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# 3.3Vector Addition and Subtraction: Analytical Methods

College Physics3.3 Vector Addition and Subtraction: Analytical Methods

Analytical methods of vector addition and subtraction employ geometry and simple trigonometry rather than the ruler and protractor of graphical methods. Part of the graphical technique is retained, because vectors are still represented by arrows for easy visualization. However, analytical methods are more concise, accurate, and precise than graphical methods, which are limited by the accuracy with which a drawing can be made. Analytical methods are limited only by the accuracy and precision with which physical quantities are known.

### Resolving a Vector into Perpendicular Components

Analytical techniques and right triangles go hand-in-hand in physics because (among other things) motions along perpendicular directions are independent. We very often need to separate a vector into perpendicular components. For example, given a vector like $AA size 12{A} {}$ in Figure 3.26, we may wish to find which two perpendicular vectors, $AxAx size 12{A rSub { size 8{x} } } {}$ and $AyAy size 12{A rSub { size 8{y} } } {}$, add to produce it.

Figure 3.26 The vector $AA size 12{A} {}$, with its tail at the origin of an x, y-coordinate system, is shown together with its x- and y-components, $AxAx size 12{A rSub { size 8{x} } } {}$ and $AyAy size 12{A rSub { size 8{y} } } {}$. These vectors form a right triangle. The analytical relationships among these vectors are summarized below.

$AxAx size 12{A rSub { size 8{x} } } {}$ and $AyAy size 12{A rSub { size 8{y} } } {}$ are defined to be the components of $AA size 12{A} {}$ along the x- and y-axes. The three vectors $AA size 12{A} {}$, $AxAx size 12{A rSub { size 8{x} } } {}$, and $AyAy size 12{A rSub { size 8{y} } } {}$ form a right triangle:

3.3

Note that this relationship between vector components and the resultant vector holds only for vector quantities (which include both magnitude and direction). The relationship does not apply for the magnitudes alone. For example, if $Ax=3 mAx=3 m size 12{A rSub { size 8{x} } } {}$ east, $Ay=4 mAy=4 m size 12{A rSub { size 8{y} } } {}$ north, and $A=5 mA=5 m size 12{A} {}$ north-east, then it is true that the vectors . However, it is not true that the sum of the magnitudes of the vectors is also equal. That is,

3.4

Thus,

$A x + A y ≠ A A x + A y ≠ A size 12{A rSub { size 8{x} } +A rSub { size 8{y} } <> A} {}$
3.5

If the vector $AA size 12{A} {}$ is known, then its magnitude $AA size 12{A} {}$ (its length) and its angle $θ θ size 12{θ} {}$ (its direction) are known. To find $AxAx size 12{A rSub { size 8{x} } } {}$ and $AyAy size 12{A rSub { size 8{y} } } {}$, its x- and y-components, we use the following relationships for a right triangle.

$A x = A cos θ A x = A cos θ size 12{A rSub { size 8{x} } =A"cos"θ} {}$
3.6

and

$Ay=Asinθ.Ay=Asinθ. size 12{A rSub { size 8{y} } =A"sin"θ"."} {}$
3.7
Figure 3.27 The magnitudes of the vector components $AxAx size 12{A rSub { size 8{x} } } {}$ and $AyAy size 12{A rSub { size 8{y} } } {}$ can be related to the resultant vector $AA size 12{A} {}$ and the angle $θ θ size 12{θ} {}$ with trigonometric identities. Here we see that $Ax=AcosθAx=Acosθ size 12{A rSub { size 8{x} } =A"cos"θ} {}$ and $Ay=AsinθAy=Asinθ size 12{A rSub { size 8{y} } =A"sin"θ} {}$.

Suppose, for example, that $AA size 12{A} {}$ is the vector representing the total displacement of the person walking in a city considered in Kinematics in Two Dimensions: An Introduction and Vector Addition and Subtraction: Graphical Methods.

Figure 3.28 We can use the relationships $Ax=AcosθAx=Acosθ size 12{A rSub { size 8{x} } =A"cos"θ} {}$ and $Ay=AsinθAy=Asinθ size 12{A rSub { size 8{y} } =A"sin"θ} {}$ to determine the magnitude of the horizontal and vertical component vectors in this example.

