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,  u,  w,   v → ,   u → ,  w → .
• Given initial point  P  and terminal point  Q,  a vector can be represented as   PQ →  .   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 )  and terminal point  ( a,b ),  a vector may be represented as ⟨ a,b ⟩.

This last symbol ⟨ a,b ⟩ has special significance. It is called the standard position. The position vector has an initial point ( 0,0 )  and a terminal point ⟨ 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   CD →   is  C( x 1 , y 1 )  and the terminal point is  D( x 2 , y 2 ),  then the position vector is found by calculating

In [link], we see the original vector   CD →   and the position vector   AB → .

Find the Position Vector

Consider the vector whose initial point is  P( 2,3 )  and terminal point is  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

v=⟨ 6−2,4−3 ⟩    =⟨ 4,1 ⟩

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

We see that the position vector is ⟨ 4,1 ⟩.

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

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

The position vector is found using the following calculation:

v=⟨ 4−(−3),5−2 ⟩   =⟨ 7,3 ⟩

Thus, the position vector begins at  ( 0,0 )  and terminates at  ( 7,3 ).  See [link].

try it feature

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

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.

a general note label

Magnitude and Direction of a Vector

Given a position vector  v =⟨ a,b ⟩, the magnitude is found by | 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 ),  as illustrated in [link].

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 )  and terminal point  Q( −2,−5 ). Draw the vector.

First, find the position vector.

u=⟨ −2,−(−8),−5−1 ⟩   =⟨ 6,−6 ⟩

We use the Pythagorean Theorem to find the magnitude.

|u|= (6) 2 + (−6) 2       = 72       =6 2

The direction is given as

tan θ= −6 6 =−1⇒θ= tan −1 (−1)             =−45°

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

Showing That Two Are Equal

Show that vector v with initial point at  ( 5,−3 )  and terminal point at  ( −1,2 )  is equal to vector u with initial point at  ( −1,−3 )  and terminal point at  ( −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 [link], draw the vector  v  starting at initial  ( 5,−3 )  and terminal point  ( −1,2 ).  Draw the vector  u  with initial point  ( −1,−3 )  and terminal point  ( −7,2 ).  Find the standard position for each.

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

v=⟨−1−5,2−(−3)⟩   =⟨−6,5⟩ u=⟨−7−(−1),2−(−3)⟩   =⟨−6,5⟩

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.

|v|= (−1−5) 2 + (2−(−3)) 2       = (−6) 2 + (5) 2       = 36+25       = 61 |u|= (−7−(−1)) 2 + (2−(−3)) 2       = (−6) 2 + (5) 2       = 36+25       = 61

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

tan θ=− 5 6 ⇒θ= tan −1 ( − 5 6 )             =−39.8°

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

Now that we understand the properties of vectors, we can perform operations involving them. While it is convenient to think of the vector u =⟨ x,y ⟩ as an arrow or directed line segment from the origin to the point  (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 [link].

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 [link] for a visual that compares vector addition and vector subtraction using parallelograms.

Adding and Subtracting

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

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

u+v=⟨ 3,−2 ⟩+⟨ −1,4 ⟩             =⟨ 3+(−1),−2+4 ⟩             =⟨ 2,2 ⟩

See [link](a).

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

u+(−v)=⟨ 3,−2 ⟩+⟨ 1,−4 ⟩                       =⟨ 3+1,−2+(−4) ⟩                       =⟨ 4,−6 ⟩

See [link](b).

(a) Sum of two vectors (b) Difference of two vectors

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.

Performing Scalar Multiplication

Given vector  v =⟨ 3,1 ⟩,  find 3v, 1 2 v,  and −v.

