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 ...
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.
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 A size 12{A} {} in [link], we may wish to find which two perpendicular vectors, Ax size 12{A rSub { size 8{x} } } {} and Ay size 12{A rSub { size 8{y} } } {}, add to produce it.
The vector A size 12{A} {}, with its tail at the origin of an x, y-coordinate system, is shown together with its x- and y-components, Ax size 12{A rSub { size 8{x} } } {} and Ay size 12{A rSub { size 8{y} } } {}. These vectors form a right triangle. The analytical relationships among these vectors are summarized below.Ax size 12{A rSub { size 8{x} } } {} and Ay size 12{A rSub { size 8{y} } } {} are defined to be the components of A size 12{A} {} along the x- and y-axes. The three vectors A size 12{A} {}, Ax size 12{A rSub { size 8{x} } } {}, and Ay size 12{A rSub { size 8{y} } } {} form a right triangle:
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 m size 12{A rSub { size 8{x} } } {} east, Ay=4 m size 12{A rSub { size 8{y} } } {} north, and A=5 m size 12{A} {} north-east, then it is true that the vectors Ax + Ay = A size 12{A rSub { size 8{x} } bold " + A" rSub { size 8{y} } bold " = A"} {}. However, it is not true that the sum of the magnitudes of the vectors is also equal. That is,
Thus,
If the vector A size 12{A} {} is known, then its magnitude A size 12{A} {} (its length) and its angle θ size 12{θ} {} (its direction) are known. To find Ax size 12{A rSub { size 8{x} } } {} and Ay size 12{A rSub { size 8{y} } } {}, its x- and y-components, we use the following relationships for a right triangle.
and
The magnitudes of the vector components Ax size 12{A rSub { size 8{x} } } {} and Ay size 12{A rSub { size 8{y} } } {} can be related to the resultant vector A size 12{A} {} and the angle
θ
size 12{θ} {}
with trigonometric identities. Here we see that Ax=Acosθ size 12{A rSub { size 8{x} } =A"cos"θ} {} and Ay=Asinθ size 12{A rSub { size 8{y} } =A"sin"θ} {}.Suppose, for example, that A 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.
We can use the relationships Ax=Acosθ size 12{A rSub { size 8{x} } =A"cos"θ} {} and Ay=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=10.3 size 12{A} {} blocks and θ = 29.1º size 12{"29.1º"} , so that
If the perpendicular components Ax size 12{A rSub { size 8{x} } } {} and Ay size 12{A rSub { size 8{y} } } {} of a vector A size 12{A} {} are known, then A size 12{A} {} can also be found analytically. To find the magnitude A size 12{A} {} and direction θ size 12{θ} {} of a vector from its perpendicular components Ax size 12{A rSub { size 8{x} } } {} and Ay size 12{A rSub { size 8{y} } } {}, we use the following relationships:
The magnitude and direction of the resultant vector can be determined once the horizontal and vertical components
A
x
size 12{A rSub { size 8{x} } } {}
and
A
y
size 12{A rSub { size 8{y} } } {}
have been determined.Note that the equation 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 Ax size 12{A rSub { size 8{x} } } {} and Ay size 12{A rSub { size 8{y} } } {} are 9 and 5 blocks, respectively, then A=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 size 12{θ="tan" rSup { size 8{–1} } ( "5/9" ) "=29" "." 1 rSup { size 8{o} } } {} , as before.
Equations Ax=Acosθ size 12{A rSub { size 8{x} } =A"cos"θ} {} and Ay=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 A size 12{A} {} and θ size 12{θ} {} to Ax size 12{A rSub { size 8{x} } } {} and Ay size 12{A rSub { size 8{y} } } {}. Equations A=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) are used to find a vector from its perpendicular components—that is, to go from Ax and Ay to A and θ . Both processes are crucial to analytical methods of vector addition and subtraction.
To see how to add vectors using perpendicular components, consider [link], in which the vectors A size 12{A} {} and B size 12{B} {} are added to produce the resultant R size 12{R} {}.