Then $A=10.3A=10.3 size 12{A} {}$ blocks and $θ = 29.1º θ = 29.1º size 12{"29.1º"}$ , so that

$Ax=Acosθ=(10.3 blocks)(cos29.1º)=9.0 blocksAx=Acosθ=(10.3 blocks)(cos29.1º)=9.0 blocks size 12{}$
3.8
$Ay=Asinθ=(10.3 blocks)(sin29.1º)=5.0 blocks.Ay=Asinθ=(10.3 blocks)(sin29.1º)=5.0 blocks. size 12{""}$
3.9

### Calculating a Resultant Vector

If the perpendicular components $AxAx size 12{A rSub { size 8{x} } } {}$ and $AyAy size 12{A rSub { size 8{y} } } {}$ of a vector $AA size 12{A} {}$ are known, then $AA size 12{A} {}$ can also be found analytically. To find the magnitude $AA size 12{A} {}$ and direction $θ θ size 12{θ} {}$ of a vector from its perpendicular components $AxAx size 12{A rSub { size 8{x} } } {}$ and $AyAy size 12{A rSub { size 8{y} } } {}$, we use the following relationships:

$A=Ax2+Ay2A=Ax2+Ay2 size 12{A= sqrt {A rSub { size 8{x} rSup { size 8{2} } } +A rSub { size 8{y} rSup { size 8{2} } } } } {}$
3.10
$θ = tan − 1 ( A y / A x ) . θ = tan − 1 ( A y / A x ) . size 12{θ="tan" rSup { size 8{ - 1} } $$A rSub { size 8{y} } /A rSub { size 8{x} }$$ } {}$
3.11
Figure 3.29 The magnitude and direction of the resultant vector can be determined once the horizontal and vertical components $A x A x size 12{A rSub { size 8{x} } } {}$ and $A y A y size 12{A rSub { size 8{y} } } {}$ have been determined.

Note that the equation $A = A x 2 + A y 2 A = A x 2 + A y 2 size 12{A= sqrt {A rSub { size 8{x} rSup { size 8{2} } } +A rSub { size 8{y} rSup { size 8{2} } } } } {}$ is just the Pythagorean theorem relating the legs of a right triangle to the length of the hypotenuse. For example, if $AxAx size 12{A rSub { size 8{x} } } {}$ and $AyAy size 12{A rSub { size 8{y} } } {}$ are 9 and 5 blocks, respectively, then $A=92+52=10.3A=92+52=10.3 size 12{A= sqrt {9 rSup { size 8{2} } "+5" rSup { size 8{2} } } "=10" "." 3} {}$ blocks, again consistent with the example of the person walking in a city. Finally, the direction is $θ = tan –1 ( 5/9 ) =29.1º θ = tan –1 ( 5/9 ) =29.1º size 12{θ="tan" rSup { size 8{–1} } $$"5/9"$$ "=29" "." 1 rSup { size 8{o} } } {}$ , as before.

### Determining Vectors and Vector Components with Analytical Methods

Equations $Ax=AcosθAx=Acosθ size 12{A rSub { size 8{x} } =A"cos"θ} {}$ and $Ay=AsinθAy=Asinθ size 12{A rSub { size 8{y} } =A"sin"θ} {}$ are used to find the perpendicular components of a vector—that is, to go from $AA size 12{A} {}$ and $θ θ size 12{θ} {}$ to $AxAx size 12{A rSub { size 8{x} } } {}$ and $AyAy size 12{A rSub { size 8{y} } } {}$. Equations $A=Ax2+Ay2A=Ax2+Ay2 size 12{A= sqrt {A rSub { size 8{x} rSup { size 8{2} } } +A rSub { size 8{y} rSup { size 8{2} } } } } {}$ and $θ=tan–1(Ay/Ax)θ=tan–1(Ay/Ax)$ are used to find a vector from its perpendicular components—that is, to go from $AxAx$ and $AyAy$ to $AA$ and $θ θ$. Both processes are crucial to analytical methods of vector addition and subtraction.

### Adding Vectors Using Analytical Methods

To see how to add vectors using perpendicular components, consider Figure 3.30, in which the vectors $AA size 12{A} {}$ and $BB size 12{B} {}$ are added to produce the resultant $RR size 12{R} {}$.

Figure 3.30 Vectors $AA size 12{A} {}$ and $BB size 12{B} {}$ are two legs of a walk, and $RR size 12{R} {}$ is the resultant or total displacement. You can use analytical methods to determine the magnitude and direction of $RR size 12{R} {}$.

