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Section10.2An Introduction to Vectors

Many quantities we think about daily can be described by a single number: temperature, speed, cost, weight and height. There are also many other concepts we encounter daily that cannot be described with just one number. For instance, a weather forecaster often describes wind with its speed and its direction (“\(\ldots\) with winds from the southeast gusting up to 30 mph \(\ldots\)”). When applying a force, we are concerned with both the magnitude and direction of that force. In both of these examples, direction is important. Because of this, we study vectors, mathematical objects that convey both magnitude and direction information.

One “bare–bones” definition of a vector is based on what we wrote above: “a vector is a mathematical object with magnitude and direction parameters.” This definition leaves much to be desired, as it gives no indication as to how such an object is to be used. Several other definitions exist; we choose here a definition rooted in a geometric visualization of vectors. It is very simplistic but readily permits further investigation.

Definition10.2.1Vector

A vector is a directed line segment.

Given points \(P\) and \(Q\) (either in the plane or in space), we denote with \(\vv{PQ}\) the vector from \(P\) to \(Q\). The point \(P\) is said to be the initial point of the vector, and the point \(Q\) is the terminal point.

The magnitude, length or norm of \(\vv{PQ}\) is the length of the line segment \(\overline{PQ}\text{:}\) \(\norm{\vv{PQ}} = \norm{\overline{PQ}}\text{.}\)

Two vectors are equal if they have the same magnitude and direction.

Figure 10.2.2 shows multiple instances of the same vector. Each directed line segment has the same direction and length (magnitude), hence each is the same vector.

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Figure10.2.2Drawing the same vector with different initial points.

We use \(\mathbb{R}^2\) (pronounced “r two”) to represent all the vectors in the plane, and use \(\mathbb{R}^3\) (pronounced “r three”) to represent all the vectors in space.

<<SVG image is unavailable, or your browser cannot render it>>

Figure10.2.3Illustrating how equal vectors have the same displacement.

Consider the vectors \(\vv{PQ}\) and \(\vv{RS}\) as shown in Figure 10.2.3. The vectors look to be equal; that is, they seem to have the same length and direction. Indeed, they are. Both vectors move 2 units to the right and 1 unit up from the initial point to reach the terminal point. One can analyze this movement to measure the magnitude of the vector, and the movement itself gives direction information (one could also measure the slope of the line passing through \(P\) and \(Q\) or \(R\) and \(S\)). Since they have the same length and direction, these two vectors are equal.

This demonstrates that inherently all we care about is displacement; that is, how far in the \(x\text{,}\) \(y\) and possibly \(z\) directions the terminal point is from the initial point. Both the vectors \(\vv{PQ}\) and \(\vv{RS}\) in Figure 10.2.3 have an \(x\)-displacement of 2 and a \(y\)-displacement of 1. This suggests a standard way of describing vectors in the plane. A vector whose \(x\)-displacement is \(a\) and whose \(y\)-displacement is \(b\) will have terminal point \((a,b)\) when the initial point is the origin, \((0,0)\text{.}\) This leads us to a definition of a standard and concise way of referring to vectors.

Definition10.2.4Component Form of a Vector
  1. The component form of a vector \(\vec{v}\) in \(\mathbb{R}^2\text{,}\) whose terminal point is \((a,b)\) when its initial point is \((0,0)\text{,}\) is \(\la a,b\ra\text{.}\)

  2. The component form of a vector \(\vec{v}\) in \(\mathbb{R}^3\text{,}\) whose terminal point is \((a,b,c)\) when its initial point is \((0,0,0)\text{,}\) is \(\la a,b,c\ra\text{.}\)

The numbers \(a\text{,}\) \(b\) (and \(c\text{,}\) respectively) are the components of \(\vec v\text{.}\)

It follows from the definition that the component form of the vector \(\vv{PQ}\text{,}\) where \(P=(x_1,y_1)\) and \(Q=(x_2,y_2)\) is \begin{equation*} \vv{PQ} = \la x_2-x_1, y_2-y_1\ra; \end{equation*} in space, where \(P=(x_1,y_1,z_1)\) and \(Q=(x_2,y_2,z_2)\text{,}\) the component form of \(\vv{PQ}\) is \begin{equation*} \vv{PQ} = \la x_2-x_1, y_2-y_1,z_2-z_1\ra. \end{equation*}

We practice using this notation in the following example.