See [link] for a geometric interpretation. If  v =⟨ 3,1 ⟩, then

  3v=⟨ 3⋅3,3⋅1 ⟩         =⟨ 9,3 ⟩   1 2 v=⟨ 1 2 ⋅3, 1 2 ⋅1 ⟩         =⟨ 3 2 , 1 2 ⟩ −v=⟨ −3,−1 ⟩

Analysis

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

In some applications involving vectors, it is helpful for us to be able to break a vector down into its components. are comprised of two components: the horizontal component is the  x  direction, and the vertical component is the  y  direction. For example, we can see in the graph in [link] that the position vector ⟨ 2,3 ⟩ comes from adding the vectors v₁ and v₂. We have v₁ with initial point  ( 0,0 )  and terminal point  ( 2,0 ). 

We also have v₂ with initial point  ( 0,0 )  and terminal point  ( 0, 3 ). 

Therefore, the position vector is

Using the Pythagorean Theorem, the magnitude of v₁ is 2, and the magnitude of v₂ is 3. To find the magnitude of v, use the formula with the position vector.

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

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

Finding the Components of the Vector

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

First find the standard position.

v=⟨ 7−3,4−2 ⟩    =⟨ 4,2 ⟩

See the illustration in [link].

The horizontal component is v 1 =⟨ 4,0 ⟩  and the vertical component is   v 2 =⟨ 0,2 ⟩.

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 =⟨ 1,0 ⟩ and is directed along the positive horizontal axis. The vertical unit vector is written as j =⟨ 0,1 ⟩ and is directed along the positive vertical axis. See [link].

Finding the Unit Vector in the Direction of v

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

First, we will find the magnitude.

|v|= (−5) 2 + (12) 2       = 25+144       = 169       =13

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

v | v | =− 5 13 i+ 12 13 j

or, in component form

v | v | =⟨ − 5 13 , 12 13 ⟩

See [link].

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

( − 5 13 ) 2 + ( 12 13 ) 2 = 25 169 + 144 169                             = 169 169 =1

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

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.

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in the Rectangular Plane

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

v=( x 2 − x 1 )i+( y 1 − y 2 )j

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

The magnitude of v = ai + bj is given as  | v |= a 2 + b 2 .  See [link].

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

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.  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 |  replacing  r.

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.

Finding the Dot Product of Two and the Angle between Them

Find the dot product of v₁ = 5i + 2j and v₂ = 3i + 7j. Then, find the angle between the two vectors.

Finding the dot product, we multiply corresponding components.

v 1 ⋅ v 2 =⟨ 5,2 ⟩⋅⟨ 3,7 ⟩                 =5⋅3+2⋅7                 =15+14                 =29

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

                         v |v| ⋅ u |u| =⟨ 5 29 + 2 29 ⟩⋅⟨ 3 58 + 7 58 ⟩                                          = 5 29 ⋅ 3 58 + 2 29 ⋅ 7 58                                          = 15 1682 + 14 1682 = 29 1682                                          =0.707107 cos −1 (0.707107)=45°

See [link].

Finding the Angle between Two

Find the angle between  u=⟨ −3,4 ⟩  and  v=⟨ 5,12 ⟩.

Using the formula, we have

                   θ= cos −1 ( u |u| ⋅ v |v| ) ( u |u| ⋅ v |v| )= −3i+4j 5 ⋅ 5i+12j 13                       =( − 3 5 ⋅ 5 13 )+( 4 5 ⋅ 12 13 )                       =− 15 65 + 48 65                       = 33 65                    θ= cos −1 ( 33 65 )                       = 59.5 ∘

See [link].

Finding Ground Speed and Bearing Using

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 [link].

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

Notice in [link], that angle  BCO  must be equal to angle  AOC  by the rule of alternating interior angles, so angle  BCO  is 140°. We can find  x  by the Law of Cosines:

x 2 = (16.2) 2 + (200) 2 −2(16.2)(200)cos(140°) x 2 =45,226.41    x= 45,226.41    x=212.7

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

                         sin α 16.2 = sin(140°) 212.7                           sin α= 16.2sin(140°) 212.7                                      =0.04896 sin −1 (0.04896)=2.8°

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