If $AA$ and $BB$ represent two legs of a walk (two displacements), then $RR$ is the total displacement. The person taking the walk ends up at the tip of $R.R.$ There are many ways to arrive at the same point. In particular, the person could have walked first in the x-direction and then in the y-direction. Those paths are the x- and y-components of the resultant, $RxRx$ and $RyRy size 12{R rSub { size 8{y} } } {}$. If we know $RxRx$ and $RyRy size 12{R rSub { size 8{y} } } {}$, we can find $RR$ and $θ θ$ using the equations $A = A x 2 + Ay 2 A = A x 2 + Ay 2$ and $θ =tan –1 (Ay /Ax )θ =tan –1 (Ay /Ax ) size 12{θ="tan" rSup { size 8{–1} } $$A rSub { size 8{y} } /A rSub { size 8{x} }$$ } {}$. When you use the analytical method of vector addition, you can determine the components or the magnitude and direction of a vector.

Step 1. Identify the x- and y-axes that will be used in the problem. Then, find the components of each vector to be added along the chosen perpendicular axes. Use the equations $Ax=AcosθAx=Acosθ size 12{A rSub { size 8{x} } =A"cos"θ} {}$ and $Ay=AsinθAy=Asinθ size 12{A rSub { size 8{y} } =A"sin"θ} {}$ to find the components. In Figure 3.31, these components are $AxAx size 12{A rSub { size 8{x} } } {}$, $AyAy size 12{A rSub { size 8{y} } } {}$, $BxBx size 12{B rSub { size 8{x} } } {}$, and $ByBy size 12{B rSub { size 8{y} } } {}$. The angles that vectors $AA size 12{A} {}$ and $BB size 12{B} {}$ make with the x-axis are $θAθA size 12{θ rSub { size 8{A} } } {}$ and $θBθB size 12{θ rSub { size 8{B} } } {}$, respectively.

Figure 3.31 To add vectors $AA size 12{A} {}$ and $BB size 12{B} {}$, first determine the horizontal and vertical components of each vector. These are the dotted vectors $AxAx size 12{A rSub { size 8{x} } } {}$, $AyAy size 12{A rSub { size 8{y} } } {}$, $BxBx size 12{B rSub { size 8{x} } } {}$ and $ByBy size 12{B rSub { size 8{y} } } {}$ shown in the image.

Step 2. Find the components of the resultant along each axis by adding the components of the individual vectors along that axis. That is, as shown in Figure 3.32,

$R x = A x + B x R x = A x + B x size 12{R rSub { size 8{x} } =A rSub { size 8{x} } +B rSub { size 8{x} } } {}$
3.12

and

$R y = A y + B y . R y = A y + B y . size 12{R rSub { size 8{y} } =A rSub { size 8{y} } +B rSub { size 8{y} } } {}$
3.13
Figure 3.32 The magnitude of the vectors $AxAx size 12{A rSub { size 8{x} } } {}$ and $BxBx size 12{B rSub { size 8{x} } } {}$ add to give the magnitude $RxRx size 12{R rSub { size 8{x} } } {}$ of the resultant vector in the horizontal direction. Similarly, the magnitudes of the vectors $AyAy size 12{A rSub { size 8{y} } } {}$ and $ByBy size 12{B rSub { size 8{y} } } {}$ add to give the magnitude $RyRy size 12{R rSub { size 8{y} } } {}$ of the resultant vector in the vertical direction.

Components along the same axis, say the x-axis, are vectors along the same line and, thus, can be added to one another like ordinary numbers. The same is true for components along the y-axis. (For example, a 9-block eastward walk could be taken in two legs, the first 3 blocks east and the second 6 blocks east, for a total of 9, because they are along the same direction.) So resolving vectors into components along common axes makes it easier to add them. Now that the components of $RR size 12{R} {}$ are known, its magnitude and direction can be found.

Step 3. To get the magnitude $RR size 12{R } {}$ of the resultant, use the Pythagorean theorem:

$R=Rx2+Ry2.R=Rx2+Ry2. size 12{R= sqrt {R rSub { size 8{x} } rSup { size 8{2} } +R rSub { size 8{y} } rSup { size 8{2} } } "."} {}$
3.14

Step 4. To get the direction of the resultant:

$θ=tan−1(Ry/Rx).θ=tan−1(Ry/Rx). size 12{θ="tan" rSup { size 8{ - 1} } $$R rSub { size 8{y} } /R rSub { size 8{x} }$$ "."} {}$
3.15

The following example illustrates this technique for adding vectors using perpendicular components.

