unit 8 vectors homework 1 introduction to vectors

8.8 Vectors

Learning objectives.

In this section you will:

• View vectors geometrically.
• Find magnitude and direction.
• Perform vector addition and scalar multiplication.
• Find the component form of a vector.
• Find the unit vector in the direction of  v v .
• Perform operations with vectors in terms of  i i  and  j j .
• Find the dot product of two vectors.

An airplane is flying at an airspeed of 200 miles per hour headed on a SE bearing of 140°. A north wind (from north to south) is blowing at 16.2 miles per hour, as shown in Figure 1 . What are the ground speed and actual bearing of the plane?

Ground speed refers to the speed of a plane relative to the ground. Airspeed refers to the speed a plane can travel relative to its surrounding air mass. These two quantities are not the same because of the effect of wind. In an earlier section, we used triangles to solve a similar problem involving the movement of boats. Later in this section, we will find the airplane’s groundspeed and bearing, while investigating another approach to problems of this type. First, however, let’s examine the basics of vectors.

A Geometric View of Vectors

A vector is a specific quantity drawn as a line segment with an arrowhead at one end. It has an initial point , where it begins, and a terminal point , where it ends. A vector is defined by its magnitude , or the length of the line, and its direction, indicated by an arrowhead at the terminal point. Thus, a vector is a directed line segment. There are various symbols that distinguish vectors from other quantities:

• Lower case, boldfaced type, with or without an arrow on top such as v , v , u , u , w , w , v → , v → , u → , w → . u → , w → .
• Given initial point P P and terminal point Q , Q , a vector can be represented as P Q → . P Q → . The arrowhead on top is what indicates that it is not just a line, but a directed line segment.
• Given an initial point of ( 0 , 0 ) ( 0 , 0 ) and terminal point ( a , b ) , ( a , b ) , a vector may be represented as 〈 a , b 〉 . 〈 a , b 〉 .

This last symbol 〈 a , b 〉 〈 a , b 〉 has special significance. It is called the standard position . The position vector has an initial point ( 0 , 0 ) ( 0 , 0 ) and a terminal point ( a , b ) . ( a , b ) . To change any vector into the position vector, we think about the change in the x -coordinates and the change in the y -coordinates. Thus, if the initial point of a vector C D → C D → is C ( x 1 , y 1 ) C ( x 1 , y 1 ) and the terminal point is D ( x 2 , y 2 ) , D ( x 2 , y 2 ) , then the position vector is found by calculating

In Figure 2 , we see the original vector C D → C D → and the position vector A b → . A b → .

Properties of Vectors

A vector is a directed line segment with an initial point and a terminal point. Vectors are identified by magnitude, or the length of the line, and direction, represented by the arrowhead pointing toward the terminal point. The position vector has an initial point at ( 0 , 0 ) ( 0 , 0 ) and is identified by its terminal point ( a , b ) . ( a , b ) .

Find the Position Vector

Consider the vector whose initial point is P ( 2 , 3 ) P ( 2 , 3 ) and terminal point is Q ( 6 , 4 ) . Q ( 6 , 4 ) . Find the position vector.

The position vector is found by subtracting one x -coordinate from the other x -coordinate, and one y -coordinate from the other y -coordinate. Thus

The position vector begins at ( 0 , 0 ) ( 0 , 0 ) and terminates at ( 4 , 1 ) . ( 4 , 1 ) . The graphs of both vectors are shown in Figure 3 .

We see that the position vector is 〈 4 , 1 〉 . 〈 4 , 1 〉 .

Drawing a Vector with the Given Criteria and Its Equivalent Position Vector

Find the position vector given that vector v v has an initial point at ( − 3 , 2 ) ( − 3 , 2 ) and a terminal point at ( 4 , 5 ) , ( 4 , 5 ) , then graph both vectors in the same plane.

The position vector is found using the following calculation:

Thus, the position vector begins at ( 0 , 0 ) ( 0 , 0 ) and terminates at ( 7 , 3 ) . ( 7 , 3 ) . See Figure 4 .

Draw a vector v v that connects from the origin to the point ( 3 , 5 ) . ( 3 , 5 ) .

Finding Magnitude and Direction

To work with a vector, we need to be able to find its magnitude and its direction. We find its magnitude using the Pythagorean Theorem or the distance formula, and we find its direction using the inverse tangent function.

Magnitude and Direction of a Vector

Given a position vector v v = 〈 a , b 〉 , = 〈 a , b 〉 , the magnitude is found by | v | = a 2 + b 2 . | v | = a 2 + b 2 . The direction is equal to the angle formed with the x -axis, or with the y -axis, depending on the application. For a position vector, the direction is found by tan θ = ( b a ) ⇒ θ = tan − 1 ( b a ) , tan θ = ( b a ) ⇒ θ = tan − 1 ( b a ) , as illustrated in Figure 5 .

Two vectors v and u are considered equal if they have the same magnitude and the same direction. Additionally, if both vectors have the same position vector, they are equal.

Finding the Magnitude and Direction of a Vector

Find the magnitude and direction of the vector with initial point P ( − 8 , 1 ) P ( − 8 , 1 ) and terminal point Q ( − 2 , − 5 ) . Q ( − 2 , − 5 ) . Draw the vector.

First, find the position vector .

We use the Pythagorean Theorem to find the magnitude.

The direction is given as

However, the angle terminates in the fourth quadrant, so we add 360° to obtain a positive angle. Thus, − 45° + 360° = 315° . − 45° + 360° = 315° . See Figure 6 .

Showing That Two Vectors Are Equal

Show that vector v with initial point at ( 5 , −3 ) ( 5 , −3 ) and terminal point at ( −1 , 2 ) ( −1 , 2 ) is equal to vector u with initial point at ( −1 , −3 ) ( −1 , −3 ) and terminal point at ( −7 , 2 ) . ( −7 , 2 ) . Draw the position vector on the same grid as v and u . Next, find the magnitude and direction of each vector.

As shown in Figure 7 , draw the vector v v starting at initial ( 5 , −3 ) ( 5 , −3 ) and terminal point ( −1 , 2 ) . ( −1 , 2 ) . Draw the vector u u with initial point ( −1 , −3 ) ( −1 , −3 ) and terminal point ( −7 , 2 ) . ( −7 , 2 ) . Find the standard position for each.

Next, find and sketch the position vector for v and u . We have

Since the position vectors are the same, v and u are the same.

An alternative way to check for vector equality is to show that the magnitude and direction are the same for both vectors. To show that the magnitudes are equal, use the Pythagorean Theorem.

As the magnitudes are equal, we now need to verify the direction. Using the tangent function with the position vector gives

However, we can see that the position vector terminates in the second quadrant, so we add 180° . 180° . Thus, the direction is − 39.8° + 180° = 140.2° . − 39.8° + 180° = 140.2° .

Performing Vector Addition and Scalar Multiplication

Now that we understand the properties of vectors, we can perform operations involving them. While it is convenient to think of the vector u u = 〈 x , y 〉 = 〈 x , y 〉 as an arrow or directed line segment from the origin to the point ( x , y ) , ( x , y ) , vectors can be situated anywhere in the plane. The sum of two vectors u and v , or vector addition , produces a third vector u + v , the resultant vector.

To find u + v , we first draw the vector u , and from the terminal end of u , we drawn the vector v . In other words, we have the initial point of v meet the terminal end of u . This position corresponds to the notion that we move along the first vector and then, from its terminal point, we move along the second vector. The sum u + v is the resultant vector because it results from addition or subtraction of two vectors. The resultant vector travels directly from the beginning of u to the end of v in a straight path, as shown in Figure 8 .

Vector subtraction is similar to vector addition. To find u − v , view it as u + (− v ). Adding − v is reversing direction of v and adding it to the end of u . The new vector begins at the start of u and stops at the end point of − v . See Figure 9 for a visual that compares vector addition and vector subtraction using parallelograms .