Vectors A size 12{A} {} and B size 12{B} {} are two legs of a walk, and R size 12{R} {} is the resultant or total displacement. You can use analytical methods to determine the magnitude and direction of R size 12{R} {}.If A and B represent two legs of a walk (two displacements), then R is the total displacement. The person taking the walk ends up at the tip of 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, Rx and Ry size 12{R rSub { size 8{y} } } {}. If we know Rx and Ry size 12{R rSub { size 8{y} } } {}, we can find R and θ using the equations A = A x 2 + Ay 2 and 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=Acosθ size 12{A rSub { size 8{x} } =A"cos"θ} {} and Ay=Asinθ size 12{A rSub { size 8{y} } =A"sin"θ} {} to find the components. In [link], these components are Ax size 12{A rSub { size 8{x} } } {}, Ay size 12{A rSub { size 8{y} } } {}, Bx size 12{B rSub { size 8{x} } } {}, and By size 12{B rSub { size 8{y} } } {}. The angles that vectors A size 12{A} {} and B size 12{B} {} make with the x-axis are θA size 12{θ rSub { size 8{A} } } {} and θB size 12{θ rSub { size 8{B} } } {}, respectively.
To add vectors A size 12{A} {} and B size 12{B} {}, first determine the horizontal and vertical components of each vector. These are the dotted vectors Ax size 12{A rSub { size 8{x} } } {}, Ay size 12{A rSub { size 8{y} } } {}, Bx size 12{B rSub { size 8{x} } } {} and By 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 [link],
and
The magnitude of the vectors Ax size 12{A rSub { size 8{x} } } {} and Bx size 12{B rSub { size 8{x} } } {} add to give the magnitude Rx size 12{R rSub { size 8{x} } } {} of the resultant vector in the horizontal direction. Similarly, the magnitudes of the vectors Ay size 12{A rSub { size 8{y} } } {} and By size 12{B rSub { size 8{y} } } {} add to give the magnitude Ry 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 R size 12{R} {} are known, its magnitude and direction can be found.
Step 3. To get the magnitude R size 12{R } {} of the resultant, use the Pythagorean theorem:
Step 4. To get the direction of the resultant:
The following example illustrates this technique for adding vectors using perpendicular components.
Add the vector A size 12{A} {} to the vector B size 12{B} {} shown in [link], 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 A size 12{A} {} represents the first leg of a walk in which a person walks 53.0 m size 12{"53" "." "0 m"} {} in a direction 20.0º size 12{"20" "." 0º } {} north of east. Vector B size 12{B} {} represents the second leg, a displacement of 34.0 m size 12{"34" "." "0 m"} {} in a direction 63.0º size 12{"63" "." 0º } {} north of east.
Vector A size 12{A} {} has magnitude 53.0 m size 12{"53" "." "0 m"} {} and direction 20.0º size 12{"20" "." 0 { size 12{ circ } } } {} north of the x-axis. Vector B size 12{B} {} has magnitude 34.0 m size 12{"34" "." "0 m"} {} and direction 63.0º size 12{"63" "." 0° } {} north of the x-axis. You can use analytical methods to determine the magnitude and direction of R size 12{R} {}.Strategy
The components of A size 12{A} {} and B 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 A size 12{A} {} and B size 12{B} {} along the x- and y-axes. Note that A=53.0 m size 12{"A" "=" "53.0 m"} {}, θA=20.0º size 12{"θ" "subA" "=" "20.0°" } {}, B=34.0 m size 12{"B" "=" "34.0" "m"} {}, and θB=63.0º size 12{θ rSub { size 8{B} } } {}. We find the x-components by using Ax=Acosθ size 12{A rSub { size 8{x} } =A"cos"θ} {}, which gives
and
Similarly, the y-components are found using Ay=AsinθA size 12{A rSub { size 8{y} } =A"sin"θ rSub { size 8{A} } } {}:
and
The x- and y-components of the resultant are thus
and
Now we can find the magnitude of the resultant by using the Pythagorean theorem:
so that
Finally, we find the direction of the resultant:
Thus,
Using analytical methods, we see that the magnitude of R size 12{R} {} is 81.2 m size 12{"81" "." "2 m"} {} and its direction is 36.6º 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) 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 are the negatives of the components of B size 12{B} {}. The x- and y-components of the resultant A−B = R size 12{A- bold "B = R"} {} are thus
and
and the rest of the method outlined above is identical to that for addition. (See [link].)