### Example 3.3

#### Adding Vectors Using Analytical Methods

Add the vector $AA size 12{A} {}$ to the vector $BB size 12{B} {}$ shown in Figure 3.33, using perpendicular components along the x- and y-axes. The x- and y-axes are along the east–west and north–south directions, respectively. Vector $AA size 12{A} {}$ represents the first leg of a walk in which a person walks $53.0 m53.0 m size 12{"53" "." "0 m"} {}$ in a direction $20.0º20.0º size 12{"20" "." 0º } {}$ north of east. Vector $BB size 12{B} {}$ represents the second leg, a displacement of $34.0 m34.0 m size 12{"34" "." "0 m"} {}$ in a direction $63.0º63.0º size 12{"63" "." 0º } {}$ north of east.

Figure 3.33 Vector $AA size 12{A} {}$ has magnitude $53.0 m53.0 m size 12{"53" "." "0 m"} {}$ and direction $20.0º20.0º size 12{"20" "." 0 { size 12{ circ } } } {}$ north of the x-axis. Vector $BB size 12{B} {}$ has magnitude $34.0 m34.0 m size 12{"34" "." "0 m"} {}$ and direction $63.0º63.0º size 12{"63" "." 0° } {}$ north of the x-axis. You can use analytical methods to determine the magnitude and direction of $RR size 12{R} {}$.

#### Strategy

The components of $AA size 12{A} {}$ and $BB size 12{B} {}$ along the x- and y-axes represent walking due east and due north to get to the same ending point. Once found, they are combined to produce the resultant.

#### Solution

Following the method outlined above, we first find the components of $AA size 12{A} {}$ and $BB size 12{B} {}$ along the x- and y-axes. Note that $A=53.0 mA=53.0 m size 12{"A" "=" "53.0 m"} {}$, $θA=20.0ºθA=20.0º size 12{"θ" "subA" "=" "20.0°" } {}$, $B=34.0 mB=34.0 m size 12{"B" "=" "34.0" "m"} {}$, and $θB=63.0ºθB=63.0º size 12{θ rSub { size 8{B} } } {}$. We find the x-components by using $Ax=AcosθAx=Acosθ size 12{A rSub { size 8{x} } =A"cos"θ} {}$, which gives

A x = A cos θ A = ( 53. 0 m ) ( cos 20.0º ) = ( 53. 0 m ) ( 0 .940 ) = 49. 8 m A x = A cos θ A = ( 53. 0 m ) ( cos 20.0º ) = ( 53. 0 m ) ( 0 .940 ) = 49. 8 m alignl { stack { size 12{A rSub { size 8{x} } =A"cos"θ rSub { size 8{A} } = $$"53" "." 0" m"$$ $$"cos""20" "." 0 { size 12{ circ } }$$ } {} # " "= $$"53" "." 0" m"$$ $$0 "." "940"$$ ="49" "." 8" m" {} } } {}
3.16

and

B x = B cos θ B = ( 34 . 0 m ) ( cos 63.0º ) = ( 34 . 0 m ) ( 0 . 454 ) = 15 . 4 m . B x = B cos θ B = ( 34 . 0 m ) ( cos 63.0º ) = ( 34 . 0 m ) ( 0 . 454 ) = 15 . 4 m . alignl { stack { size 12{B rSub { size 8{x} } =B"cos"θ rSub { size 8{B} } = $$"34" "." 0" m"$$ $$"cos""63" "." 0 { size 12{ circ } }$$ } {} # " "= $$"34" "." 0" m"$$ $$0 "." "454"$$ ="15" "." 4" m" {} } } {}
3.17

Similarly, the y-components are found using $Ay=AsinθAAy=AsinθA size 12{A rSub { size 8{y} } =A"sin"θ rSub { size 8{A} } } {}$:

A y = A sin θ A = ( 53 . 0 m ) ( sin 20.0º ) = ( 53 . 0 m ) ( 0 . 342 ) = 18 . 1 m A y = A sin θ A = ( 53 . 0 m ) ( sin 20.0º ) = ( 53 . 0 m ) ( 0 . 342 ) = 18 . 1 m alignl { stack { size 12{A rSub { size 8{y} } =A"sin"θ rSub { size 8{A} } = $$"53" "." 0" m"$$ $$"sin""20" "." 0 { size 12{ circ } }$$ } {} # " "= $$"53" "." 0" m"$$ $$0 "." "342"$$ ="18" "." 1" m" {} } } {}
3.18

and

B y = B sin θ B = ( 34 . 0 m ) ( sin 63 . 0 º ) = ( 34 . 0 m ) ( 0 . 891 ) = 30 . 3 m . B y = B sin θ B = ( 34 . 0 m ) ( sin 63 . 0 º ) = ( 34 . 0 m ) ( 0 . 891 ) = 30 . 3 m . alignl { stack { size 12{B rSub { size 8{y} } =B"sin"θ rSub { size 8{B} } = $$"34" "." 0" m"$$ $$"sin""63" "." 0 { size 12{ circ } }$$ } {} # " "= $$"34" "." 0" m"$$ $$0 "." "891"$$ ="30" "." 3" m" "." {} } } {}
3.19