Adding and Subtracting Vectors

Given u u = 〈 3 , − 2 〉 = 〈 3 , − 2 〉 and v v = 〈 −1 , 4 〉 , = 〈 −1 , 4 〉 , find two new vectors u + v , and u − v .

To find the sum of two vectors, we add the components. Thus,

See Figure 10 (a) .

To find the difference of two vectors, add the negative components of v v to u . u . Thus,

See Figure 10 (b).

Multiplying By a Scalar

While adding and subtracting vectors gives us a new vector with a different magnitude and direction, the process of multiplying a vector by a scalar , a constant, changes only the magnitude of the vector or the length of the line. Scalar multiplication has no effect on the direction unless the scalar is negative, in which case the direction of the resulting vector is opposite the direction of the original vector.

Scalar Multiplication

Scalar multiplication involves the product of a vector and a scalar. Each component of the vector is multiplied by the scalar. Thus, to multiply v v = 〈 a , b 〉 = 〈 a , b 〉 by k k , we have

Only the magnitude changes, unless k k is negative, and then the vector reverses direction.

Performing Scalar Multiplication

Given vector v v = 〈 3 , 1 〉 , = 〈 3 , 1 〉 , find 3 v , 1 2 1 2 v , v , and − v .

See Figure 11 for a geometric interpretation. If v v = 〈 3 , 1 〉 , = 〈 3 , 1 〉 , then

Notice that the vector 3 v is three times the length of v , 1 2 1 2 v v is half the length of v , and – v is the same length of v , but in the opposite direction.

Find the scalar multiple 3 u u given u u = 〈 5 , 4 〉 . = 〈 5 , 4 〉 .

Using Vector Addition and Scalar Multiplication to Find a New Vector

Given u u = 〈 3 , − 2 〉 = 〈 3 , − 2 〉 and v v = 〈 − 1 , 4 〉 , = 〈 − 1 , 4 〉 , find a new vector w = 3 u + 2 v .

First, we must multiply each vector by the scalar.

Then, add the two together.

So, w w = 〈 7 , 2 〉 . = 〈 7 , 2 〉 .

Finding Component Form

In some applications involving vectors, it is helpful for us to be able to break a vector down into its components. Vectors are comprised of two components: the horizontal component is the x x direction, and the vertical component is the y y direction. For example, we can see in the graph in Figure 12 that the position vector 〈 2 , 3 〉 〈 2 , 3 〉 comes from adding the vectors v 1 and v 2 . We have v 1 with initial point ( 0 , 0 ) ( 0 , 0 ) and terminal point ( 2 , 0 ) . ( 2 , 0 ) .

We also have v 2 with initial point ( 0 , 0 ) ( 0 , 0 ) and terminal point ( 0 , 3 ) . ( 0 , 3 ) .

Therefore, the position vector is

Using the Pythagorean Theorem, the magnitude of v 1 is 2, and the magnitude of v 2 is 3. To find the magnitude of v , use the formula with the position vector.

The magnitude of v is 13 . 13 . To find the direction, we use the tangent function tan θ = y x . tan θ = y x .

Thus, the magnitude of v v is 13 13 and the direction is 56.3 ∘ 56.3 ∘ off the horizontal.

Finding the Components of the Vector

Find the components of the vector v v with initial point ( 3 , 2 ) ( 3 , 2 ) and terminal point ( 7 , 4 ) . ( 7 , 4 ) .

First find the standard position.

See the illustration in Figure 13 .

The horizontal component is v 1 v 1 = 〈 4 , 0 〉 = 〈 4 , 0 〉 and the vertical component is v 2 v 2 = 〈 0 , 2 〉. = 〈 0 , 2 〉.

Finding the Unit Vector in the Direction of v

In addition to finding a vector’s components, it is also useful in solving problems to find a vector in the same direction as the given vector, but of magnitude 1. We call a vector with a magnitude of 1 a unit vector . We can then preserve the direction of the original vector while simplifying calculations.

Unit vectors are defined in terms of components. The horizontal unit vector is written as i i = 〈 1 , 0 〉 = 〈 1 , 0 〉 and is directed along the positive horizontal axis. The vertical unit vector is written as j j = 〈 0 , 1 〉 = 〈 0 , 1 〉 and is directed along the positive vertical axis. See Figure 14 .

The Unit Vectors

If v v is a nonzero vector, then v | v | v | v | is a unit vector in the direction of v . v . Any vector divided by its magnitude is a unit vector. Notice that magnitude is always a scalar, and dividing by a scalar is the same as multiplying by the reciprocal of the scalar.

Find a unit vector in the same direction as v v = 〈 −5 , 12 〉. = 〈 −5 , 12 〉.

First, we will find the magnitude.

Then we divide each component by | v |, | v |, which gives a unit vector in the same direction as v :

or, in component form

See Figure 15 .

Verify that the magnitude of the unit vector equals 1. The magnitude of − 5 13 i + 12 13 j − 5 13 i + 12 13 j is given as

The vector u = 5 13 = 5 13 i + 12 13 + 12 13 j is the unit vector in the same direction as v = 〈 − 5 , 12 〉 . = 〈 − 5 , 12 〉 .

Performing Operations with Vectors in Terms of i and j

So far, we have investigated the basics of vectors: magnitude and direction, vector addition and subtraction, scalar multiplication, the components of vectors, and the representation of vectors geometrically. Now that we are familiar with the general strategies used in working with vectors, we will represent vectors in rectangular coordinates in terms of i and j .

Vectors in the Rectangular Plane

Given a vector v v with initial point P = ( x 1 , y 1 ) P = ( x 1 , y 1 ) and terminal point Q = ( x 2 , y 2 ), Q = ( x 2 , y 2 ), v is written as

The position vector from ( 0 , 0 ) ( 0 , 0 ) to ( a , b ) , ( a , b ) , where ( x 2 − x 1 ) = a ( x 2 − x 1 ) = a and ( y 2 − y 1 ) = b , ( y 2 − y 1 ) = b , is written as v = a i + b j . This vector sum is called a linear combination of the vectors i and j .

The magnitude of v = a i + b j is given as | v | = a 2 + b 2 . | v | = a 2 + b 2 . See Figure 16 .

Writing a Vector in Terms of i and j

Given a vector v v with initial point P = ( 2 , −6 ) P = ( 2 , −6 ) and terminal point Q = ( −6 , 6 ) , Q = ( −6 , 6 ) , write the vector in terms of i i and j . j .

Begin by writing the general form of the vector. Then replace the coordinates with the given values.

Writing a Vector in Terms of i and j Using Initial and Terminal Points

Given initial point P 1 = ( − 1 , 3 ) P 1 = ( − 1 , 3 ) and terminal point P 2 = ( 2 , 7 ) , P 2 = ( 2 , 7 ) , write the vector v v in terms of i i and j . j .

Write the vector u u with initial point P = ( − 1 , 6 ) P = ( − 1 , 6 ) and terminal point Q = ( 7 , − 5 ) Q = ( 7 , − 5 ) in terms of i i and j . j .

Performing Operations on Vectors in Terms of i and j

When vectors are written in terms of i i and j , j , we can carry out addition, subtraction, and scalar multiplication by performing operations on corresponding components.

Adding and Subtracting Vectors in Rectangular Coordinates

Given v = a i + b j and u = c i + d j , then

Finding the Sum of the Vectors

Find the sum of v 1 = 2 i − 3 j v 1 = 2 i − 3 j and v 2 = 4 i + 5 j . v 2 = 4 i + 5 j .

According to the formula, we have

Calculating the Component Form of a Vector: Direction

We have seen how to draw vectors according to their initial and terminal points and how to find the position vector. We have also examined notation for vectors drawn specifically in the Cartesian coordinate plane using i and j . i and j . For any of these vectors, we can calculate the magnitude. Now, we want to combine the key points, and look further at the ideas of magnitude and direction.

