# The Magnitude of a Vector

## Homework Statement

For any vector in 2D space, it can be broken down into its horizontal and vertical components.

## Homework Equations

In one of my engineering classes, we are using the following equation to determine the magnitude of a vector:

$$u=v_1 \cdot cos\theta +u_2 \cdot sin\theta$$

Where $\theta$ is the angle with respect to the horizontal, v1 is the horizontal component and v2 is the vertical component of the vector.

I know this equation works but I don't understand why.
I feel like I am missing a fundamental concept, because to determine the magnitude of a vector, I would use Pythagoras theorem, and I cannot derive the above equation from Pythagoras's equation.

## Answers and Replies

gneill
Mentor
Can you provide some context for where this equation is applied? Perhaps give a specific example.

In general, this equation will not work for a single vector whose x and y components are ##u_1## and ##u_2##. Perhaps they are summing the horizontal components of two different vectors to obtain a net horizontal resultant?

vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
That equation doesn't give the magnitude of the vector. It gives you the component of the vector in the direction of ##\hat n = \cos\theta\,\hat i + \sin\theta\,\hat j##.

CWatters
Science Advisor
Homework Helper
Gold Member
In one of my engineering classes, we are using the following equation to determine the magnitude of a vector:
u=v1⋅cosθ + u2⋅sinθ​

Where $\theta$ is the angle with respect to the horizontal, v1 is the horizontal component and v2 is the vertical component of the vector.

I know this equation works but I don't understand why.

It comes from geometry... See this diagram... If that's not clear do say and I will explain some more. #### Attachments

• CivilSigma
Chestermiller
Mentor
I think you meant to write the equation as $$u=u_1\cos{\theta}+u_2\sin{\theta}\tag{1}$$where $$u_1=u\cos{\theta}\tag{2}$$and$$u_2=u\sin{\theta}\tag{3}$$If you substitute Eqns. 2 and 3 into Eqn. 1, you get:
$$u=u\cos^2{\theta}+u\sin^2{\theta}=u$$

• CivilSigma and SammyS