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.
The subtraction of the two vectors shown in [link]. The components of –B size 12{B} {} are the negatives of the components of B size 12{B} {}. The method of subtraction is the same as that for 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.
Vector Addition- The analytical method of vector addition and subtraction involves using the Pythagorean theorem and trigonometric identities to determine the magnitude and direction of a resultant vector.
- The steps to add vectors A size 12{A} {} and B size 12{B} {} using the analytical method are as follows:
Step 1: Determine the coordinate system for the vectors. Then, determine the horizontal and vertical components of each vector using the equations
A x = A cos θ B x = B cos θ alignl { stack { size 12{A rSub { size 8{x} } =A"cos"θ} {} # B rSub { size 8{x} } =B"cos"θ {} } } {}and
A y = A sin θ By = B sin θ . alignl { stack { size 12{A rSub { size 8{y} } =A" sin"θ} {} # B=B suby " sin "θ {} } } {}Step 2: Add the horizontal and vertical components of each vector to determine the components Rx size 12{R rSub { size 8{x} } } {} and Ry size 12{R rSub { size 8{y} } } {} of the resultant vector, R size 12{R} {}:
R x = A x + B x size 12{R rSub { size 8{x} } =A rSub { size 8{x} } +B rSub { size 8{x} } } {}and
Ry=Ay+By. size 12{R rSub { size 8{y} } =A rSub { size 8{y} } +B rSub { size 8{y} } } {}Step 3: Use the Pythagorean theorem to determine the magnitude, R size 12{R} {}, of the resultant vector R size 12{R} {}:
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} } } } {}Step 4: Use a trigonometric identity to determine the direction, θ size 12{θ} {}, of R size 12{R} {}:
θ=tan−1(Ry/Rx). size 12{θ="tan" rSup { size 8{ - 1} } ( R rSub { size 8{y} } /R rSub { size 8{x} } ) } {}
Suppose you add two vectors A size 12{A} {} and B size 12{B} {}. What relative direction between them produces the resultant with the greatest magnitude? What is the maximum magnitude? What relative direction between them produces the resultant with the smallest magnitude? What is the minimum magnitude?
Give an example of a nonzero vector that has a component of zero.
Explain why a vector cannot have a component greater than its own magnitude.
If the vectors A size 12{A} {} and B size 12{B} {} are perpendicular, what is the component of A size 12{A} {} along the direction of B size 12{B} {}? What is the component of B size 12{B} {} along the direction of A size 12{A} {}?
Find the following for path C in [link]: (a) the total distance traveled and (b) the magnitude and direction of the displacement from start to finish. In this part of the problem, explicitly show how you follow the steps of the analytical method of vector addition.
The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side.
(a) 1.56 km
(b) 120 m east
Find the following for path D in [link]: (a) the total distance traveled and (b) the magnitude and direction of the displacement from start to finish. In this part of the problem, explicitly show how you follow the steps of the analytical method of vector addition.
Find the north and east components of the displacement from San Francisco to Sacramento shown in [link].
North-component 87.0 km, east-component 87.0 km
Solve the following problem using analytical techniques: Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements A size 12{A} {} and B size 12{B} {}, as in [link], then this problem asks you to find their sum R=A+B size 12{R=A+B} {}.)
The two displacements A size 12{A} {} and B size 12{B} {} add to give a total displacement R size 12{R} {} having magnitude R size 12{R} {} and direction θ size 12{θ} {}.Note that you can also solve this graphically. Discuss why the analytical technique for solving this problem is potentially more accurate than the graphical technique.
Repeat [link] using analytical techniques, but reverse the order of the two legs of the walk and show that you get the same final result. (This problem shows that adding them in reverse order gives the same result—that is, B + A = A + B .) Discuss how taking another path to reach the same point might help to overcome an obstacle blocking you other path.