Calculating direction follows the same straightforward process we used for polar coordinates. We find the direction of the vector by finding the angle to the horizontal. We do this by using the basic trigonometric identities, but with | v | | v | replacing r . r .

Vector Components in Terms of Magnitude and Direction

Given a position vector v = 〈 x , y 〉 v = 〈 x , y 〉 and a direction angle θ , θ ,

Thus, v = x i + y j = | v | cos θ i + | v | sin θ j , v = x i + y j = | v | cos θ i + | v | sin θ j , and magnitude is expressed as | v | = x 2 + y 2 . | v | = x 2 + y 2 .

Writing a Vector in Component Form When It Is Given in Magnitude and Direction Form

Given a vector with length 7 and an angle of 135°, write it in component form.

Using the conversion formulas x = | v | cos θ x = | v | cos θ and y = | v | sin θ , y = | v | sin θ , we find that

This vector can be written as v = 7 cos ( 135° ) + 7 sin ( 135° ) v = 7 cos ( 135° ) + 7 sin ( 135° ) or simplified as

A vector travels from the origin to the point ( 3 , 5 ) . ( 3 , 5 ) . Write the vector in terms of magnitude and direction.

Finding the Dot Product of Two Vectors

As we discussed earlier in the section, scalar multiplication involves multiplying a vector by a scalar, and the result is a vector. As we have seen, multiplying a vector by a number is called scalar multiplication. If we multiply a vector by a vector, there are two possibilities: the dot product and the cross product . We will only examine the dot product here; you may encounter the cross product in more advanced mathematics courses.

The dot product of two vectors involves multiplying two vectors together, and the result is a scalar.

Dot Product

The dot product of two vectors v = 〈 a , b 〉 v = 〈 a , b 〉 and u = 〈 c , d 〉 u = 〈 c , d 〉 is the sum of the product of the horizontal components and the product of the vertical components.

To find the angle between the two vectors, use the formula below.

Find the dot product of v = 〈 5 , 12 〉 v = 〈 5 , 12 〉 and u = 〈 −3 , 4 〉 . u = 〈 −3 , 4 〉 .

Using the formula, we have

Finding the Dot Product of Two Vectors and the Angle between Them

Find the dot product of v 1 = 5 i + 2 j and v 2 = 3 i + 7 j . Then, find the angle between the two vectors.

Finding the dot product, we multiply corresponding components.

To find the angle between them, we use the formula cos θ = v | v | ⋅ u | u | . cos θ = v | v | ⋅ u | u | .

See Figure 17 .

Finding the Angle between Two Vectors

Find the angle between u = 〈 − 3 , 4 〉 u = 〈 − 3 , 4 〉 and v = 〈 5 , 12 〉 . v = 〈 5 , 12 〉 .

See Figure 18 .

Finding Ground Speed and Bearing Using Vectors

We now have the tools to solve the problem we introduced in the opening of the section.

An airplane is flying at an airspeed of 200 miles per hour headed on a SE bearing of 140°. A north wind (from north to south) is blowing at 16.2 miles per hour. What are the ground speed and actual bearing of the plane? See Figure 19 .

The ground speed is represented by x x in the diagram, and we need to find the angle α α in order to calculate the adjusted bearing, which will be 140° + α . 140° + α .

Notice in Figure 19 , that angle b C O b C O must be equal to angle A O C A O C by the rule of alternating interior angles, so angle b C O b C O is 140°. We can find x x by the Law of Cosines:

The ground speed is approximately 213 miles per hour. Now we can calculate the bearing using the Law of Sines.

Therefore, the plane has a SE bearing of 140°+2.8°=142.8°. The ground speed is 212.7 miles per hour.

Access these online resources for additional instruction and practice with vectors.

• Introduction to Vectors
• Vector Operations
• The Unit Vector

8.8 Section Exercises

What are the characteristics of the letters that are commonly used to represent vectors?

How is a vector more specific than a line segment?

What are i i and j , j , and what do they represent?

What is component form?

When a unit vector is expressed as 〈 a , b 〉 , 〈 a , b 〉 , which letter is the coefficient of the i i and which the j ? j ?

Given a vector with initial point ( 5 , 2 ) ( 5 , 2 ) and terminal point ( − 1 , − 3 ) , ( − 1 , − 3 ) , find an equivalent vector whose initial point is ( 0 , 0 ) . ( 0 , 0 ) . Write the vector in component form 〈 a , b 〉 . 〈 a , b 〉 .

Given a vector with initial point ( − 4 , 2 ) ( − 4 , 2 ) and terminal point ( 3 , − 3 ) , ( 3 , − 3 ) , find an equivalent vector whose initial point is ( 0 , 0 ) . ( 0 , 0 ) . Write the vector in component form 〈 a , b 〉 . 〈 a , b 〉 .

Given a vector with initial point ( 7 , − 1 ) ( 7 , − 1 ) and terminal point ( − 1 , − 7 ) , ( − 1 , − 7 ) , find an equivalent vector whose initial point is ( 0 , 0 ) . ( 0 , 0 ) . Write the vector in component form 〈 a , b 〉 . 〈 a , b 〉 .

For the following exercises, determine whether the two vectors u u and v v are equal, where u u has an initial point P 1 P 1 and a terminal point P 2 P 2 and v v has an initial point P 3 P 3 and a terminal point P 4 P 4 .

P 1 = ( 5 , 1 ) , P 2 = ( 3 , − 2 ) , P 3 = ( − 1 , 3 ) , P 1 = ( 5 , 1 ) , P 2 = ( 3 , − 2 ) , P 3 = ( − 1 , 3 ) , and P 4 = ( 9 , − 4 ) P 4 = ( 9 , − 4 )

P 1 = ( 2 , − 3 ) , P 2 = ( 5 , 1 ) , P 3 = ( 6 , − 1 ) , P 1 = ( 2 , − 3 ) , P 2 = ( 5 , 1 ) , P 3 = ( 6 , − 1 ) , and P 4 = ( 9 , 3 ) P 4 = ( 9 , 3 )

P 1 = ( − 1 , − 1 ) , P 2 = ( − 4 , 5 ) , P 3 = ( − 10 , 6 ) , P 1 = ( − 1 , − 1 ) , P 2 = ( − 4 , 5 ) , P 3 = ( − 10 , 6 ) , and P 4 = ( − 13 , 12 ) P 4 = ( − 13 , 12 )

P 1 = ( 3 , 7 ) , P 2 = ( 2 , 1 ) , P 3 = ( 1 , 2 ) , P 1 = ( 3 , 7 ) , P 2 = ( 2 , 1 ) , P 3 = ( 1 , 2 ) , and P 4 = ( − 1 , − 4 ) P 4 = ( − 1 , − 4 )

P 1 = ( 8 , 3 ) , P 2 = ( 6 , 5 ) , P 3 = ( 11 , 8 ) , P 1 = ( 8 , 3 ) , P 2 = ( 6 , 5 ) , P 3 = ( 11 , 8 ) , and P 4 = ( 9 , 10 ) P 4 = ( 9 , 10 )

Given initial point P 1 = ( − 3 , 1 ) P 1 = ( − 3 , 1 ) and terminal point P 2 = ( 5 , 2 ) , P 2 = ( 5 , 2 ) , write the vector v v in terms of i i and j . j .

Given initial point P 1 = ( 6 , 0 ) P 1 = ( 6 , 0 ) and terminal point P 2 = ( − 1 , − 3 ) , P 2 = ( − 1 , − 3 ) , write the vector v v in terms of i i and j . j .

For the following exercises, use the vectors u = i + 5 j , v = −2 i − 3 j ,  and w = 4 i − j .