The x- and y-components of the resultant are thus

$R x = A x + B x = 49 . 8 m + 15 . 4 m = 65 . 2 m R x = A x + B x = 49 . 8 m + 15 . 4 m = 65 . 2 m size 12{R rSub { size 8{x} } =A rSub { size 8{x} } +B rSub { size 8{x} } ="49" "." 8" m"+"15" "." 4" m"="65" "." 2" m"} {}$
3.20

and

$Ry=Ay+By=18.1 m+30.3 m=48.4 m.Ry=Ay+By=18.1 m+30.3 m=48.4 m. size 12{R rSub { size 8{y} } =A rSub { size 8{y} } +B rSub { size 8{y} } ="18" "." 1" m"+"30" "." 3" m"="48" "." 4" m."} {}$
3.21

Now we can find the magnitude of the resultant by using the Pythagorean theorem:

$R = R x 2 + R y 2 = ( 65 . 2 ) 2 + ( 48 . 4 ) 2 m R = R x 2 + R y 2 = ( 65 . 2 ) 2 + ( 48 . 4 ) 2 m size 12{R= sqrt {R rSub { size 8{x} } rSup { size 8{2} } +R rSub { size 8{y} } rSup { size 8{2} } } = sqrt { $$"65" "." 2$$ rSup { size 8{2} } + $$"48" "." 4$$ rSup { size 8{2} } } " m"} {}$
3.22

so that

$R = 81.2 m. R = 81.2 m. size 12{R ="81.2" "m."} {}$
3.23

Finally, we find the direction of the resultant:

$θ=tan−1(Ry/Rx)=+tan−1(48.4/65.2).θ=tan−1(Ry/Rx)=+tan−1(48.4/65.2). size 12{θ="tan" rSup { size 8{ - 1} } $$R rSub { size 8{y} } /R rSub { size 8{x} }$$ "=+""tan" rSup { size 8{ - 1} } $$"48" "." 4/"65" "." 2$$ "."} {}$
3.24

Thus,

$θ=tan−1(0.742)=36.6º.θ=tan−1(0.742)=36.6º. size 12{θ="tan" rSup { size 8{ - 1} } $$0 "." "742"$$ ="36" "." 6 { size 12{ circ } } "."} {}$
3.25
Figure 3.34 Using analytical methods, we see that the magnitude of $RR size 12{R} {}$ is $81.2 m81.2 m size 12{"81" "." "2 m"} {}$ and its direction is $36.6º36.6º size 12{"36" "." 6°} {}$ north of east.

#### Discussion

This example illustrates the addition of vectors using perpendicular components. Vector subtraction using perpendicular components is very similar—it is just the addition of a negative vector.

Subtraction of vectors is accomplished by the addition of a negative vector. That is, $A−B≡A+(–B)A−B≡A+(–B) size 12{A – B equiv A+ $$- B$$ } {}$. Thus, the method for the subtraction of vectors using perpendicular components is identical to that for addition. The components of $–B–B$ are the negatives of the components of $BB size 12{B} {}$. The x- and y-components of the resultant $A−B = RA−B = R size 12{A- bold "B = R"} {}$ are thus

$R x = A x + (– B x ) R x = A x + (– B x ) size 12{R rSub { size 8{x} } =A rSub { size 8{x} } +-B rSub { size 8{x} } } {}$
3.26

and

$R y = A y + (– B y ) R y = A y + (– B y ) size 12{R rSub { size 8{y} } =A rSub { size 8{y} } +-B rSub { size 8{y} } } {}$
3.27

and the rest of the method outlined above is identical to that for addition. (See Figure 3.35.)

Analyzing vectors using perpendicular components is very useful in many areas of physics, because perpendicular quantities are often independent of one another. The next module, Projectile Motion, is one of many in which using perpendicular components helps make the picture clear and simplifies the physics.

Figure 3.35 The subtraction of the two vectors shown in Figure 3.30. The components of $–B–B size 12{B} {}$ are the negatives of the components of $BB size 12{B} {}$. The method of subtraction is the same as that for addition.

### PhET Explorations

#### Vector Addition

Learn how to add vectors. Drag vectors onto a graph, change their length and angle, and sum them together. The magnitude, angle, and components of each vector can be displayed in several formats.

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