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Calculus Volume 3

2.1 Vectors in the Plane

Calculus Volume 32.1 Vectors in the Plane
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  1. Preface
  2. 1 Parametric Equations and Polar Coordinates
    1. Introduction
    2. 1.1 Parametric Equations
    3. 1.2 Calculus of Parametric Curves
    4. 1.3 Polar Coordinates
    5. 1.4 Area and Arc Length in Polar Coordinates
    6. 1.5 Conic Sections
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  3. 2 Vectors in Space
    1. Introduction
    2. 2.1 Vectors in the Plane
    3. 2.2 Vectors in Three Dimensions
    4. 2.3 The Dot Product
    5. 2.4 The Cross Product
    6. 2.5 Equations of Lines and Planes in Space
    7. 2.6 Quadric Surfaces
    8. 2.7 Cylindrical and Spherical Coordinates
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  4. 3 Vector-Valued Functions
    1. Introduction
    2. 3.1 Vector-Valued Functions and Space Curves
    3. 3.2 Calculus of Vector-Valued Functions
    4. 3.3 Arc Length and Curvature
    5. 3.4 Motion in Space
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  5. 4 Differentiation of Functions of Several Variables
    1. Introduction
    2. 4.1 Functions of Several Variables
    3. 4.2 Limits and Continuity
    4. 4.3 Partial Derivatives
    5. 4.4 Tangent Planes and Linear Approximations
    6. 4.5 The Chain Rule
    7. 4.6 Directional Derivatives and the Gradient
    8. 4.7 Maxima/Minima Problems
    9. 4.8 Lagrange Multipliers
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Chapter Review Exercises
  6. 5 Multiple Integration
    1. Introduction
    2. 5.1 Double Integrals over Rectangular Regions
    3. 5.2 Double Integrals over General Regions
    4. 5.3 Double Integrals in Polar Coordinates
    5. 5.4 Triple Integrals
    6. 5.5 Triple Integrals in Cylindrical and Spherical Coordinates
    7. 5.6 Calculating Centers of Mass and Moments of Inertia
    8. 5.7 Change of Variables in Multiple Integrals
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  7. 6 Vector Calculus
    1. Introduction
    2. 6.1 Vector Fields
    3. 6.2 Line Integrals
    4. 6.3 Conservative Vector Fields
    5. 6.4 Green’s Theorem
    6. 6.5 Divergence and Curl
    7. 6.6 Surface Integrals
    8. 6.7 Stokes’ Theorem
    9. 6.8 The Divergence Theorem
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Chapter Review Exercises
  8. 7 Second-Order Differential Equations
    1. Introduction
    2. 7.1 Second-Order Linear Equations
    3. 7.2 Nonhomogeneous Linear Equations
    4. 7.3 Applications
    5. 7.4 Series Solutions of Differential Equations
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  9. A | Table of Integrals
  10. B | Table of Derivatives
  11. C | Review of Pre-Calculus
  12. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
  13. Index

Learning Objectives

  • 2.1.1. Describe a plane vector, using correct notation.
  • 2.1.2. Perform basic vector operations (scalar multiplication, addition, subtraction).
  • 2.1.3. Express a vector in component form.
  • 2.1.4. Explain the formula for the magnitude of a vector.
  • 2.1.5. Express a vector in terms of unit vectors.
  • 2.1.6. Give two examples of vector quantities.

When describing the movement of an airplane in flight, it is important to communicate two pieces of information: the direction in which the plane is traveling and the plane’s speed. When measuring a force, such as the thrust of the plane’s engines, it is important to describe not only the strength of that force, but also the direction in which it is applied. Some quantities, such as or force, are defined in terms of both size (also called magnitude) and direction. A quantity that has magnitude and direction is called a vector. In this text, we denote vectors by boldface letters, such as v.

Definition

A vector is a quantity that has both magnitude and direction.

Vector Representation

A vector in a plane is represented by a directed line segment (an arrow). The endpoints of the segment are called the initial point and the terminal point of the vector. An arrow from the initial point to the terminal point indicates the direction of the vector. The length of the line segment represents its magnitude. We use the notation vv to denote the magnitude of the vector v.v. A vector with an initial point and terminal point that are the same is called the zero vector, denoted 0.0. The zero vector is the only vector without a direction, and by convention can be considered to have any direction convenient to the problem at hand.

Vectors with the same magnitude and direction are called equivalent vectors. We treat equivalent vectors as equal, even if they have different initial points. Thus, if vv and ww are equivalent, we write

v=w.v=w.

Definition

Vectors are said to be equivalent vectors if they have the same magnitude and direction.

The arrows in Figure 2.2(b) are equivalent. Each arrow has the same length and direction. A closely related concept is the idea of parallel vectors. Two vectors are said to be parallel if they have the same or opposite directions. We explore this idea in more detail later in the chapter. A vector is defined by its magnitude and direction, regardless of where its initial point is located.

This figure has two images. The first is labeled “a” and has a line segment representing vector v. The line segment begins at the initial point and goes to the terminal point. There is an arrowhead at the terminal point. The second image is labeled “b” and is five vectors, each labeled v sub 1, v sub 2, v sub 3, v sub 4, v sub 5. They all are pointing in the same direction and have the same length.
Figure 2.2 (a) A vector is represented by a directed line segment from its initial point to its terminal point. (b) Vectors v1v1 through v5v5 are equivalent.

The use of boldface, lowercase letters to name vectors is a common representation in print, but there are alternative notations. When writing the name of a vector by hand, for example, it is easier to sketch an arrow over the variable than to simulate boldface type: v.v. When a vector has initial point PP and terminal point Q,Q, the notation PQPQ is useful because it indicates the direction and location of the vector.

Example 2.1

Sketching Vectors

Sketch a vector in the plane from initial point P(1,1)P(1,1) to terminal point Q(8,5).Q(8,5).

Solution

See Figure 2.3. Because the vector goes from point PP to point Q,Q, we name it PQ.PQ.