Find u + ( v − w )

Find 4 v + 2 u

For the following exercises, use the given vectors to compute u + v , u − v , and 2 u − 3 v .

u = 〈 2 , − 3 〉 , v = 〈 1 , 5 〉 u = 〈 2 , − 3 〉 , v = 〈 1 , 5 〉

u = 〈 − 3 , 4 〉 , v = 〈 − 2 , 1 〉 u = 〈 − 3 , 4 〉 , v = 〈 − 2 , 1 〉

Let v = −4 i + 3 j . Find a vector that is half the length and points in the same direction as v . v .

Let v = 5 i + 2 j . Find a vector that is twice the length and points in the opposite direction as v . v .

For the following exercises, find a unit vector in the same direction as the given vector.

a = 3 i + 4 j

b = −2 i + 5 j

c = 10 i – j

d = − 1 3 i + 5 2 j d = − 1 3 i + 5 2 j

u = 100 i + 200 j

u = −14 i + 2 j

For the following exercises, find the magnitude and direction of the vector, 0 ≤ θ < 2 π . 0 ≤ θ < 2 π .

〈 0 , 4 〉 〈 0 , 4 〉

〈 6 , 5 〉 〈 6 , 5 〉

〈 2 , −5 〉 〈 2 , −5 〉

〈 −4 , −6 〉 〈 −4 , −6 〉

Given u = 3 i − 4 j and v = −2 i + 3 j , calculate u ⋅ v . u ⋅ v .

Given u = − i − j and v = i + 5 j , calculate u ⋅ v . u ⋅ v .

Given u = 〈 − 2 , 4 〉 u = 〈 − 2 , 4 〉 and v = 〈 − 3 , 1 〉 , v = 〈 − 3 , 1 〉 , calculate u ⋅ v . u ⋅ v .

Given u = 〈 − 1 , 6 〉 = 〈 − 1 , 6 〉 and v = 〈 6 , − 1 〉 , = 〈 6 , − 1 〉 , calculate u ⋅ v . u ⋅ v .

For the following exercises, given v , v , draw v , v , 3 v and 1 2 v . 1 2 v .

〈 2 , −1 〉 〈 2 , −1 〉

〈 −1 , 4 〉 〈 −1 , 4 〉

〈 −3 , −2 〉 〈 −3 , −2 〉

For the following exercises, use the vectors shown to sketch u + v , u − v , and 2 u .

For the following exercises, use the vectors shown to sketch 2 u + v .

For the following exercises, use the vectors shown to sketch u − 3 v .

For the following exercises, write the vector shown in component form.

Given initial point P 1 = ( 2 , 1 ) P 1 = ( 2 , 1 ) and terminal point P 2 = ( − 1 , 2 ) , P 2 = ( − 1 , 2 ) , write the vector v v in terms of i i and j , j , then draw the vector on the graph.

Given initial point P 1 = ( 4 , − 1 ) P 1 = ( 4 , − 1 ) and terminal point P 2 = ( − 3 , 2 ) , P 2 = ( − 3 , 2 ) , write the vector v v in terms of i i and j . j . Draw the points and the vector on the graph.

Given initial point P 1 = ( 3 , 3 ) P 1 = ( 3 , 3 ) and terminal point P 2 = ( − 3 , 3 ) , P 2 = ( − 3 , 3 ) , write the vector v v in terms of i i and j . j . Draw the points and the vector on the graph.

For the following exercises, use the given magnitude and direction in standard position, write the vector in component form.

| v | = 6 , θ = 45 ° | v | = 6 , θ = 45 °

| v | = 8 , θ = 220° | v | = 8 , θ = 220°

| v | = 2 , θ = 300° | v | = 2 , θ = 300°

| v | = 5 , θ = 135° | v | = 5 , θ = 135°

A 60-pound box is resting on a ramp that is inclined 12°. Rounding to the nearest tenth,

• ⓐ Find the magnitude of the normal (perpendicular) component of the force.
• ⓑ Find the magnitude of the component of the force that is parallel to the ramp.

A 25-pound box is resting on a ramp that is inclined 8°. Rounding to the nearest tenth,

Find the magnitude of the horizontal and vertical components of a vector with magnitude 8 pounds pointed in a direction of 27° above the horizontal. Round to the nearest hundredth.

Find the magnitude of the horizontal and vertical components of the vector with magnitude 4 pounds pointed in a direction of 127° above the horizontal. Round to the nearest hundredth.

Find the magnitude of the horizontal and vertical components of a vector with magnitude 5 pounds pointed in a direction of 55° above the horizontal. Round to the nearest hundredth.

Find the magnitude of the horizontal and vertical components of the vector with magnitude 1 pound pointed in a direction of 8° above the horizontal. Round to the nearest hundredth.

Real-World Applications

A woman leaves home and walks 3 miles west, then 2 miles southwest. How far from home is she, and in what direction must she walk to head directly home?

A boat leaves the marina and sails 6 miles north, then 2 miles northeast. How far from the marina is the boat, and in what direction must it sail to head directly back to the marina?

A man starts walking from home and walks 4 miles east, 2 miles southeast, 5 miles south, 4 miles southwest, and 2 miles east. How far has he walked? If he walked straight home, how far would he have to walk?

A woman starts walking from home and walks 4 miles east, 7 miles southeast, 6 miles south, 5 miles southwest, and 3 miles east. How far has she walked? If she walked straight home, how far would she have to walk?

A man starts walking from home and walks 3 miles at 20° north of west, then 5 miles at 10° west of south, then 4 miles at 15° north of east. If he walked straight home, how far would he have to the walk, and in what direction?

A woman starts walking from home and walks 6 miles at 40° north of east, then 2 miles at 15° east of south, then 5 miles at 30° south of west. If she walked straight home, how far would she have to walk, and in what direction?

An airplane is heading north at an airspeed of 600 km/hr, but there is a wind blowing from the southwest at 80 km/hr. How many degrees off course will the plane end up flying, and what is the plane’s speed relative to the ground?

An airplane is heading north at an airspeed of 500 km/hr, but there is a wind blowing from the northwest at 50 km/hr. How many degrees off course will the plane end up flying, and what is the plane’s speed relative to the ground?

An airplane needs to head due north, but there is a wind blowing from the southwest at 60 km/hr. The plane flies with an airspeed of 550 km/hr. To end up flying due north, how many degrees west of north will the pilot need to fly the plane?

An airplane needs to head due north, but there is a wind blowing from the northwest at 80 km/hr. The plane flies with an airspeed of 500 km/hr. To end up flying due north, how many degrees west of north will the pilot need to fly the plane?

As part of a video game, the point ( 5 , 7 ) ( 5 , 7 ) is rotated counterclockwise about the origin through an angle of 35°. Find the new coordinates of this point.

As part of a video game, the point ( 7 , 3 ) ( 7 , 3 ) is rotated counterclockwise about the origin through an angle of 40°. Find the new coordinates of this point.

Two children are throwing a ball back and forth straight across the back seat of a car. The ball is being thrown 10 mph relative to the car, and the car is traveling 25 mph down the road. If one child doesn't catch the ball, and it flies out the window, in what direction does the ball fly (ignoring wind resistance)?

Two children are throwing a ball back and forth straight across the back seat of a car. The ball is being thrown 8 mph relative to the car, and the car is traveling 45 mph down the road. If one child doesn't catch the ball, and it flies out the window, in what direction does the ball fly (ignoring wind resistance)?

A 50-pound object rests on a ramp that is inclined 19°. Find the magnitude of the components of the force parallel to and perpendicular to (normal) the ramp to the nearest tenth of a pound.

Suppose a body has a force of 10 pounds acting on it to the right, 25 pounds acting on it upward, and 5 pounds acting on it 45° from the horizontal. What single force is the resultant force acting on the body?

Suppose a body has a force of 10 pounds acting on it to the right, 25 pounds acting on it ─135° from the horizontal, and 5 pounds acting on it directed 150° from the horizontal. What single force is the resultant force acting on the body?

