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

A x = A cos θ = ( 10.3 blocks ) ( cos 29.1º ) = 9.0 blocks size 12{}
A y = A sin θ = ( 10.3 blocks ) ( sin 29.1º ) = 5.0 blocks . size 12{""}

Calculating a resultant vector

If the perpendicular components A x size 12{A rSub { size 8{x} } } {} and A y 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 A x size 12{A rSub { size 8{x} } } {} and A y size 12{A rSub { size 8{y} } } {} , we use the following relationships:

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} } } } } {}
θ = tan 1 ( A y / A x ) . size 12{θ="tan" rSup { size 8{ - 1} } \( A rSub { size 8{y} } /A rSub { size 8{x} } \) } {}
Vector A is shown with its horizontal and vertical components A sub x and A sub y respectively. The magnitude of vector A is equal to the square root of A sub x squared plus A sub y squared. The angle theta of the vector A with the x axis is equal to inverse tangent of A sub y over A sub x
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 A x size 12{A rSub { size 8{x} } } {} and A y size 12{A rSub { size 8{y} } } {} are 9 and 5 blocks, respectively, then A = 9 2 +5 2 =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.

Determining vectors and vector components with analytical methods

Equations A x = A cos θ size 12{A rSub { size 8{x} } =A"cos"θ} {} and A y = A sin θ 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 A x size 12{A rSub { size 8{x} } } {} and A y size 12{A rSub { size 8{y} } } {} . Equations 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} } } } } {} and θ = tan –1 ( A y / A x ) are used to find a vector from its perpendicular components—that is, to go from A x and A y to A and 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 [link] , in which the vectors A size 12{A} {} and B size 12{B} {} are added to produce the resultant R size 12{R} {} .

Two vectors A and B are shown. The tail of vector B is at the head of vector A and the tail of the vector A is at origin. Both the vectors are in the first quadrant. The resultant R of these two vectors extending from the tail of vector A to the head of vector B is also shown.
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, R x and R y size 12{R rSub { size 8{y} } } {} . If we know R x and R y size 12{R rSub { size 8{y} } } {} , we can find R and θ using the equations A = A x 2 + A y 2 and θ = tan –1 ( A y / A x ) size 12{θ="tan" rSup { size 8{–1} } \( A rSub { size 8{y} } /A rSub { size 8{x} } \) } {} . 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 A x = A cos θ size 12{A rSub { size 8{x} } =A"cos"θ} {} and A y = A sin θ size 12{A rSub { size 8{y} } =A"sin"θ} {} to find the components. In [link] , these components are A x size 12{A rSub { size 8{x} } } {} , A y size 12{A rSub { size 8{y} } } {} , B x size 12{B rSub { size 8{x} } } {} , and B y 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 θ A size 12{θ rSub { size 8{A} } } {} and θ B size 12{θ rSub { size 8{B} } } {} , respectively.

Two vectors A and B are shown. The tail of the vector B is at the head of vector A and the tail of the vector A is at origin. Both the vectors are in the first quadrant. The resultant R of these two vectors extending from the tail of vector A to the head of vector B is also shown. The horizontal and vertical components of the vectors A and B are shown with the help of dotted lines. The vectors labeled as A sub x and A sub y are the components of vector A, and B sub x and B sub y as the components of vector B..
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 A x size 12{A rSub { size 8{x} } } {} , A y size 12{A rSub { size 8{y} } } {} , B x size 12{B rSub { size 8{x} } } {} and B y 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] ,

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

and

R y = A y + B y . size 12{R rSub { size 8{y} } =A rSub { size 8{y} } +B rSub { size 8{y} } } {}
Two vectors A and B are shown. The tail of vector B is at the head of vector A and the tail of the vector A is at origin. Both the vectors are in the first quadrant. The resultant R of these two vectors extending from the tail of vector A to the head of vector B is also shown. The vectors A and B are resolved into the horizontal and vertical components shown as dotted lines parallel to x axis and y axis respectively. The horizontal components of vector A and vector B are labeled as A sub x and B sub x and the horizontal component of the resultant R is labeled at R sub x and is equal to A sub x plus B sub x. The vertical components of vector A and vector B are labeled as A sub y and B sub y and the vertical components of the resultant R is labeled as R sub y is equal to A sub y plus B sub y.
The magnitude of the vectors A x size 12{A rSub { size 8{x} } } {} and B x size 12{B rSub { size 8{x} } } {} add to give the magnitude R x size 12{R rSub { size 8{x} } } {} of the resultant vector in the horizontal direction. Similarly, the magnitudes of the vectors A y size 12{A rSub { size 8{y} } } {} and B y size 12{B rSub { size 8{y} } } {} add to give the magnitude R y size 12{R rSub { size 8{y} } } {} of the resultant vector in the vertical direction.
Practice Key Terms 1

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Source:  OpenStax, Sample chapters: openstax college physics for ap® courses. OpenStax CNX. Oct 23, 2015 Download for free at http://legacy.cnx.org/content/col11896/1.9
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