This figure is a graph of the first quadrant. There is a line segment beginning at the ordered pair (1, 1). Also, this point is labeled “P.” The line segment ends at the ordered pair (8, 5) and is labeled “Q.” The line segment is labeled “PQ.”
Figure 2.3 The vector with initial point (1,1)(1,1) and terminal point (8,5)(8,5) is named PQ.PQ.
Checkpoint 2.1

Sketch the vector STST where SS is point (3,−1)(3,−1) and TT is point (−2,3).(−2,3).

Combining Vectors

Vectors have many real-life applications, including situations involving force or velocity. For example, consider the forces acting on a boat crossing a river. The boat’s motor generates a force in one direction, and the current of the river generates a force in another direction. Both forces are vectors. We must take both the magnitude and direction of each force into account if we want to know where the boat will go.

A second example that involves vectors is a quarterback throwing a football. The quarterback does not throw the ball parallel to the ground; instead, he aims up into the air. The velocity of his throw can be represented by a vector. If we know how hard he throws the ball (magnitude—in this case, speed), and the angle (direction), we can tell how far the ball will travel down the field.

A real number is often called a scalar in mathematics and physics. Unlike vectors, scalars are generally considered to have a magnitude only, but no direction. Multiplying a vector by a scalar changes the vector’s magnitude. This is called scalar multiplication. Note that changing the magnitude of a vector does not indicate a change in its direction. For example, wind blowing from north to south might increase or decrease in speed while maintaining its direction from north to south.

Definition

The product kvkv of a vector v and a scalar k is a vector with a magnitude that is |k||k| times the magnitude of v,v, and with a direction that is the same as the direction of vv if k>0,k>0, and opposite the direction of vv if k<0.k<0. This is called scalar multiplication. If k=0k=0 or v=0,v=0, then kv=0.kv=0.

As you might expect, if k=−1,k=−1, we denote the product kvkv as

kv=(−1)v=v.kv=(−1)v=v.

Note that vv has the same magnitude as v,v, but has the opposite direction (Figure 2.4).

This graphic has 4 figures. The first figure is a vector labeled “v.” The second figure is a vector twice as long as the first vector and is labeled “2 v.” The third figure is half as long as the first and is labeled “1/2 v.” The fourth figure is a vector in the opposite direction as the first. It is labeled “-v.”
Figure 2.4 (a) The original vector v has length n units. (b) The length of 2v2v equals 2n2n units. (c) The length of v/2v/2 is n/2n/2 units. (d) The vectors vv and vv have the same length but opposite directions.

Another operation we can perform on vectors is to add them together in vector addition, but because each vector may have its own direction, the process is different from adding two numbers. The most common graphical method for adding two vectors is to place the initial point of the second vector at the terminal point of the first, as in Figure 2.5(a). To see why this makes sense, suppose, for example, that both vectors represent displacement. If an object moves first from the initial point to the terminal point of vector v,v, then from the initial point to the terminal point of vector w,w, the overall displacement is the same as if the object had made just one movement from the initial point to the terminal point of the vector v+w.v+w. For obvious reasons, this approach is called the triangle method. Notice that if we had switched the order, so that ww was our first vector and v was our second vector, we would have ended up in the same place. (Again, see Figure 2.5(a).) Thus, v+w=w+v.v+w=w+v.

A second method for adding vectors is called the parallelogram method. With this method, we place the two vectors so they have the same initial point, and then we draw a parallelogram with the vectors as two adjacent sides, as in Figure 2.5(b). The length of the diagonal of the parallelogram is the sum. Comparing Figure 2.5(b) and Figure 2.5(a), we can see that we get the same answer using either method. The vector v+wv+w is called the vector sum.

Definition

The sum of two vectors vv and ww can be constructed graphically by placing the initial point of ww at the terminal point of v.v. Then, the vector sum, v+w,v+w, is the vector with an initial point that coincides with the initial point of vv and has a terminal point that coincides with the terminal point of w.w. This operation is known as vector addition.

This image has two figures. The first has two vectors, v and w with the same initial point. A parallelogram is formed by sketching broken lines parallel to the two vectors. A diagonal line is drawn from the same initial point to the opposite corner. It is labeled “v + w.” The second has two vectors, v and w. Vector v begins at the terminal point of vector w. A parallelogram is formed by sketching broken lines parallel to the two vectors. A diagonal line is drawn from the same initial point as vector w to the opposite corner. It is labeled “v + w.”
Figure 2.5 (a) When adding vectors by the triangle method, the initial point of ww is the terminal point of v.v. (b) When adding vectors by the parallelogram method, the vectors vv and ww have the same initial point.

It is also appropriate here to discuss vector subtraction. We define vwvw as v+(w)=v+(−1)w.v+(w)=v+(−1)w. The vector vwvw is called the vector difference. Graphically, the vector vwvw is depicted by drawing a vector from the terminal point of ww to the terminal point of vv (Figure 2.6).

This image has two figures. The first figure has two vectors, one labeled “v” and the other labeled “w.” Both vectors have the same initial point. A third vector is drawn between the terminal points of v and w. It is labeled “v – w.” The second figure has two vectors, one labeled “v” and the other labeled “-w.” The vector “-w” has its initial point at the terminal point of “v.” A parallelogram is created with broken lines where “v” is the diagonal and “w” is the top side.
Figure 2.6 (a) The vector difference vwvw is depicted by drawing a vector from the terminal point of ww to the terminal point of v.v. (b) The vector vwvw is equivalent to the vector v+(w).v+(w).

In Figure 2.5(a), the initial point of v+wv+w is the initial point of v.v. The terminal point of v+wv+w is the terminal point of w.w. These three vectors form the sides of a triangle. It follows that the length of any one side is less than the sum of the lengths of the remaining sides. So we have

v+wv+w.v+wv+w.

This is known more generally as the triangle inequality. There is one case, however, when the resultant vector u+vu+v has the same magnitude as the sum of the magnitudes of uu and v.v. This happens only when uu and vv have the same direction.