The condition of equilibrium is when the sum of the forces acting on a body is the zero vector. Suppose a body has a force of 2 pounds acting on it to the right, 5 pounds acting on it upward, and 3 pounds acting on it 45° from the horizontal. What single force is needed to produce a state of equilibrium on the body?

Suppose a body has a force of 3 pounds acting on it to the left, 4 pounds acting on it upward, and 2 pounds acting on it 30° from the horizontal. What single force is needed to produce a state of equilibrium on the body? Draw the vector.

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10.2: An Introduction to Vectors

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• Page ID 4214

• Gregory Hartman et al.
• Virginia Military Institute

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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.

Definition 51 Vector

A vector is a directed line segment.

Given points \(P\) and \(Q\) (either in the plane or in space), we denote with \(\vec{PQ}\) 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 \(\vec{PQ}\) is the length of the line segment \(\overline{PQ}\):

\[\norm{\vec{PQ}} = \norm{\overline{PQ}}.\]

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

Figure 10.18 shows multiple instances of the same vector. Each directed line segment has the same direction and length (magnitude), hence each is the same vector.

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.

Consider the vectors \(\vec{PQ}\) and \(\vec{RS}\) as shown in Figure 10.19. 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 \(x\), \(y\) and possibly \(z\) directions the terminal point is from the initial point. Both the vectors \(\vec{PQ}\) and \(\vec{RS}\) in Figure 10.19 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, \((0,0)\). This leads us to a definition of a standard and concise way of referring to vectors.

Definition 52 Component Form of a Vector

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

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

It follows from the definition that the component form of the vector \(\vec{PQ}\), where \(P=(x_1,y_1)\) and \(Q=(x_2,y_2)\) is

\[\vec{PQ} = \langle x_2-x_1, y_2-y_1\rangle;\]

in space, where \(P=(x_1,y_1,z_1)\) and \(Q=(x_2,y_2,z_2)\), the component form of \(\vec{PQ}\) is

\[\vec{PQ} = \langle x_2-x_1, y_2-y_1,z_2-z_1\rangle.\]

We practice using this notation in the following example.

Example \(\PageIndex{1}\): Using component form notation for vectors

• Sketch the vector \(\vec v=\langle 2,-1\rangle\) starting at \(P=(3,2)\) and find its magnitude.
• Find the component form of the vector \(\vec w\) whose initial point is \(R=(-3,-2)\) and whose terminal point is \(S=(-1,2)\).
• Sketch the vector \(\vec u = \langle 2,-1,3\rangle\) starting at the point \(Q = (1,1,1)\) and find its magnitude.
• Using \(P\) as the initial point, we move 2 units in the positive \(x\)-direction and \(-1\) units in the positive \(y\)-direction to arrive at the terminal point \(P\,'=(5,1)\), as drawn in Figure 10.20(a). The magnitude of \(\vec v\) is determined directly from the component form: \[\norm{\vec v} =\sqrt{2^2+(-1)^2} = \sqrt{5}. \]
• Using the note following Definition 52, we have \[\vec{RS} = \langle -1-(-3), 2-(-2)\rangle = \langle 2,4\rangle.\] One can readily see from Figure 10.20(a) that the \(x\)- and \(y\)-displacement of \(\vec{RS}\) is 2 and 4, respectively, as the component form suggests.
• Using \(Q\) as the initial point, we move 2 units in the positive \(x\)-direction, \(-1\) unit in the positive \(y\)-direction, and 3 units in the positive \(z\)-direction to arrive at the terminal point \(Q' = (3,0,4)\), illustrated in Figure 10.20(b). The magnitude of \(\vec u\) is: \[\norm{\vec u} = \sqrt{2^2+(-1)^2+3^2} = \sqrt{14}.\]

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.

Definition 53 VECTOR ALGEBRA

• Let \(\vec u = \langle u_1,u_2\rangle\) and \(\vec v = \langle v_1,v_2\rangle\) be vectors in \(\mathbb{R}^2\), and let \(c\) be a scalar. (a) The addition, or sum, of the vectors \(\vec u\) and \(\vec v\) is the vector \[\vec u+\vec v = \langle u_1+v_1, u_2+v_2\rangle.\] (b) The scalar product of \(c\) and \(\vec v\) is the vector \[c\vec v = c\langle v_1,v_2\rangle = \langle cv_1,cv_2\rangle.\]
• Let \(\vec u = \langle u_1,u_2,u_3\rangle\) and \(\vec v = \langle v_1,v_2,v_3\rangle\) be vectors in \(\mathbb{R}^3\), and let \(c\) be a scalar. (a) The addition, or sum, of the vectors \(\vec u\) and \(\vec v\) is the vector \[\vec u+\vec v = \langle u_1+v_1, u_2+v_2, u_3+v_3\rangle.\] (b) The scalar product of \(c\) and \(\vec v\) is the vector \[c\vec v = c\langle v_1,v_2,v_3\rangle = \langle cv_1,cv_2,cv_3\rangle.\]

In short, we say addition and scalar multiplication are computed "component--wise.''

Example \(\PageIndex{2}\): Adding vectors

Sketch the vectors \(\vec u = \langle1,3\rangle\), \(\vec v = \langle 2,1\rangle\) and \(\vec u+\vec v\) all with initial point at the origin. Solution

We first compute \(\vec u +\vec v\).

\[\begin{align*} \vec u+\vec v &= \langle 1,3\rangle + \langle 2,1\rangle\\ &= \langle 3,4\rangle. \end{align*}\] These are all sketched in Figure 10.21.

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:

\[\text{"Starting at an initial point, go out \(\vec u\), then go out \(\vec v\)."}\]

This idea is sketched in Figure 10.22, where the initial point of \(\vec v\) is the terminal point of \(\vec u\). 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.''

Analytically, it is easy to see that \(\vec u+\vec v = \vec v+\vec u\). Figure 10.22 also gives a graphical representation of this, using gray vectors. Note that the vectors \(\vec u\) and \(\vec v\), 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 \(\vec v\); 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 53 that \[\vec u-\vec v = \vec u + (-1)\vec v.\] The Parallelogram Law gives us a good way to visualize this subtraction. We demonstrate this in the following example.

Example \(\PageIndex{3}\): Vector Subtraction

Let \(\vec u = \langle 3,1\rangle\) and \(\vec v=\langle 1,2\rangle.\) Compute and sketch \(\vec u-\vec v\).

The computation of \(\vec u-\vec v\) is straightforward, and we show all steps below. Usually the formal step of multiplying by \((-1)\) is omitted and we "just subtract.''

\[\begin{align*} \vec u-\vec v &= \vec u + (-1)\vec v \\ &= \langle 3,1\rangle + \langle -1,-2\rangle\\ &= \langle 2,-1\rangle. \end{align*}\]

Figure 10.23 illustrates, using the Head to Tail Rule, how the subtraction can be viewed as the sum \(\vec u + (-\vec v)\). The figure also illustrates how \(\vec u-\vec v\) can be obtained by looking only at the terminal points of \(\vec u\) and \(\vec v\) (when their initial points are the same).

Example \(\PageIndex{4}\): Scaling vectors

• Sketch the vectors \(\vec v = \langle 2,1\rangle\) and \(2\vec v\) with initial point at the origin.
• Compute the magnitudes of \(\vec v\) and \(2\vec v\).
• We compute \(2\vec v\): \[\begin{align*}2\vec v &= 2\langle 2,1\rangle\\&= \langle 4,2\rangle.\end{align*}\] Both \(\vec v\) and \(2\vec v\) are sketched in Figure10.24. Make note that \(2\vec v\) does not start at the terminal point of \(\vec v\); rather, its initial point is also the origin.
• The figure suggests that \(2\vec v\) is twice as long as \(\vec v\). We compute their magnitudes to confirm this. \[\begin{align*}\norm{\vec v} &= \sqrt{2^2+1^2}\\&= \sqrt{5}.\\\norm{2\vec v}&=\sqrt{4^2+2^2} \\&= \sqrt{20}\\&= \sqrt{4\cdot 5} = 2\sqrt{5}.\end{align*}\] As we suspected, \(2\vec v\) is twice as long as \(\vec v\).