30.8 m, 35.8 west of north
You drive 7.50 km size 12{7 "." "50 km"} {} in a straight line in a direction 15º size 12{"15º"} {} east of north. (a) Find the distances you would have to drive straight east and then straight north to arrive at the same point. (This determination is equivalent to find the components of the displacement along the east and north directions.) (b) Show that you still arrive at the same point if the east and north legs are reversed in order.
Do [link] again using analytical techniques and change the second leg of the walk to 25.0 m straight south. (This is equivalent to subtracting B size 12{B} {} from A size 12{A} {} —that is, finding R′=A – B) (b) Repeat again, but now you first walk 25.0 m size 12{"25" "." "0 m"} {} north and then 18.0 m size 12{"18" "." "0 m"} {} east. (This is equivalent to subtract A size 12{A} {} from B size 12{B} {} —that is, to find A=B+C size 12{A=B+C} {}. Is that consistent with your result?)
(a) 30.8 m size 12{"30" "." "8 m"} {}, 54.2º size 12{"54" "." 2°} {} south of west
(b) 30.8 m size 12{"30" "." "8 m"} {}, 54.2º size 12{"54" "." 2°} {} north of east
A new landowner has a triangular piece of flat land she wishes to fence. Starting at the west corner, she measures the first side to be 80.0 m long and the next to be 105 m. These sides are represented as displacement vectors A size 12{A} {} from B size 12{B} {} in [link]. She then correctly calculates the length and orientation of the third side C size 12{C} {}. What is her result?
You fly 32.0 km size 12{"32" "." "0 km"} {} in a straight line in still air in the direction 35.0º size 12{"35"°} {} south of west. (a) Find the distances you would have to fly straight south and then straight west to arrive at the same point. (This determination is equivalent to finding the components of the displacement along the south and west directions.) (b) Find the distances you would have to fly first in a direction 45.0º size 12{"45.0º"} {} south of west and then in a direction 45.0º size 12{"45.0º"} {} west of north. These are the components of the displacement along a different set of axes—one rotated 45º size 12{"45"°} {}.
18.4 km south, then 26.2 km west(b) 31.5 km at 45.0º size 12{"45.0º"} {} south of west, then 5.56 km at 45.0º size 12{"45.0º"} {} west of north
A farmer wants to fence off his four-sided plot of flat land. He measures the first three sides, shown as A , size 12{A,} {} B , size 12{B,} {} and C size 12{C} {} in [link], and then correctly calculates the length and orientation of the fourth side D size 12{D} {}. What is his result?
In an attempt to escape his island, Gilligan builds a raft and sets to sea. The wind shifts a great deal during the day, and he is blown along the following straight lines: 2.50 km size 12{2 "." "50 km"} {} 45.0º size 12{"45.0º"} {} north of west; then 4.70 km size 12{4 "." "70 km"} {} 60.0º size 12{"60"°} {} south of east; then 1.30 km size 12{"1.30" " km"} {} 25.0º size 12{"25"°} {} south of west; then 5.10 km size 12{5 "." "10 km"} {} straight east; then 1.70 km size 12{"1.70" " km"} {} 5.00º size 12{5 rSup { size 8{ circ } } } {} east of north; then 7.20 km size 12{7 "." "20 km"} {} 55.0º size 12{"55"°} {} south of west; and finally 2.80 km size 12{2 "." "80 km"} {} 10.0º size 12{"10"°} {} north of east. What is his final position relative to the island?
7.34 km size 12{2 "." "97 km"} {}, 63.5º size 12{"22" "." 2°} {} south of east
Suppose a pilot flies 40.0 km size 12{"40" "." "0 km"} {} in a direction 60º size 12{"60"°} {} north of east and then flies 30.0 km size 12{"30" "." "0 km"} {} in a direction 15º size 12{"15"°} {} north of east as shown in [link]. Find her total distance R size 12{R} {} from the starting point and the direction θ size 12{θ} {} of the straight-line path to the final position. Discuss qualitatively how this flight would be altered by a wind from the north and how the effect of the wind would depend on both wind speed and the speed of the plane relative to the air mass.