Example 2.2

Combining Vectors

Given the vectors vv and ww shown in Figure 2.7, sketch the vectors

  1. 3w3w
  2. v+wv+w
  3. 2vw2vw
    This figure has two vectors. They are vector v and vector w. They are not connected.
    Figure 2.7 Vectors vv and ww lie in the same plane.

Solution

  1. The vector 3w3w has the same direction as w;w; it is three times as long as w.w.
    This figure has two vectors. The first is labeled “w.” The second one is parallel to “w” and is labeled “3w.” It is three times as long as w in the same direction.
    Vector 3w3w has the same direction as ww and is three times as long.
  2. Use either addition method to find v+w.v+w.
    This image has two figures. The first has two vectors, labeled “v” and “w.” They both have the same initial point. A third vector is drawn, labeled “v + w.” It is the diagonal of the parallelogram formed by having sides parallel to vectors v and w. The second figure is a triangle formed by having vector v on one side and vector w adjacent to v. The terminal point of v is the initial point of w. The third side is labeled “v + w.”
    Figure 2.8 To find v+w,v+w, align the vectors at their initial points or place the initial point of one vector at the terminal point of the other. (a) The vector v+wv+w is the diagonal of the parallelogram with sides vv and ww (b) The vector v+wv+w is the third side of a triangle formed with ww placed at the terminal point of v.v.
  3. To find 2vw,2vw, we can first rewrite the expression as 2v+(w).2v+(w). Then we can draw the vector w,w, then add it to the vector 2v.2v.
    This figure is a triangle formed by having vector 2v on one side and vector -w adjacent to 2v. The terminal point of 2v is the initial point of -w. The third side is labeled “2v - w.”
    Figure 2.9 To find 2vw,2vw, simply add 2v+(w).2v+(w).
Checkpoint 2.2

Using vectors vv and ww from Example 2.2, sketch the vector 2wv.2wv.

Vector Components

Working with vectors in a plane is easier when we are working in a coordinate system. When the initial points and terminal points of vectors are given in Cartesian coordinates, computations become straightforward.

Example 2.3

Comparing Vectors

Are vv and ww equivalent vectors?

  1. vv has initial point (3,2)(3,2) and terminal point (7,2)(7,2)
    ww has initial point (1,−4)(1,−4) and terminal point (1,0)(1,0)
  2. vv has initial point (0,0)(0,0) and terminal point (1,1)(1,1)
    ww has initial point (−2,2)(−2,2) and terminal point (−1,3)(−1,3)

Solution

  1. The vectors are each 44 units long, but they are oriented in different directions. So vv and ww are not equivalent (Figure 2.10).
    This figure is a Cartesian coordinate system with two vectors. The first vector labeled “v” has initial point at (3, 2) and terminal point (7, 2). It is parallel to the x-axis. The second vector is labeled “w” and has initial point (1, -4) and terminal point (1, 0). It is parallel to the y-axis.
    Figure 2.10 These vectors are not equivalent.
  2. Based on Figure 2.11, and using a bit of geometry, it is clear these vectors have the same length and the same direction, so vv and ww are equivalent.
    This figure is a Cartesian coordinate system with two vectors. The first vector labeled “v” has initial point at (0, 0) and terminal point (1, 1). The second vector is labeled “w” and has initial point (-2, 2) and terminal point (-1, 3).
    Figure 2.11 These vectors are equivalent.
Checkpoint 2.3

Which of the following vectors are equivalent?

This figure is a coordinate system with 6 vectors, each labeled a through f. Three of the vectors, “a,” “b,” and “e” have the same length and are pointing in the same direction.

We have seen how to plot a vector when we are given an initial point and a terminal point. However, because a vector can be placed anywhere in a plane, it may be easier to perform calculations with a vector when its initial point coincides with the origin. We call a vector with its initial point at the origin a standard-position vector. Because the initial point of any vector in standard position is known to be (0,0),(0,0), we can describe the vector by looking at the coordinates of its terminal point. Thus, if vector v has its initial point at the origin and its terminal point at (x,y),(x,y), we write the vector in component form as

v=x,y.v=x,y.

When a vector is written in component form like this, the scalars x and y are called the components of v.v.

Definition

The vector with initial point (0,0)(0,0) and terminal point (x,y)(x,y) can be written in component form as

v=x,y.v=x,y.

The scalars xx and yy are called the components of v.v.

Recall that vectors are named with lowercase letters in bold type or by drawing an arrow over their name. We have also learned that we can name a vector by its component form, with the coordinates of its terminal point in angle brackets. However, when writing the component form of a vector, it is important to distinguish between x,yx,y and (x,y).(x,y). The first ordered pair uses angle brackets to describe a vector, whereas the second uses parentheses to describe a point in a plane. The initial point of x,yx,y is (0,0);(0,0); the terminal point of x,yx,y is (x,y).(x,y).

When we have a vector not already in standard position, we can determine its component form in one of two ways. We can use a geometric approach, in which we sketch the vector in the coordinate plane, and then sketch an equivalent standard-position vector. Alternatively, we can find it algebraically, using the coordinates of the initial point and the terminal point. To find it algebraically, we subtract the x-coordinate of the initial point from the x-coordinate of the terminal point to get the x component, and we subtract the y-coordinate of the initial point from the y-coordinate of the terminal point to get the y component.

Rule: Component Form of a Vector

Let v be a vector with initial point (xi,yi)(xi,yi) and terminal point (xt,yt).(xt,yt). Then we can express v in component form as v=xtxi,ytyi.v=xtxi,ytyi.

Example 2.4

Expressing Vectors in Component Form

Express vector vv with initial point (−3,4)(−3,4) and terminal point (1,2)(1,2) in component form.