The zero vector is the vector whose initial point is also its terminal point. It is denoted by \(\vec 0\). Its component form, in \(\mathbb{R}^2\), is \(\langle 0,0\rangle\); in \(\mathbb{R}^3\), it is \(\langle 0,0,0\rangle\). 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.

THEOREM 84 PROPERTIES OF VECTOR OPERATIONS

The following are true for all scalars \(c\) and \(d\), and for all vectors \(\vec u\), \(\vec v\) and \(\vec w\), where \(\vec u\), \(\vec v\) and \(\vec w\) are all in \(\mathbb{R}^2\) or where \(\vec u\), \(\vec v\) and \(\vec w\) are all in \(\mathbb{R}^3\):

• \(\underbrace{\vec u+\vec v = \vec v+\vec u}_{Commutative Property}\)
• \(\underbrace{\vec u+\vec v)+\vec w = \vec u+(\vec v+\vec w)}_{Associative Property}\)
• \(\underbrace{\vec v+\vec 0 = \vec v}_{Additive Identity}\)
• \((cd)\vec v= c(d\vec v)\)
• \(\underbrace{c(\vec u+\vec v) = c\vec u+c\vec v}_{Distributive Property}\)
• \(\underbrace{(c+d)\vec v = c\vec v+d\vec v}_{Distributive Property}\)
• \(0\vec v = \vec 0\)
• \(\norm{c\vec v} = |c|\cdot\norm{\vec v}\)
• \(\norm u = 0\) if, and only if, \(\vec u = \vec 0\).

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

Definition 54 Unit Vector

A unit vector is a vector \(\vec v\) with a magnitude of 1; that is,

\[\norm{\vec v}=1.\]

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

Property 8 of Theorem 84 holds the key. If we divide \(\vec v\) by its magnitude, it becomes a vector of length 1. Consider:

\[\begin{align*} \Big\| \frac{1}{\norm{\vec v}}\vec v\Big\| &= \frac{1}{\norm{\vec v}}\norm{\vec v} & \text{ (we can pull out \( \frac{1}{\norm{\vec v}}\) as it is a scalar)}\\ &= 1. \end{align*} \]

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

Example \(\PageIndex{5}\): Using Unit Vectors

Let \(\vec v= \langle 3,1\rangle\) and let \(\vec w = \langle 1,2,2\rangle\).

• Find the unit vector in the direction of \(\vec v\).
• Find the unit vector in the direction of \(\vec w\).
• Find the vector in the direction of \(\vec v\) with magnitude 5.
• We find \(\norm{\vec v} = \sqrt{10}\). So the unit vector \(\vec u\) in the direction of \(\vec v\) is \[\vec u = \frac{1}{\sqrt{10}}\vec v = \langle \frac{3}{\sqrt{10}},\frac{1}{\sqrt{10}}\rangle.\]
• We find \(\norm{\vec w} = 3\), so the unit vector \(\vec z\) in the direction of \(\vec w\) is \[\vec u = \frac13\vec w = \langle \frac13,\frac23,\frac23\rangle.\]
• To create a vector with magnitude 5 in the direction of \(\vec v\), we multiply the unit vector \(\vec u\) by 5. Thus \(5\vec u = \langle 15/\sqrt{10},5/\sqrt{10}\rangle\) is the vector we seek. This is sketched in Figure 10.25.

The basic formation of the unit vector \(\vec u\) in the direction of a vector \(\vec v\) leads to a interesting equation. It is: \[\vec v = \norm{\vec v}\frac{1}{\norm{\vec v}}\vec v.\] We rewrite the equation with parentheses to make a point:

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 .

Definition 55 Parallel Vectors

• Unit vectors \(\vec u_1\) and \(\vec u_2\) are parallel if \(\vec u_1 = \pm \vec u_2\).
• 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).

Note : \(\vec 0\) is directionless; because \(\norm{0}=0\), there is no unit vector in the "direction'' of \(\vec 0\).

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 \(\vec v\).

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.

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 \(\langle \cos\theta,\sin\theta\rangle\) for some angle \(\theta\).

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 \(\mathbb{R}^2\). Important concepts about unit vectors are given in the following Key Idea.

KEY IDEA 48 UNIT VECTORS

• The unit vector in the direction of \(\vec v\) is \[ \vec u = \frac1{\norm{\vec v}} \vec v.\]
• A vector \(\vec u\) in \(\mathbb{R}^2\) is a unit vector if, and only if, its component form is \(\langle \cos\theta,\sin\theta\rangle\) for some angle \(\theta\).
• A vector \(\vec u\) in \(\mathbb{R}^3\) is a unit vector if, and only if, its component form is \(\langle \sin\theta\cos\varphi,\sin\theta\sin\varphi,\cos\theta\rangle\) for some angles \(\theta\) and \(\varphi\).

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

Example \(\PageIndex{6}\): Finding Component Forces

Consider a weight of 50lb hanging from two chains, as shown in Figure 10.26. One chain makes an angle of \(30^\circ\) with the vertical, and the other an angle of \(45^\circ\). Find the force applied to each chain.

Solution Knowing that gravity is pulling the 50lb weight straight down, we can create a vector \(\vec F\) to represent this force. \[\vec F = 50\langle 0,-1\rangle = \langle 0,-50\rangle.\]

We can view each chain as "pulling'' the weight up, preventing it from falling. We can represent the force from each chain with a vector. Let \(\vec F_1\) represent the force from the chain making an angle of \(30^\circ\) with the vertical, and let \(\vec F_2\) represent the force form the other chain. Convert all angles to be measured from the horizontal (as shown in Figure 10.27), and apply Key Idea 48. As we do not yet know the magnitudes of these vectors, (that is the problem at hand), we use \(m_1\) and \(m_2\) to represent them.

\[\vec F_1 = m_1\langle \cos 120^\circ,\sin120^\circ\rangle\]

\[\vec F_2 = m_2\langle \cos 45^\circ,\sin45^\circ\rangle\]

As the weight is not moving, we know the sum of the forces is \(\vec 0\). This gives:

\[\begin{align*} \vec F + \vec F_1 + \vec F_2 & = \vec 0\\ \langle 0,-50\rangle + m_1\langle \cos 120^\circ,\sin120^\circ\rangle + m_2\langle \cos 45^\circ,\sin45^\circ\rangle &=\vec 0 \end{align*}\]

The sum of the entries in the first component is 0, and the sum of the entries in the second component is also 0. This leads us to the following two equations: \[\begin{align*} m_1\cos120^\circ + m_2\cos45^\circ &=0 \\ m_1\sin120^\circ + m_2\sin45^\circ &=50 \end{align*}\] This is a simple 2-equation, 2-unkown system of linear equations. We leave it to the reader to verify that the solution is \[m_1=50(\sqrt{3}-1) \approx 36.6;\qquad m_2=\frac{50\sqrt{2}}{1+\sqrt{3}} \approx 25.88.\]

It might seem odd that the sum of the forces applied to the chains is more than 50lb. We leave it to a physics class to discuss the full details, but offer this short explanation. Our equations were established so that the vertical components of each force sums to 50lb, thus supporting the weight. Since the chains are at an angle, they also pull against each other, creating an "additional'' horizontal force while holding the weight in place.

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.