Example10.2.5Using component form notation for vectors
  1. Sketch the vector \(\vec v=\la 2,-1\ra\) starting at \(P=(3,2)\) and find its magnitude.

  2. Find the component form of the vector \(\vec w\) whose initial point is \(R=(-3,-2)\) and whose terminal point is \(S=(-1,2)\text{.}\)

  3. Sketch the vector \(\vec u = \la 2,-1,3\ra\) starting at the point \(Q = (1,1,1)\) and find its magnitude.

Solution

Now that we have defined vectors, and have created a nice notation by which to describe them, we start considering how vectors interact with each other. That is, we define an algebra on vectors.

Definition10.2.9Vector Algebra
  1. Let \(\vec u = \la u_1,u_2\ra\) and \(\vec v = \la v_1,v_2\ra\) be vectors in \(\mathbb{R}^2\text{,}\) and let \(c\) be a scalar.

    1. The addition, or sum, of the vectors \(\vec u\) and \(\vec v\) is the vector \begin{equation*} \vec u+\vec v = \la u_1+v_1, u_2+v_2\ra. \end{equation*}

    2. The scalar product of \(c\) and \(\vec v\) is the vector \begin{equation*} c\vec v = c\la v_1,v_2\ra = \la cv_1,cv_2\ra. \end{equation*}

  2. Let \(\vec u = \la u_1,u_2,u_3\ra\) and \(\vec v = \la v_1,v_2,v_3\ra\) be vectors in \(\mathbb{R}^3\text{,}\) and let \(c\) be a scalar.

    1. The addition, or sum, of the vectors \(\vec u\) and \(\vec v\) is the vector \begin{equation*} \vec u+\vec v = \la u_1+v_1, u_2+v_2, u_3+v_3\ra. \end{equation*}

    2. The scalar product of \(c\) and \(\vec v\) is the vector \begin{equation*} c\vec v = c\la v_1,v_2,v_3\ra = \la cv_1,cv_2,cv_3\ra. \end{equation*}

In short, we say addition and scalar multiplication are computed “component–wise.”

Example10.2.10Adding vectors

Sketch the vectors \(\vec u = \la1,3\ra\text{,}\) \(\vec v = \la 2,1\ra\) and \(\vec u+\vec v\) all with initial point at the origin.

Solution

As vectors convey magnitude and direction information, the sum of vectors also convey length and magnitude information. Adding \(\vec u+\vec v\) suggests the following idea:

“Starting at an initial point, go out \(\vec u\text{,}\) then go out \(\vec v\text{.}\)”

This idea is sketched in Figure 10.2.12, where the initial point of \(\vec v\) is the terminal point of \(\vec u\text{.}\) This is known as the “Head to Tail Rule” of adding vectors. Vector addition is very important. For instance, if the vectors \(\vec u\) and \(\vec v\) represent forces acting on a body, the sum \(\vec u+\vec v\) gives the resulting force. Because of various physical applications of vector addition, the sum \(\vec u+\vec v\) is often referred to as the resultant vector, or just the “resultant.”

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Figure10.2.12Illustrating how to add vectors using the Head to Tail Rule and Parallelogram Law.

Analytically, it is easy to see that \(\vec u+\vec v = \vec v+\vec u\text{.}\) Figure 10.2.12 also gives a graphical representation of this, using gray vectors. Note that the vectors \(\vec u\) and \(\vec v\text{,}\) when arranged as in the figure, form a parallelogram. Because of this, the Head to Tail Rule is also known as the Parallelogram Law: the vector \(\vec u+\vec v\) is defined by forming the parallelogram defined by the vectors \(\vec u\) and \(\vec v\text{;}\) the initial point of \(\vec u+\vec v\) is the common initial point of parallelogram, and the terminal point of the sum is the common terminal point of the parallelogram.

While not illustrated here, the Head to Tail Rule and Parallelogram Law hold for vectors in \(\mathbb{R}^3\) as well.

It follows from the properties of the real numbers and Definition 10.2.9 that \begin{equation*} \vec u-\vec v = \vec u + (-1)\vec v. \end{equation*}

The Parallelogram Law gives us a good way to visualize this subtraction. We demonstrate this in the following example.