Solution

  1. Geometric
    1. Sketch the vector in the coordinate plane (Figure 2.12).
    2. The terminal point is 4 units to the right and 2 units down from the initial point.
    3. Find the point that is 4 units to the right and 2 units down from the origin.
    4. In standard position, this vector has initial point (0,0)(0,0) and terminal point (4,−2):(4,−2):
      v=4,−2.v=4,−2.
      This figure is a coordinate system. There are two vectors on the graph. The first vector has initial point at the origin and terminal point at (4, -2). The horizontal distance from the initial to the terminal point for the vector is labeled as “4 units.” The vertical distance from the initial to the terminal point is labeled as “2 units.” The second vector has initial point at (-3, 4) and terminal point at (1, 2). The horizontal distance from the initial to the terminal point for the vector is labeled as “4 units.” The vertical distance from the initial to the terminal point is labeled as “2 units.”
      Figure 2.12 These vectors are equivalent.
  2. Algebraic
    In the first solution, we used a sketch of the vector to see that the terminal point lies 4 units to the right. We can accomplish this algebraically by finding the difference of the x-coordinates:
    xtxi=1(−3)=4.xtxi=1(−3)=4.

    Similarly, the difference of the y-coordinates shows the vertical length of the vector.
    ytyi=24=−2.ytyi=24=−2.

    So, in component form,
    v=xtxi,ytyi=1(−3),24=4,−2.v=xtxi,ytyi=1(−3),24=4,−2.
Checkpoint 2.4

Vector ww has initial point (−4,−5)(−4,−5) and terminal point (−1,2).(−1,2). Express ww in component form.

To find the magnitude of a vector, we calculate the distance between its initial point and its terminal point. The magnitude of vector v=x,yv=x,y is denoted v,v, or |v|,|v|, and can be computed using the formula

v=x2+y2.v=x2+y2.

Note that because this vector is written in component form, it is equivalent to a vector in standard position, with its initial point at the origin and terminal point (x,y).(x,y). Thus, it suffices to calculate the magnitude of the vector in standard position. Using the distance formula to calculate the distance between initial point (0,0)(0,0) and terminal point (x,y),(x,y), we have

v=(x0)2+(y0)2=x2+y2.v=(x0)2+(y0)2=x2+y2.

Based on this formula, it is clear that for any vector v,v, v0,v0, and v=0v=0 if and only if v=0.v=0.

The magnitude of a vector can also be derived using the Pythagorean theorem, as in the following figure.

This figure is a right triangle. The two sides are labeled “x” and “y.” The hypotenuse is represented as a vector and is labeled “square root (x^2 + y^2).”
Figure 2.13 If you use the components of a vector to define a right triangle, the magnitude of the vector is the length of the triangle’s hypotenuse.

We have defined scalar multiplication and vector addition geometrically. Expressing vectors in component form allows us to perform these same operations algebraically.

Definition

Let v=x1,y1v=x1,y1 and w=x2,y2w=x2,y2 be vectors, and let kk be a scalar.

Scalar multiplication:kv=kx1,ky1kv=kx1,ky1

Vector addition:v+w=x1,y1+x2,y2=x1+x2,y1+y2v+w=x1,y1+x2,y2=x1+x2,y1+y2

Example 2.5

Performing Operations in Component Form

Let vv be the vector with initial point (2,5)(2,5) and terminal point (8,13),(8,13), and let w=−2,4.w=−2,4.

  1. Express vv in component form and find v.v. Then, using algebra, find
  2. v+w,v+w,
  3. 3v,3v, and
  4. v2w.v2w.

Solution

  1. To place the initial point of vv at the origin, we must translate the vector 22 units to the left and 55 units down (Figure 2.15). Using the algebraic method, we can express vv as v=82,135=6,8:v=82,135=6,8:
    v=62+82=36+64=100=10.v=62+82=36+64=100=10.

    This figure is the first quadrant of a coordinate system. It has two vectors. The first vector has initial point at (2, 5) and terminal point (8, 13). The second vector has initial point at the origin and terminal point at (6, 8).
    Figure 2.14 In component form, v=6,8.v=6,8.
  2. To find v+w,v+w, add the x-components and the y-components separately:
    v+w=6,8+−2,4=4,12.v+w=6,8+−2,4=4,12.
  3. To find 3v,3v, multiply vv by the scalar k=3:k=3:
    3v=3·6,8=3·6,3·8=18,24.3v=3·6,8=3·6,3·8=18,24.
  4. To find v2w,v2w, find −2w−2w and add it to v:v:
    v2w=6,82·−2,4=6,8+4,−8=10,0.v2w=6,82·−2,4=6,8+4,−8=10,0.
Checkpoint 2.5

Let a=7,1a=7,1 and let bb be the vector with initial point (3,2)(3,2) and terminal point (−1,−1).(−1,−1).

  1. Find a.a.
  2. Express bb in component form.
  3. Find 3a4b.3a4b.

Now that we have established the basic rules of vector arithmetic, we can state the properties of vector operations. We will prove two of these properties. The others can be proved in a similar manner.

Theorem 2.1

Properties of Vector Operations

Let u,v,andwu,v,andw be vectors in a plane. Let r and sr and s be scalars.

i.u+v=v+uCommutative propertyii.(u+v)+w=u+(v+w)Associative propertyiii.u+0=uAdditive identity propertyiv.u+(u)=0Additive inverse propertyv.r(su)=(rs)uAssociativity of scalar multiplicationvi.(r+s)u=ru+suDistributive propertyvii.r(u+v)=ru+rvDistributive propertyviii.1u=u,0u=0Identity and zero propertiesi.u+v=v+uCommutative propertyii.(u+v)+w=u+(v+w)Associative propertyiii.u+0=uAdditive identity propertyiv.u+(u)=0Additive inverse propertyv.r(su)=(rs)uAssociativity of scalar multiplicationvi.(r+s)u=ru+suDistributive propertyvii.r(u+v)=ru+rvDistributive propertyviii.1u=u,0u=0Identity and zero properties

Proof of Commutative Property

Let u=x1,y1u=x1,y1 and v=x2,y2.v=x2,y2. Apply the commutative property for real numbers:

u+v=x1+x2,y1+y2=x2+x1,y2+y1=v+u.u+v=x1+x2,y1+y2=x2+x1,y2+y1=v+u.

