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$\textbf{A}$in Figure 1, we may wish to find which two perpendicular vectors,$\boldsymbol{\textbf{A}_x}$and$\boldsymbol{\textbf{A}_y}$, add to produce it.

$\boldsymbol{\textbf{A}_x}$and$\boldsymbol{\textbf{A}_y}$are defined to be the components of$\textbf{A}$the x- and y-axes. The three vectors$\textbf{A},\:\boldsymbol{\textbf{A}_x},$and$\boldsymbol{\textbf{A}_y}$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$\boldsymbol{\textbf{A}_x=3\textbf{ m}}$east,$\boldsymbol{\textbf{A}_y=4\textbf{ m}}$north, and$\boldsymbol{\textbf{A}=5\textbf{ m}}$north-east, then it is true that the vectors$\boldsymbol{\textbf{A}_x+\textbf{A}_y=\textbf{A}}.$However, it is not true that the sum of the magnitudes of the vectors is also equal. That is,

Thus,

If the vector$\textbf{A}$is known, then its magnitude$\textbf{A}$and its angle$\boldsymbol{\theta}$(its direction) are known. To find$\boldsymbol{\textbf{A}_x}$and$\boldsymbol{A_y},$ its x- and y-components, we use the following relationships for a right triangle.

and

Suppose, for example, that$\textbf{A}$is the vector representing the total displacement of the person walking in a city considered in Chapter 3.1 Kinematics in Two Dimensions: An Introduction and Chapter 3.2 Vector Addition and Subtraction: Graphical Methods.

Then$\boldsymbol{\textbf{A}=10.3\textbf{ blocks}}$and$\boldsymbol{\theta=29.1^o},$ so that

Calculating a Resultant Vector

If the perpendicular components$\boldsymbol{\textbf{A}_x}$and$\boldsymbol{\textbf{A}_y}$of a vector$\textbf{A}$are known, then$\textbf{A}$can also be found analytically. To find the magnitude$\textbf{A}$and direction$\boldsymbol{\theta}$of a vector from its perpendicular components$\boldsymbol{\textbf{A}_x}$and$\boldsymbol{\textbf{A}_y},$we use the following relationships:

Note that the equation$\boldsymbol{\textbf{A}=\sqrt{A_x\:^2\:+\:A_y\:^2}}$is just the Pythagorean theorem relating the legs of a right triangle to the length of the hypotenuse. For example, if$\boldsymbol{\textbf{A}_x}$and$\boldsymbol{\textbf{A}_y}$are 9 and 5 blocks, respectively, then$\boldsymbol{\textbf{A}=\sqrt{9^2\:+\:5^2}=10.3}$blocks, again consistent with the example of the person walking in a city. Finally, the direction is$\boldsymbol{\theta=\textbf{tan}^{-1}(5/9)=29.1^o},$as before.

Adding Vectors Using Analytical Methods

To see how to add vectors using perpendicular components, consider Figure 5, in which the vectors$\textbf{A}$and$\textbf{B}$are added to produce the resultant$\textbf{R}.$

If$\textbf{A}$and$\textbf{B}$represent two legs of a walk (two displacements), then$\textbf{R}$is the total displacement. The person taking the walk ends up at the tip of$\textbf{R}.$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,$\boldsymbol{\textbf{R}_x}$and$\boldsymbol{\textbf{R}_y}.$If we know$\boldsymbol{\textbf{R}_x}$and$\boldsymbol{\textbf{R}_y},$we can find$\textbf{R}$and$\boldsymbol{\theta}$using the equations$\boldsymbol{\textbf{A}=\sqrt{{\textbf{A}_x}^2\:+\:{\textbf{A}_y}^2}}$and$\boldsymbol{\theta=\textbf{tan}^{-1}(\textbf{A}_y\:/\:\textbf{A}_x)}.$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$\boldsymbol{\textbf{A}_x=\textbf{A cos}\:\theta}$and$\boldsymbol{\textbf{A}_y=\textbf{A sin}\:\theta}$to find the components. In Figure 6, these components are$\boldsymbol{\textbf{A}_x},\boldsymbol{\textbf{A}_x},\boldsymbol{\textbf{B}_x},$and$\boldsymbol{\textbf{B}_y}.$The angles that vectors$\textbf{A}$and$\textbf{B}$make with the x-axis are$\boldsymbol{\theta_{A}}$and$\boldsymbol{\theta_{B}},$respectively.

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 7,

and

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$\textbf{R}$are known, its magnitude and direction can be found.

Step 3. To get the magnitude$\textbf{R}$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.

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, Chapter 3.4 Projectile Motion, is one of many in which using perpendicular components helps make the picture clear and simplifies the physics.