Definition 56 STANDARD UNIT VECTORS

• In \(\mathbb{R}^2\), the standard unit vectors are \[\vec i = \langle 1,0\rangle \quad \text{and}\quad \vec j = \langle 0,1\rangle.\]
• In \(\mathbb{R}^3\), the standard unit vectors are \[\vec i = \langle 1,0,0\rangle \quad \text{and}\quad \vec j = \langle 0,1,0\rangle \quad \text{and}\quad \vec k = \langle 0,0,1\rangle.\]

Example \(\PageIndex{7}\): Using standard unit vectors

• Rewrite \(\vec v = \langle 2,-3\rangle\) using the standard unit vectors.
• Rewrite \(\vec w = 4\vec i - 5\vec j +2\vec k\) in component form.
• \[\begin{align}\vec v &= \langle 2,-3\rangle \\&= \langle 2,0\rangle + \langle 0,-3\rangle \\&= 2\langle 1,0\rangle -3\langle 0,1\rangle\\&= 2\vec i - 3\vec j\end{align}\]
• \[\begin{align}\vec w &= 4\vec i - 5\vec j +2\vec k\\&= \langle 4,0,0\rangle +\langle 0,-5,0\rangle + \langle 0,0,2\rangle \\&= \langle 4,-5,2\rangle\end{align}\]

These two examples demonstrate that converting between component form and the standard unit vectors is rather straightforward. Many mathematicians prefer component form, and it is the preferred notation in this text. Many engineers prefer using the standard unit vectors, and many engineering text use that notation.

Example \(\PageIndex{8}\): Finding 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 10.28. What angle will the chain make with the vertical as a result of the wind's pushing? How much higher will the weight be?

Solution The force of the wind is represented by the vector \(\vec F_w = 5\vec i\). The force of gravity on the weight is represented by \(\vec F_g = -25\vec j\). The direction and magnitude of the vector representing the force on the chain are both unknown. We represent this force with \[\vec F_c = m\langle \cos\varphi,\sin\varphi\rangle = m\cos\varphi\, \vec i + m\sin\varphi\,\vec j\] for some magnitude \(m\) and some angle with the horizontal \(\varphi\). (Note: \(\theta\) is the angle the chain makes with the vertical ; \(\varphi\) is the angle with the horizontal .)

As the weight is at equilibrium, the sum of the forces is \(\vec0\): \[\begin{align*} \vec F_c + \vec F_w + \vec F_g &= \vec 0\\ m\cos\varphi\, \vec i + m\sin\varphi\,\vec j + 5\vec i - 25\vec j &=\vec 0 \end{align*}\]

Thus the sum of the \(\vec i\) and \(\vec j\) components are 0, leading us to the following system of equations: \[5+m\cos\varphi = 0\] \[-25+m\sin\varphi = 0\label{eq:vect8}\]

This is enough to determine \(\vec F_c\) already, as we know \(m\cos \varphi = -5\) and \(m\sin\varphi =25\). Thus \(F_c = \langle -5,25\rangle.\) We can use this to find the magnitude \(m\): \[m = \sqrt{(-5)^2+25^2} = 5\sqrt{26}\approx 25.5\text{lb}.\] We can then use either equality from Equation \ref{eq:vect8} to solve for \(\varphi\). We choose the first equality as using arccosine will return an angle in the \(2^\text{nd}\) quadrant: \[5 + 5\sqrt{26}\cos \varphi = 0 \quad \Rightarrow \quad \varphi = \cos^{-1}\left(\frac{-5}{5\sqrt{26}}\right) \approx 1.7682\approx 101.31^\circ.\]

Subtracting \(90^\circ\) from this angle gives us an angle of \(11.31^\circ\) with the vertical.

We can now use trigonometry to find out how high the weight is lifted. The diagram shows that a right triangle is formed with the 2ft chain as the hypotenuse with an interior angle of \(11.31^\circ\). The length of the adjacent side (in the diagram, the dashed vertical line) is \(2\cos 11.31^\circ \approx 1.96\)ft. Thus the weight is lifted by about \(0.04\)ft, almost 1/2in.

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.

Contributors and Attributions

Gregory Hartman (Virginia Military Institute). Contributions were made by Troy Siemers and Dimplekumar Chalishajar of VMI and Brian Heinold of Mount Saint Mary's University. This content is copyrighted by a Creative Commons Attribution - Noncommercial (BY-NC) License.  http://www.apexcalculus.com/

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Pre-Calculus  Unit 8: Vectors

This unit includes 47 pages of guided notes, homework assignments, two quizzes, a study guide, and a unit test that cover the topics listed in the description below.

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This unit contains the following topics:

• Naming Vectors • Types of Vectors (Equivalent, Parallel, Opposite) • Proving Vectors are Equivalent • Component Form of a Vector • Finding Magnitude and Direction • Operations with Vectors • Unit Vectors • Standard Unit Vectors; Writing Vectors as a Linear Combination • Writing a Vector in Trigonometric Form • Applications: Resultant Force and Velocity Vectors • Horizontal and Vertical Components of a Vector • Dot Products • Orthogonal Vectors • Finding the Angle Between Vectors • Vector Projections • Vector Projections Applications: Force and Work • Three-Dimensional Coordinate System • Vectors in 3D Space: Component Form, Linear Combinations, Magnitude, Operations, Dot Products, and Angle between Vectors

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Unit 8 - Vectors

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• Math Review 31m
• Introduction to Units 23m
• Unit Conversions 18m
• Solving Density Problems 13m
• Dimensional Analysis 10m
• Counting Significant Figures 5m
• Operations with Significant Figures 11m
• Vectors, Scalars, & Displacement 13m
• Average Velocity 32m
• Intro to Acceleration 7m
• Position-Time Graphs & Velocity 26m
• Conceptual Problems with Position-Time Graphs 22m
• Velocity-Time Graphs & Acceleration 5m
• Calculating Displacement from Velocity-Time Graphs 15m
• Conceptual Problems with Velocity-Time Graphs 10m
• Calculating Change in Velocity from Acceleration-Time Graphs 10m
• Graphing Position, Velocity, and Acceleration Graphs 11m
• Kinematics Equations 37m
• Vertical Motion and Free Fall 19m
• Catch/Overtake Problems 23m
• Review of Vectors vs. Scalars 1m
• Introduction to Vectors 7m
• Adding Vectors Graphically 22m
• Vector Composition & Decomposition 11m
• Adding Vectors by Components 13m
• Trig Review 23m
• Unit Vectors 15m
• Introduction to Dot Product (Scalar Product) 12m
• Calculating Dot Product Using Components 12m
• Intro to Cross Product (Vector Product) 23m
• Calculating Cross Product Using Components 17m
• Intro to Motion in 2D: Position & Displacement 20m
• Velocity in 2D 27m
• Acceleration in 2D 12m
• Kinematics in 2D 18m
• Intro to Relative Velocity 23m
• Intro to Projectile Motion: Horizontal Launch 35m
• Negative (Downward) Launch 24m
• Symmetrical Launch 25m
• Projectiles Launched From Moving Vehicles 15m
• Special Equations in Symmetrical Launches 16m
• Positive (Upward) Launch 50m
• Using Equation Substitution 17m
• Newton's First & Second Laws 16m
• Types Of Forces & Free Body Diagrams 20m
• Forces & Kinematics 12m
• Vertical Forces & Acceleration 23m
• Vertical Equilibrium & The Normal Force 18m
• Forces in 2D 36m
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• Newton's Third Law & Action-Reaction Pairs 11m
• Forces in Connected Systems of Objects 38m
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• Uniform Circular Motion 7m
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• Newton's Law of Gravity 30m
• Gravitational Forces in 2D 25m
• Acceleration Due to Gravity 13m
• Satellite Motion: Intro 5m
• Satellite Motion: Speed & Period 35m
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• Overview of Kepler's Laws 5m
• Kepler's First Law 11m
• Kepler's Third Law 16m
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• Mass Distribution with Calculus 45m
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• Work On Inclined Planes 16m
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• Types of Motion & Energy 24m
• Conservation of Energy with Rotation 35m
• Torque with Kinematic Equations 56m
• Rotational Dynamics with Two Motions 50m
• Rotational Dynamics of Rolling Motion 27m
• Torque & Acceleration (Rotational Dynamics) 15m
• How to Solve: Energy vs Torque 10m
• Torque Due to Weight 23m
• Intro to Torque 26m
• Net Torque & Sign of Torque 13m
• Torque on Discs & Pulleys 35m
• Equilibrium with Multiple Objects 30m
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• Center of Mass & Simple Balance 29m
• Equilibrium in 2D - Ladder Problems 40m
• Beam / Shelf Against a Wall 53m
• More 2D Equilibrium Problems 14m
• Review: Center of Mass 14m
• Torque & Equilibrium 22m
• Opening/Closing Arms on Rotating Stool 18m
• Conservation of Angular Momentum 46m
• Angular Momentum & Newton's Second Law 10m
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• Gauss' Law 32m
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• Relationships Between Force, Field, Energy, Potential 25m
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• Equipotential Surfaces 13m
• Capacitors & Capacitance 8m
• Parallel Plate Capacitors 19m
• Energy Stored by Capacitor 15m
• Capacitance Using Calculus 7m
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• Resistors and Ohm's Law 14m
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• Toroidal Solenoids aka Toroids 12m
• Biot-Savart Law (Calculus) 18m
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• Intro to Induction 5m
• Magnetic Flux 12m
• Faraday's Law 28m
• Lenz's Law 22m
• Motional EMF 18m
• Transformers 9m
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• Self Inductance 18m
• Inductors 7m
• LR Circuits 22m
• LC Circuits 35m
• LRC Circuits 14m
• Alternating Voltages and Currents 18m
• RMS Current and Voltage 9m
• Phasors 20m
• Resistors in AC Circuits 9m
• Phasors for Resistors 7m
• Capacitors in AC Circuits 16m
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• Inductors in AC Circuits 13m
• Phasors for Inductors 7m
• Impedance in AC Circuits 18m
• Series LRC Circuits 11m
• Resonance in Series LRC Circuits 10m
• Power in AC Circuits 5m
• Intro to Electromagnetic (EM) Waves 21m
• The Electromagnetic Spectrum 7m
• Intensity of EM Waves 22m
• Wavefunctions of EM Waves 19m
• Radiation Pressure 24m
• Polarization & Polarization Filters 30m
• The Doppler Effect of Light 6m
• Ray Nature Of Light 10m
• Reflection of Light 9m
• Index of Refraction 16m
• Refraction of Light & Snell's Law 22m
• Total Internal Reflection 5m
• Ray Diagrams For Mirrors 36m
• Mirror Equation 19m
• Refraction At Spherical Surfaces 9m
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• Thin Lens And Lens Maker Equations 24m
• Diffraction 8m
• Diffraction with Huygen's Principle 14m
• Young's Double Slit Experiment 24m
• Single Slit Diffraction 27m
• Inertial Reference Frames 14m
• Special Vs. Galilean Relativity 17m
• Consequences of Relativity 52m
• Lorentz Transformations 45m