Proof of Distributive Property

Apply the distributive property for real numbers:

r(u+v)=r·x1+x2,y1+y2=r(x1+x2),r(y1+y2)=rx1+rx2,ry1+ry2=rx1,ry1+rx2,ry2=ru+rv.r(u+v)=r·x1+x2,y1+y2=r(x1+x2),r(y1+y2)=rx1+rx2,ry1+ry2=rx1,ry1+rx2,ry2=ru+rv.

Checkpoint 2.6

Prove the additive inverse property.

We have found the components of a vector given its initial and terminal points. In some cases, we may only have the magnitude and direction of a vector, not the points. For these vectors, we can identify the horizontal and vertical components using trigonometry (Figure 2.15).

This figure is a right triangle. There is an angle labeled theta. The two sides are labeled “magnitude of v times cosine theta” and “magnitude of v times sine theta.” The hypotenuse is labeled “magnitude of v.”
Figure 2.15 The components of a vector form the legs of a right triangle, with the vector as the hypotenuse.

Consider the angle θθ formed by the vector v and the positive x-axis. We can see from the triangle that the components of vector vv are vcosθ,vsinθ.vcosθ,vsinθ. Therefore, given an angle and the magnitude of a vector, we can use the cosine and sine of the angle to find the components of the vector.

Example 2.6

Finding the Component Form of a Vector Using Trigonometry

Find the component form of a vector with magnitude 4 that forms an angle of −45°−45° with the x-axis.

Solution

Let xx and yy represent the components of the vector (Figure 2.16). Then x=4cos(−45°)=22x=4cos(−45°)=22 and y=4sin(−45°)=−22.y=4sin(−45°)=−22. The component form of the vector is 22,−22.22,−22.

This figure is a right triangle. The two sides are labeled “x” and “y.” The hypotenuse is labeled “4.” There is also an angle labeled “45 degrees.” The hypotenuse is represented as a vector.
Figure 2.16 Use trigonometric ratios, x=vcosθx=vcosθ and y=vsinθ,y=vsinθ, to identify the components of the vector.

Checkpoint 2.7

Find the component form of vector vv with magnitude 1010 that forms an angle of 120°120° with the positive x-axis.

Unit Vectors

A unit vector is a vector with magnitude 1.1. For any nonzero vector v,v, we can use scalar multiplication to find a unit vector uu that has the same direction as v.v. To do this, we multiply the vector by the reciprocal of its magnitude:

u=1vv.u=1vv.

Recall that when we defined scalar multiplication, we noted that kv=|k|·v.kv=|k|·v. For u=1vv,u=1vv, it follows that u=1v(v)=1.u=1v(v)=1. We say that uu is the unit vector in the direction ofvv (Figure 2.17). The process of using scalar multiplication to find a unit vector with a given direction is called normalization.

This image has two figures. The first is a vector labeled “v.” The second figure is a vector in the same direction labeled “u.” This vector has a length of 1 unit.
Figure 2.17 The vector vv and associated unit vector u=1vv.u=1vv. In this case, v>1.v>1.

Example 2.7

Finding a Unit Vector

Let v=1,2.v=1,2.

  1. Find a unit vector with the same direction as v.v.
  2. Find a vector ww with the same direction as vv such that w=7.w=7.

Solution

  1. First, find the magnitude of v,v, then divide the components of vv by the magnitude:
    v=12+22=1+4=5v=12+22=1+4=5

    u=1vv=151,2=15,25.u=1vv=151,2=15,25.
  2. The vector uu is in the same direction as vv and u=1.u=1. Use scalar multiplication to increase the length of uu without changing direction:
    w=7u=715,25=75,145.w=7u=715,25=75,145.
Checkpoint 2.8

Let v=9,2.v=9,2. Find a vector with magnitude 55 in the opposite direction as v.v.

We have seen how convenient it can be to write a vector in component form. Sometimes, though, it is more convenient to write a vector as a sum of a horizontal vector and a vertical vector. To make this easier, let’s look at standard unit vectors. The standard unit vectors are the vectors i=1,0i=1,0 and j=0,1j=0,1 (Figure 2.18).

This figure has the x and y axes of a coordinate system in the first quadrant. On the x-axis there is a vector labeled “i,” which equals <1,0>. The second vector is on the y-axis and is labeled “j” which equals <0,1>.
Figure 2.18 The standard unit vectors ii and j.j.

By applying the properties of vectors, it is possible to express any vector in terms of ii and jj in what we call a linear combination:

v=x,y=x,0+0,y=x1,0+y0,1=xi+yj.v=x,y=x,0+0,y=x1,0+y0,1=xi+yj.

Thus, vv is the sum of a horizontal vector with magnitude x,x, and a vertical vector with magnitude y,y, as in the following figure.

This figure is a right triangle. The horizontal side is labeled “xi.” The vertical side is labeled “yj.” The hypotenuse is a vector labeled “v.”
Figure 2.19 The vector vv is the sum of xixi and yj.yj.

Example 2.8

Using Standard Unit Vectors

  1. Express the vector w=3,−4w=3,−4 in terms of standard unit vectors.
  2. Vector uu is a unit vector that forms an angle of 60°60° with the positive x-axis. Use standard unit vectors to describe u.u.

Solution

  1. Resolve vector ww into a vector with a zero y-component and a vector with a zero x-component:
    w=3,−4=3i4j.w=3,−4=3i4j.
  2. Because uu is a unit vector, the terminal point lies on the unit circle when the vector is placed in standard position (Figure 2.20).
    u=cos60°,sin60°=12,32=12i+32j.u=cos60°,sin60°=12,32=12i+32j.

    This figure is a unit circle. It is a circle centered at the origin. It has a vector with initial point at the origin and terminal point on the circle. The terminal point is labeled (cos(theta), sin(theta)). The length of the vector is 1 unit. There is also a right triangle formed with the vector as the hypotenuse. The horizontal side is labeled “cos(theta)” and the vertical side is labeled “sin(theta).”
    Figure 2.20 The terminal point of uu lies on the unit circle (cosθ,sinθ).(cosθ,sinθ).
Checkpoint 2.9

Let a=16,−11a=16,−11 and let bb be a unit vector that forms an angle of 225°225° with the positive x-axis. Express aa and bb in terms of the standard unit vectors.

