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## Algebra (all content)

### Course: Algebra (all content)>Unit 19

Lesson 7: Magnitude & direction form of vectors

# Vector forms review

Review all the different ways in which we can represent vectors: components, magnitude & direction, and unit vectors.

## What are the different vector forms?

Component form$\left(a,b\right)$
Unit vectors$a\stackrel{^}{i}+b\stackrel{^}{j}$
Magnitude and direction$\mid \mid \phantom{\rule{-0.167em}{0ex}}\phantom{\rule{-0.167em}{0ex}}\phantom{\rule{-0.167em}{0ex}}\stackrel{\to }{u}\phantom{\rule{-0.167em}{0ex}}\phantom{\rule{-0.167em}{0ex}}\mid \mid ,\phantom{\rule{0.167em}{0ex}}\theta$

## Component form

In component form, we treat the vector as a point on the coordinate plane, or as a directed line segment on the plane. The components are the vector's $x$- and $y$-coordinates.

## Unit vector form

These are the unit vectors in their component form:
$\stackrel{^}{i}=\left(1,0\right)$
$\stackrel{^}{j}=\left(0,1\right)$
Using vector addition and scalar multiplication, we can represent any vector as a combination of the unit vectors. For example, $\left(3,4\right)$ can be written as $3\stackrel{^}{i}+4\stackrel{^}{j}$.

## Magnitude and direction form

Considering the vector graphically, we can also describe it by its $\text{magnitude}$ (the length of the line segment) and its $\text{direction}$ (the angle the line forms with the positive $x$-axis).

## Want to join the conversation?

• Are degrees or radians more commonly used for the direction of vectors? And why? • I know I can easily transform Component form to Magnitude and direction form.But how can I use Magnitude and direction form to get Component form? •  Let's use vector (3, 4) as an example. That in magnitude direction form is ||5||, 53.130102°.

Let's think of this vector as a triangle on the unit circle

First, evaluate what quadrant you're in. This information will judge which sides are negative and which are positive. Now, we need to find the reference angle of our angle. In this case, it's 53.130102°.

Now, we do some simple trig to find the two sides using our angle and the hypotenuse (the magnitude):

sin(53.130102) = Opposite / 55 * sin(53.130102) = (Opposite / 5) * 55sin(53.130102) = OppositeOpposite = 3.99999998146Opposite = 4The answers are a bit off because we rounded our angle a couple decimal placescos(53.130102) = Adjacent / 55 * cos(53.130102) = (Adjacent / 5) * 55cos(53.130102) = AdjacentAdjacent = 3.00000002472Adjacent = 3Our vector in component form is (3, 4) •  