overrightarrow {v}\) is a unit vector we write it as \(\hat {v}\) as opposed to \(\overrightarrow {v}\) because we have reserved the ‘\(\hat {~}\)’ notation for unit vectors. The process of multiplying a nonzero vector by the factor \(\frac {1}{\| \overrightarrow {v} \|}\) to produce a unit vector is called ‘normalizing the vector.’

The terminal points of unit vectors, when plotted in standard position, lie on the Unit Circle. (You should take the time to show this.) As a result, we visualize normalizing a nonzero vector \(\overrightarrow {v}\) as shrinking ^[17] (…if \(\| \overrightarrow {v} \| > 1\) …) its terminal point, when plotted in standard position, back to the Unit Circle.

Of all of the unit vectors, two deserve special mention.

Geometrically, in the \(xy\)-plane, the vector \(\hat {\text {i}}\) as represents the positive \(x\)-direction, whereas the vector \(\hat {\text {j}}\) represents the positive \(y\)-direction. We have the following ‘decomposition’ theorem. ^[18] (We will see a generalization of Theorem ijdecomp in Section TheDotProduct. Stay tuned!)

The proof of Theorem ijdecomp is straightforward. Since \(\hat {\text {i}} = \left <1,0\right >\) and \(\hat {\text {j}} = \left < 0,1\right >\), we have from the definition of scalar multiplication and vector addition that

Geometrically, the situation looks like this:

We conclude this section with a classic example which demonstrates how vectors are used in physics to study forces. A ‘force’ is defined as a ‘push’ or a ‘pull.’ The intensity of the push or pull is the magnitude of the force, and is measured in Netwons (N) in the SI system or pounds (lbs.)\(\!\) in the English system. ^[19] (See also Section harmomicmotion.)

The following example uses all of the concepts in this section, and should be studied in great detail.

Note that the sum of the tensions on the wires in Example forceex exceed the \(50\) pounds of the speaker. Explaining why this happens is a good exercise and gets at the heart of the concept of vectors and resolution of forces. Speaking of exercises …