Applications of Vectors

Because vectors have both direction and magnitude, they are valuable tools for solving problems involving such applications as motion and force. Recall the boat example and the quarterback example we described earlier. Here we look at two other examples in detail.

Example 2.9

Finding Resultant Force

Jane’s car is stuck in the mud. Lisa and Jed come along in a truck to help pull her out. They attach one end of a tow strap to the front of the car and the other end to the truck’s trailer hitch, and the truck starts to pull. Meanwhile, Jane and Jed get behind the car and push. The truck generates a horizontal force of 300300 lb on the car. Jane and Jed are pushing at a slight upward angle and generate a force of 150150 lb on the car. These forces can be represented by vectors, as shown in Figure 2.21. The angle between these vectors is 15°.15°. Find the resultant force (the vector sum) and give its magnitude to the nearest tenth of a pound and its direction angle from the positive x-axis.

This image is the side view of an automobile. From the front of the automobile there is a horizontal vector labeled “300 pounds.” Also, from the front of the automobile there is another vector labeled “150 pounds.” The angle between the two vectors is 15 degrees.
Figure 2.21 Two forces acting on a car in different directions.

Solution

To find the effect of combining the two forces, add their representative vectors. First, express each vector in component form or in terms of the standard unit vectors. For this purpose, it is easiest if we align one of the vectors with the positive x-axis. The horizontal vector, then, has initial point (0,0)(0,0) and terminal point (300,0).(300,0). It can be expressed as 300,0300,0 or 300i.300i.

The second vector has magnitude 150150 and makes an angle of 15°15° with the first, so we can express it as 150cos(15°),150sin(15°),150cos(15°),150sin(15°), or 150cos(15°)i+150sin(15°)j.150cos(15°)i+150sin(15°)j. Then, the sum of the vectors, or resultant vector, is r=300,0+150cos(15°),150sin(15°),r=300,0+150cos(15°),150sin(15°), and we have

r=(300+150cos(15°))2+(150sin(15°))2446.6.r=(300+150cos(15°))2+(150sin(15°))2446.6.

The angle θθ made by rr and the positive x-axis has tanθ=150sin15°(300+150cos15°)0.09,tanθ=150sin15°(300+150cos15°)0.09, so θtan−1(0.09)5°,θtan−1(0.09)5°, which means the resultant force rr has an angle of 5°5° above the horizontal axis.

Example 2.10

Finding Resultant Velocity

An airplane flies due west at an airspeed of 425425 mph. The wind is blowing from the northeast at 4040 mph. What is the ground speed of the airplane? What is the bearing of the airplane?

Solution

Let’s start by sketching the situation described (Figure 2.22).

This figure is the image of an airplane. Coming out of the front of the airplane are two vectors. The first vector is labeled “425” and the second vector is labeled “40.” The angle between the vectors is 45 degrees.
Figure 2.22 Initially, the plane travels due west. The wind is from the northeast, so it is blowing to the southwest. The angle between the plane’s course and the wind is 45°.45°. (Figure not drawn to scale.)

Set up a sketch so that the initial points of the vectors lie at the origin. Then, the plane’s velocity vector is p=−425i.p=−425i. The vector describing the wind makes an angle of 225°225° with the positive x-axis:

w=40cos(225°),40sin(225°)=402,402=402i402j.w=40cos(225°),40sin(225°)=402,402=402i402j.

When the airspeed and the wind act together on the plane, we can add their vectors to find the resultant force:

p+w=−425i+(402i402j)=(−425402)i402j.p+w=−425i+(402i402j)=(−425402)i402j.

The magnitude of the resultant vector shows the effect of the wind on the ground speed of the airplane:

p+w=(−425402)2+(402)2454.17mphp+w=(−425402)2+(402)2454.17mph

As a result of the wind, the plane is traveling at approximately 454454 mph relative to the ground.

To determine the bearing of the airplane, we want to find the direction of the vector p+w:p+w:

tanθ=402(−425402)0.06θ3.57°.tanθ=402(−425402)0.06θ3.57°.

The overall direction of the plane is 3.57°3.57° south of west.

Checkpoint 2.10

An airplane flies due north at an airspeed of 550550 mph. The wind is blowing from the northwest at 5050 mph. What is the ground speed of the airplane?

Section 2.1 Exercises

For the following exercises, consider points P(−1,3),P(−1,3), Q(1,5),Q(1,5), and R(−3,7).R(−3,7). Determine the requested vectors and express each of them a. in component form and b. by using the standard unit vectors.

1.

PQPQ

2.

PRPR

3.

QPQP

4.

RPRP

5.

PQ+PRPQ+PR

6.

PQPRPQPR

7.

2PQ2PR2PQ2PR

8.

2PQ+12PR2PQ+12PR

9.

The unit vector in the direction of PQPQ

10.

The unit vector in the direction of PRPR

11.

A vector vv has initial point (−1,−3)(−1,−3) and terminal point (2,1).(2,1). Find the unit vector in the direction of v.v. Express the answer in component form.

12.

A vector vv has initial point (−2,5)(−2,5) and terminal point (3,−1).(3,−1). Find the unit vector in the direction of v.v. Express the answer in component form.

13.

The vector vv has initial point P(1,0)P(1,0) and terminal point QQ that is on the y-axis and above the initial point. Find the coordinates of terminal point QQ such that the magnitude of the vector vv is 5.5.

14.

The vector vv has initial point P(1,1)P(1,1) and terminal point QQ that is on the x-axis and left of the initial point. Find the coordinates of terminal point QQ such that the magnitude of the vector vv is 10.10.

For the following exercises, use the given vectors aa and b.b.

  1. Determine the vector sum a+ba+b and express it in both the component form and by using the standard unit vectors.
  2. Find the vector difference abab and express it in both the component form and by using the standard unit vectors.
  3. Verify that the vectors a,a, b,b, and a+b,a+b, and, respectively, a,a, b,b, and abab satisfy the triangle inequality.
  4. Determine the vectors 2a,2a, b,b, and 2ab.2ab. Express the vectors in both the component form and by using standard unit vectors.
15.

a=2i+j,a=2i+j, b=i+3jb=i+3j

16.

a=2i,a=2i, b=−2i+2jb=−2i+2j

17.