Example10.2.13Vector Subtraction

Let \(\vec u = \la 3,1\ra\) and \(\vec v=\la 1,2\ra\text{.}\) Compute and sketch \(\vec u-\vec v\text{.}\)

Solution
Example10.2.15Scaling vectors
  1. Sketch the vectors \(\vec v = \la 2,1\ra\) and \(2\vec v\) with initial point at the origin.

  2. Compute the magnitudes of \(\vec v\) and \(2\vec v\text{.}\)

Solution

The zero vector is the vector whose initial point is also its terminal point. It is denoted by \(\vec 0\text{.}\) Its component form, in \(\mathbb{R}^2\text{,}\) is \(\la 0,0\ra\text{;}\) in \(\mathbb{R}^3\text{,}\) it is \(\la 0,0,0\ra\text{.}\) Usually the context makes is clear whether \(\vec 0\) is referring to a vector in the plane or in space.

Our examples have illustrated key principles in vector algebra: how to add and subtract vectors and how to multiply vectors by a scalar. The following theorem states formally the properties of these operations.

As stated before, each vector \(\vec v\) conveys magnitude and direction information. We have a method of extracting the magnitude, which we write as \(\norm{\vec v}\text{.}\) Unit vectors are a way of extracting just the direction information from a vector.

Definition10.2.18Unit Vector

A unit vector is a vector \(\vec v\) with a magnitude of 1; that is, \begin{equation*} \norm{\vec v}=1. \end{equation*}

Consider this scenario: you are given a vector \(\vec v\) and are told to create a vector of length 10 in the direction of \(\vec v\text{.}\) How does one do that? If we knew that \(\vec u\) was the unit vector in the direction of \(\vec v\text{,}\) the answer would be easy: \(10\vec u\text{.}\) So how do we find \(\vec u\) ?

Property 8 of Theorem 10.2.17 holds the key. If we divide \(\vec v\) by its magnitude, it becomes a vector of length 1. Consider: \begin{align*} \snorm{\frac{1}{\norm{\vec v}}\vec v} \amp = \frac{1}{\norm{\vec v}}\norm{\vec v} \amp \text{ (we can pull out \(\ds \frac{1}{\norm{\vec v}}\) as it is a scalar)}\\ \amp = 1. \end{align*}

So the vector of length 10 in the direction of \(\vec v\) is \(\ds 10\frac{1}{\norm{\vec v}}\vec v\text{.}\) An example will make this more clear.

Example10.2.19Using Unit Vectors

Let \(\vec v= \la 3,1\ra\) and let \(\vec w = \la 1,2,2\ra\text{.}\)

  1. Find the unit vector in the direction of \(\vec v\text{.}\)

  2. Find the unit vector in the direction of \(\vec w\text{.}\)

  3. Find the vector in the direction of \(\vec v\) with magnitude 5.

Solution

The basic formation of the unit vector \(\vec u\) in the direction of a vector \(\vec v\) leads to a interesting equation. It is: \begin{equation*} \vec v = \norm{\vec v}\frac{1}{\norm{\vec v}}\vec v. \end{equation*}

We rewrite the equation with parentheses to make a point: \begin{equation*} \vec v = \underbrace{\norm{\vec v}}_{\text{ \scriptsize magnitude } }\cdot\underbrace{\left(\frac{1}{\norm{\vec v}}\vec v\right)}_{\text{ \scriptsize direction } }. \end{equation*}

This equation illustrates the fact that a vector has both magnitude and direction, where we view a unit vector as supplying only direction information. Identifying unit vectors with direction allows us to define parallel vectors.

Definition10.2.21Parallel Vectors
  1. Unit vectors \(\vec u_1\) and \(\vec u_2\) are parallel if \(\vec u_1 = \pm \vec u_2\text{.}\)

  2. Nonzero vectors \(\vec v_1\) and \(\vec v_2\) are parallel if their respective unit vectors are parallel.

It is equivalent to say that vectors \(\vec v_1\) and \(\vec v_2\) are parallel if there is a scalar \(c\neq 0\) such that \(\vec v_1 = c\vec v_2\) (see marginal note).

\(\vec 0\) is directionless; because \(\vnorm{0}=0\text{,}\) there is no unit vector in the “direction” of \(\vec 0\text{.}\)

Some texts define two vectors as being parallel if one is a scalar multiple of the other. By this definition, \(\vec 0\) is parallel to all vectors as \(\vec 0 = 0\vec v\) for all \(\vec v\text{.}\)

We prefer the given definition of parallel as it is grounded in the fact that unit vectors provide direction information. One may adopt the convention that \(\vec 0\) is parallel to all vectors if they desire. (See also the marginal note on <<Unresolved xref, reference "note_crossp"; check spelling or use "provisional" attribute>>.)

