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# Resolving Vectors into Axial Components

## Identify perpendicular components of vectors

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Resolving Vectors into Axial Components

### Resolving Vectors into Axial Components

We know that when two vectors are in the same dimension, they can be added arithmetically.  Suppose we have two vectors that are on a north-south, east-west grid, as shown below.  One of the methods we can use to add these vectors is to resolve each one into a pair of vectors that lay on the north-south and east-west axes.

Credit: Richard Parsons
Source: CK-12 Foundation

The two vectors we are to add is a force of 65 N at 30° north of east and a force of 35 N at 60° north of west.

Credit: Richard Parsons
Source: CK-12 Foundation

We can resolve each of the vectors into two components on the axes lines. Each vector is resolved into a component on the north-south axis and a component on the east-west axis.

Using trigonometry, we can resolve (break down) each of these vectors into a pair of vectors that lay on the axial lines (shown in red above).

The east-west component of the first vector is (65 N)(cos 30° ) = (65 N)(0.866) = 56.3 N east

The north-south component of the first vector is (65 N)(sin 30°) = (65 N)(0.500) = 32.5 N north

The east-west component of the 2nd vector is (35 N)(cos 60°) = (35 N)(0.500) = 17.5 N west

The north-south component of the 2nd vector is (35 N)(sin 60°) = (35 N)(0.866) = 30.3 N north

#### Summary

• Vectors can be resolved into component vectors that lie on the axes lines.

#### Practice

Questions

1. What does SohCahToa mean?
2. Why is SohCahToa relevant to resolving a vector into components?
3. Why is the sum of the components larger than the resultant vector?

#### Review

Questions

1. A force of 150. N is exerted 22° north of east.  Find the northward and eastward components of this force.
2. An automobile travels a displacement of 75 km 45° north of east.  How far east does it travel and how far north does it travel?

1. [1]^ Credit: Richard Parsons; Source: CK-12 Foundation; License: CC BY-NC 3.0
2. [2]^ Credit: Richard Parsons; Source: CK-12 Foundation; License: CC BY-NC 3.0

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