Let aa be a standard-position vector with terminal point (−2,−4).(−2,−4). Let bb be a vector with initial point (1,2)(1,2) and terminal point (−1,4).(−1,4). Find the magnitude of vector −3a+b4i+j.−3a+b4i+j.

18.

Let aa be a standard-position vector with terminal point at (2,5).(2,5). Let bb be a vector with initial point (−1,3)(−1,3) and terminal point (1,0).(1,0). Find the magnitude of vector a3b+14i14j.a3b+14i14j.

19.

Let uu and vv be two nonzero vectors that are nonequivalent. Consider the vectors a=4u+5va=4u+5v and b=u+2vb=u+2v defined in terms of uu and v.v. Find the scalar λλ such that vectors a+λba+λb and uvuv are equivalent.

20.

Let uu and vv be two nonzero vectors that are nonequivalent. Consider the vectors a=2u4va=2u4v and b=3u7vb=3u7v defined in terms of uu and v.v. Find the scalars αα and ββ such that vectors αa+βbαa+βb and uvuv are equivalent.

21.

Consider the vector a(t)=cost,sinta(t)=cost,sint with components that depend on a real number t.t. As the number tt varies, the components of a(t)a(t) change as well, depending on the functions that define them.

  1. Write the vectors a(0)a(0) and a(π)a(π) in component form.
  2. Show that the magnitude a(t)a(t) of vector a(t)a(t) remains constant for any real number t.t.
  3. As tt varies, show that the terminal point of vector a(t)a(t) describes a circle centered at the origin of radius 1.1.
22.

Consider vector a(x)=x,1x2a(x)=x,1x2 with components that depend on a real number x[−1,1].x[−1,1]. As the number xx varies, the components of a(x)a(x) change as well, depending on the functions that define them.

  1. Write the vectors a(0)a(0) and a(1)a(1) in component form.
  2. Show that the magnitude a(x)a(x) of vector a(x)a(x) remains constant for any real number xx
  3. As xx varies, show that the terminal point of vector a(x)a(x) describes a circle centered at the origin of radius 1.1.
23.

Show that vectors a(t)=cost,sinta(t)=cost,sint and a(x)=x,1x2a(x)=x,1x2 are equivalent for x=rx=r and t=2kπ,t=2kπ, where kk is an integer.

24.

Show that vectors a(t)=cost,sinta(t)=cost,sint and a(x)=x,1x2a(x)=x,1x2 are opposite for x=rx=r and t=π+2kπ,t=π+2kπ, where kk is an integer.

For the following exercises, find vector vv with the given magnitude and in the same direction as vector u.u.

25.

v=7,u=3,4v=7,u=3,4

26.

v=3,u=−2,5v=3,u=−2,5

27.

v=7,u=3,−5v=7,u=3,−5

28.

v=10,u=2,−1v=10,u=2,−1

For the following exercises, find the component form of vector u,u, given its magnitude and the angle the vector makes with the positive x-axis. Give exact answers when possible.

29.

u=2,u=2, θ=30°θ=30°

30.

u=6,u=6, θ=60°θ=60°

31.

u=5,u=5, θ=π2θ=π2

32.

u=8,u=8, θ=πθ=π

33.

u=10,u=10, θ=5π6θ=5π6

34.

u=50,u=50, θ=3π4θ=3π4

For the following exercises, vector uu is given. Find the angle θ[0,2π)θ[0,2π) that vector uu makes with the positive direction of the x-axis, in a counter-clockwise direction.

35.

u=52i52ju=52i52j

36.

u=3iju=3ij

37.

Let a=a1,a2,a=a1,a2, b=b1,b2,b=b1,b2, and c=c1,c2c=c1,c2 be three nonzero vectors. If a1b2a2b10,a1b2a2b10, then show there are two scalars, αα and β,β, such that c=αa+βb.c=αa+βb.

38.

Consider vectors a=2,−4,a=2,−4, b=−1,2,b=−1,2, and c = 0 Determine the scalars αα and ββ such that c=αa+βb.c=αa+βb.

39.

Let P(x0,f(x0))P(x0,f(x0)) be a fixed point on the graph of the differential function ff with a domain that is the set of real numbers.

  1. Determine the real number z0z0 such that point Q(x0+1,z0)Q(x0+1,z0) is situated on the line tangent to the graph of ff at point P.P.
  2. Determine the unit vector uu with initial point PP and terminal point Q.Q.
40.

Consider the function f(x)=x4,f(x)=x4, where x.x.

  1. Determine the real number z0z0 such that point Q(2,z0)Q(2,z0) s situated on the line tangent to the graph of ff at point P(1,1).P(1,1).
  2. Determine the unit vector uu with initial point PP and terminal point Q.Q.
41.

Consider ff and gg two functions defined on the same set of real numbers D.D. Let a=x,f(x)a=x,f(x) and b=x,g(x)b=x,g(x) be two vectors that describe the graphs of the functions, where xD.xD. Show that if the graphs of the functions ff and gg do not intersect, then the vectors aa and bb are not equivalent.

42.

Find xx such that vectors a=x,sinxa=x,sinx and b=x,cosxb=x,cosx are equivalent.

43.

Calculate the coordinates of point DD such that ABCDABCD is a parallelogram, with A(1,1),A(1,1), B(2,4),B(2,4), and C(7,4).C(7,4).

44.

Consider the points A(2,1),A(2,1), B(10,6),B(10,6), C(13,4),C(13,4), and D(16,−2).D(16,−2). Determine the component form of vector AD.AD.

45.

The speed of an object is the magnitude of its related velocity vector. A football thrown by a quarterback has an initial speed of 7070 mph and an angle of elevation of 30°.30°. Determine the velocity vector in mph and express it in component form. (Round to two decimal places.)

This figure shows a football being thrown. The path of the football is represented by an arcing curve. At the beginning of the curve is a vector indicating the initial velocity.
46.