If one graphed all unit vectors in \(\mathbb{R}^2\) with the initial point at the origin, then the terminal points would all lie on the unit circle. Based on what we know from trigonometry, we can then say that the component form of all unit vectors in \(\mathbb{R}^2\) is \(\la \cos(\theta) ,\sin(\theta) \ra\) for some angle \(\theta\text{.}\)

A similar construction in \(\mathbb{R}^3\) shows that the terminal points all lie on the unit sphere. These vectors also have a particular component form, but its derivation is not as straightforward as the one for unit vectors in \(\mathbb{R}^2\text{.}\) Important concepts about unit vectors are given in the following Key Idea.

Key Idea10.2.22Unit Vectors
  1. The unit vector in the direction of \(\vec v\) is \begin{equation*} \vec u = \frac1{\norm{\vec v}} \vec v. \end{equation*}

  2. A vector \(\vec u\) in \(\mathbb{R}^2\) is a unit vector if, and only if, its component form is \(\la \cos(\theta) ,\sin(\theta) \ra\) for some angle \(\theta\text{.}\)

  3. A vector \(\vec u\) in \(\mathbb{R}^3\) is a unit vector if, and only if, its component form is \(\la \sin(\theta) \cos(\varphi) ,\sin(\theta) \sin(\varphi) ,\cos(\theta) \ra\) for some angles \(\theta\) and \(\varphi\text{.}\)

These formulas can come in handy in a variety of situations, especially the formula for unit vectors in the plane.

Example10.2.23Finding Component Forces

Consider a weight of 50lb hanging from two chains, as shown in Figure 10.2.24. One chain makes an angle of \(30^\circ\) with the vertical, and the other an angle of \(45^\circ\text{.}\) Find the force applied to each chain.

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Figure10.2.24A diagram of a weight hanging from 2 chains in Example 10.2.23.
Solution

Unit vectors were very important in the previous calculation; they allowed us to define a vector in the proper direction but with an unknown magnitude. Our computations were then computed component–wise. Because such calculations are often necessary, the standard unit vectors can be useful.

Definition10.2.26Standard Unit Vectors
  1. In \(\mathbb{R}^2\text{,}\) the standard unit vectors are \begin{equation*} \vec i = \la 1,0\ra \text{ and } \vec j = \la 0,1\ra. \end{equation*}

  2. In \(\mathbb{R}^3\text{,}\) the standard unit vectors are \begin{equation*} \vec i = \la 1,0,0\ra \text{ and } \vec j = \la 0,1,0\ra \text{ and } \vec k = \la 0,0,1\ra. \end{equation*}

Example10.2.27Using standard unit vectors
  1. Rewrite \(\vec v = \la 2,-3\ra\) using the standard unit vectors.

  2. Rewrite \(\vec w = 4\vec i - 5\vec j +2\vec k\) in component form.

Solution
Example10.2.28Finding Component Force

A weight of 25lb is suspended from a chain of length 2ft while a wind pushes the weight to the right with constant force of 5lb as shown in Figure 10.2.29. What angle will the chain make with the vertical as a result of the wind's pushing? How much higher will the weight be?

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Figure10.2.29A figure of a weight being pushed by the wind in Example 10.2.28.
Solution

The algebra we have applied to vectors is already demonstrating itself to be very useful. There are two more fundamental operations we can perform with vectors, the dot product and the cross product. The next two sections explore each in turn.

Subsection10.2.1Exercises

In the following exercises, points \(P\) and \(Q\) are given. Write the vector \(\overrightarrow{PQ}\) in component form and using the standard unit vectors.

In the following exercises, sketch \(\vec u\text{,}\) \(\vec v\text{,}\) \(\vec u+\vec v\) and \(\vec u-\vec v\) on the same axes.

In the following exercises, find \(\norm{\vec u}\text{,}\) \(\norm{\vec v}\text{,}\) \(\norm{\vec u+\vec v}\) and \(\norm{\vec u-\vec v}\text{.}\)

In the following exercises, find the unit vector \(\vec u\) in the direction of \(\vec v\text{.}\)