Introduction to Vectors - Video Tutorials & Practice Problems

Introduction to vector math.

Hey, guys, We're gonna be working with vectors a lot in physics, so you're gonna have to get very good at how we manipulate and combine them. So in this video, I want to introduce to you what vector math is and what it's all about. But I want to take a minute to describe what the next few videos they're gonna be about. We're gonna be working with vectors with all these diagrams with these little axes and these grids and boxes because it's a great way to understand, very visually what's going on vectors. Then, later on, what we're gonna see is we're gonna see all of the math equations that describe vectors. So let's get to it. So adding and subtracting scale er's is pretty easy to remember. Scaler Zehr Just simple numbers, But vectors have direction. So Mathis sometimes not a straightforward. For example, if you were to combine scaler, let's say you're moving and you're combining a 3 kg in 4 kg box, Then the way that you add these boxes together, if you were a lump them together in a single box is you just add three and four straight up, so three plus four is just seven. So combining scaler, XYZ just simple addition where things get a little bit more tricky is when you start combining vectors and it's really just two cases you're going to see you combine parallel vectors. Then there would be like walking 3 m to the right in 4 m to the right. So because vectors have direction, they're drawn as arrows. So this 3 m to the right and 4 m to the right would look like this. So this is 3 m and this is 4 m. Now, if you were to combine these two vectors together, it would basically just be as if you walked 7 m. So when you combine parallel vectors, the total displacement, if you will, is three plus four, which is 7 m. And so when they're parallel, they just add together like normal numbers. By the way, the same thing would work. If you actually had these two vectors together, let's say you had three and then four like this. Well, that would also just form a total displacement of seven. So as long as these things are parallel in whatever direction will always add together where things get a little bit. Trickier is when you're combining perpendicular vectors. So now let's say I walk 3 m to the right and then 4 m up. So we got this grid that's gonna help us visualize what's going on here. So I only walked 3 m to the right. So 123 and then I'm gonna walk 4 m up like this. So this is four, and this is three. Now, you might think the total displacement is just three plus four, which is seven. But it's actually not because your total displacement is really just the shortest path from where you start to where you end. So you were three and four. You're actually starting over here and you're ending over here. So what happens is we make a little triangle like this. So this is my total displacement. So how do I calculate this? Well, I can't just use simple addition. I'm gonna have to use a different kind of math. And this, actually, if you'll notice here, makes just a triangle. So vectors just make a bunch of triangles. The way we saw for this is by using an old idea an old math equation from algebra or trigonometry called the Pythagorean Theorem. So this is the Pythagorean theorem, which says that a squared plus B squared equals C squared. So as long as you know, two sides of a triangle, you can always figure out the third one. So I'm gonna call this a this one b, and this one is C. So if I want to figure out the sea, all I have to do is just c equals the square roots of a squared plus B squared. So that means my total displacement is this square roots of three squared in four squared, and that's actually equal to 5 m. So even though I went three and four, this actually is five in this direction and basically so we can see is that vectors just form a bunch of triangles. So that means that the way that we're going to solve vectors is just by using a lot of triangle math. That's really all there is to vector math. Let's go ahead and get some more examples. So I'm gonna walk 10 m to the right and then 6 m to the left. So I'm gonna walk 10 m to the rights so there's gonna be my 10 m and then m to the left. So something like this. So my total displacement, where to draw the displacement vectors which we are did and calculate the total displacement while if I walked 10 to the rights and then six to the left, then my total displacement is actually justify started here, and I ended here. And so this is just 10 minus six, which is just 4 m. So that's pretty straightforward. It's because these things air parallel or technically, they're anti parallel, but they sort of all long lie along the same plane or the same line like this is just Some of them are forwards and backwards. So now we're gonna walk 6 m to the right and 6 m and 8 m down for part B. So now I'm gonna walk 6 m to the right. Let's say it's like that, and then 8 m like this doesn't have to necessarily be to scale. So what is the total displacement? Well, just like before now we have perpendicular vectors, so we're gonna draw this line here that connects these two points, and this is going to be a triangle so notice how it just forms a triangle like this. And I wanna I wanna figure out with this, uh, see, or the high pot news of this triangle is so I'm just gonna use the Pythagorean theorem, so C is equal to six squared plus eight squared. Notice how you don't have to plug in. You know, you have to worry about signs or anything like that. So you're just gonna do six and eight and this equals 10 m. So this is the displacement over here. It's 10. Alright, guys, that's it for this one. Let me know if you have any questions.

Two perpendicular forces act on a box, one pushing to the right and one pushing up. An instrument tells you the magnitude of the total force is 13N. You measure the force pushing to the right is 12N. Calculate the force pushing up.

Do you want more practice, your physics tutor.

• Trace the vectors in FIGURE EX3.1 onto your paper. Then find (b) A - B
• Trace the vectors in FIGURE EX3.1 onto your paper. Then find (a) A + B
• (II) V (→ above V) is a vector 21.8 units in magnitude and points at an angle of 23.4° above the negative 𝓍 a...