A baseball player throws a baseball at an angle of 30°30° with the horizontal. If the initial speed of the ball is 100100 mph, find the horizontal and vertical components of the initial velocity vector of the baseball. (Round to two decimal places.)

47.

A bullet is fired with an initial velocity of 15001500 ft/sec at an angle of 60°60° with the horizontal. Find the horizontal and vertical components of the velocity vector of the bullet. (Round to two decimal places.)

This figure is the first quadrant of a coordinate system. There is a vector from the origin that is labeled “v sub 0 = 1500 feet per second.” The angle between the x-axis and the vector is 60 degrees.
48.

[T] A 65-kg sprinter exerts a force of 798798 N at a 19°19° angle with respect to the ground on the starting block at the instant a race begins. Find the horizontal component of the force. (Round to two decimal places.)

49.

[T] Two forces, a horizontal force of 4545 lb and another of 5252 lb, act on the same object. The angle between these forces is 25°.25°. Find the magnitude and direction angle from the positive x-axis of the resultant force that acts on the object. (Round to two decimal places.)

This figure has two vectors with the same initial point. The first vector is labeled “52 lb” and the second vector is labeled “45 lb.” The angle between the vectors is 25 degrees.
50.

[T] Two forces, a vertical force of 2626 lb and another of 4545 lb, act on the same object. The angle between these forces is 55°.55°. Find the magnitude and direction angle from the positive x-axis of the resultant force that acts on the object. (Round to two decimal places.)

51.

[T] Three forces act on object. Two of the forces have the magnitudes 5858 N and 2727 N, and make angles 53°53° and 152°,152°, respectively, with the positive x-axis. Find the magnitude and the direction angle from the positive x-axis of the third force such that the resultant force acting on the object is zero. (Round to two decimal places.)

52.

Three forces with magnitudes 8080 lb, 120120 lb, and 6060 lb act on an object at angles of 45°,45°, 60°60° and 30°,30°, respectively, with the positive x-axis. Find the magnitude and direction angle from the positive x-axis of the resultant force. (Round to two decimal places.)

This figure is the first quadrant of a coordinate system. There are three vectors, all with the origin as the initial point. The first vector is labeled “60 lb” and makes an angle of 30 degrees with the x-axis. The second vector is labeled “80 lb” and makes an angle of 45 degrees with the x-axis. The third vector is labeled “120 lb” and makes a 60 degree angle with the x-axis.
53.

[T] An airplane is flying in the direction of 43°43° east of north (also abbreviated as N43E)N43E) at a speed of 550550 mph. A wind with speed 2525 mph comes from the southwest at a bearing of N15E.N15E. What are the ground speed and new direction of the airplane?

This figure is the first quadrant of a coordinate system. There are two vectors both of which have the origin as the initial point. The first vector is labeled “550 miles per hour” and has an angle of 43 degrees from the y-axis. There is also an image of an airplane at the end of the vector. The second vector is labeled “25 miles per hour” and has an angle of 15 degrees from the y-axis.
54.

[T] A boat is traveling in the water at 3030 mph in a direction of N20EN20E (that is, 20°20° east of north). A strong current is moving at 1515 mph in a direction of N45E.N45E. What are the new speed and direction of the boat?

This figure is the first quadrant of a coordinate system. There are two vectors both of which have the origin as the initial point. The first vector is labeled “15” and has an angle of 45 degrees from the y-axis. The second vector is labeled “30” and has an angle of 20 degrees from the y-axis. There is also an image of a boat at the end of the vector.
55.

[T] A 50-lb weight is hung by a cable so that the two portions of the cable make angles of 40°40° and 53°,53°, respectively, with the horizontal. Find the magnitudes of the forces of tension T1T1 and T2T2 in the cables if the resultant force acting on the object is zero. (Round to two decimal places.)

This figure is a horizontal line with two vectors below the line formina a triangle with the line. The vectors point toward the ends of the line. The angle between the horizontal line and the first vector is 40 degrees. The angle between the horizontal line and the second vector is 53 degrees. At the point where the two vectors meet is a rectangle labeled “50 lb.”
56.

[T] A 62-lb weight hangs from a rope that makes the angles of 29°29° and 61°,61°, respectively, with the horizontal. Find the magnitudes of the forces of tension T1T1 and T2T2 in the cables if the resultant force acting on the object is zero. (Round to two decimal places.)

57.

[T] A 1500-lb boat is parked on a ramp that makes an angle of 30°30° with the horizontal. The boat’s weight vector points downward and is a sum of two vectors: a horizontal vector v1v1 that is parallel to the ramp and a vertical vector v2v2 that is perpendicular to the inclined surface. The magnitudes of vectors v1v1 and v2v2 are the horizontal and vertical component, respectively, of the boat’s weight vector. Find the magnitudes of v1v1 and v2.v2. (Round to the nearest integer.)

This figure shows a right triangle. The angle between the horizontal base and the hypotenuse is 30 degrees. On the hypotenuse is the image of a boat. From center of the boat there are three vectors. Two of the vectors are orthogonal with one in the direction of the boat and the other below the boat. The third vector is down, perpendicular to the vertical side.
58.

[T] An 85-lb box is at rest on a 26°26° incline. Determine the magnitude of the force parallel to the incline necessary to keep the box from sliding. (Round to the nearest integer.)

59.

A guy-wire supports a pole that is 7575 ft high. One end of the wire is attached to the top of the pole and the other end is anchored to the ground 5050 ft from the base of the pole. Determine the horizontal and vertical components of the force of tension in the wire if its magnitude is 5050 lb. (Round to the nearest integer.)

This figure is a tower with a guy wire from the top to the ground. The tower, guy wire, and the ground form a right triangle. The base is labeled “50 feet” and is horizontal. The other side is labeled “75 feet” and is vertical. This side is the tower.
60.

A telephone pole guy-wire has an angle of elevation of 35°35° with respect to the ground. The force of tension in the guy-wire is 120120 lb. Find the horizontal and vertical components of the force of tension. (Round to the nearest integer.)

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