Finding the magnitude (length) and direction (angle) of a vector

  • #1
Ineedhelpwithphysics
43
7
Homework Statement
In the Picture
Relevant Equations
Pythagoras theorem, inverse tan function
1697566568581.png

So i found the magnitude which is
(-1)^2 + (-2)^2 = P^2 =
Sqrt(5)

Then I used the inverse tan function to find the angle (direction)
theta = arctan (-2/-1) = 63.8 degrees

Im confused with my 63.8 degrees since the angle in the graph looks greater than 63.4 degrees

I subtracted 180 by 63.8 and got 116.6

Since it's going clock wise it's -116.6

Am i right?
 
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  • #2
Ineedhelpwithphysics said:
Homework Statement: In the Picture
Relevant Equations: Pythagoras theorem, inverse tan function

View attachment 333724
So i found the magnitude which is
(-1)^2 + (-2)^2 = P^2 =
Sqrt(5)

Then I used the inverse tan function to find the angle (direction)
theta = arctan (-2/-1) = 63.8 degrees

Im confused with my 63.8 degrees since the angle in the graph looks greater than 63.4 degrees

I subtracted 180 by 63.8 and got 116.6

Since it's going clock wise it's -116.6

Am i right?
You are not right. Remember ##~\text{tan} =\dfrac{\text{opposite}}{\text{adjacent}}.##
 
  • #3
Ineedhelpwithphysics said:
Homework Statement: In the Picture
Relevant Equations: Pythagoras theorem, inverse tan function

View attachment 333724
So i found the magnitude which is
(-1)^2 + (-2)^2 = P^2 =
Sqrt(5)

Then I used the inverse tan function to find the angle (direction)
theta = arctan (-2/-1) = 63.8 degrees

Im confused with my 63.8 degrees since the angle in the graph looks greater than 63.4 degrees

I subtracted 180 by 63.8 and got 116.6

Since it's going clock wise it's -116.6

Am i right?

Hi,
1698138797220.png

In your case, + or - π will give you the right answer ;)
 

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FAQ: Finding the magnitude (length) and direction (angle) of a vector

How do you find the magnitude of a vector?

To find the magnitude of a vector, you use the Pythagorean theorem. For a vector \( \mathbf{v} = \langle x, y \rangle \) in 2D space, the magnitude is given by \( |\mathbf{v}| = \sqrt{x^2 + y^2} \). For a vector \( \mathbf{v} = \langle x, y, z \rangle \) in 3D space, the magnitude is \( |\mathbf{v}| = \sqrt{x^2 + y^2 + z^2} \).

How do you find the direction angle of a vector in 2D?

The direction angle \( \theta \) of a vector \( \mathbf{v} = \langle x, y \rangle \) in 2D can be found using the arctangent function: \( \theta = \tan^{-1}\left(\frac{y}{x}\right) \). Make sure to consider the quadrant in which the vector lies to determine the correct angle.

How do you find the direction angles of a vector in 3D?

In 3D, the direction angles \( \alpha, \beta, \) and \( \gamma \) with respect to the x, y, and z axes can be found using the dot product. For a vector \( \mathbf{v} = \langle x, y, z \rangle \), the direction cosines are \( \cos(\alpha) = \frac{x}{|\mathbf{v}|} \), \( \cos(\beta) = \frac{y}{|\mathbf{v}|} \), and \( \cos(\gamma) = \frac{z}{|\mathbf{v}|} \). The angles are then \( \alpha = \cos^{-1}\left(\frac{x}{|\mathbf{v}|}\right) \), \( \beta = \cos^{-1}\left(\frac{y}{|\mathbf{v}|}\right) \), and \( \gamma = \cos^{-1}\left(\frac{z}{|\mathbf{v}|}\right) \).

What tools can I use to calculate the magnitude and direction of a vector?

You can use a scientific calculator or software tools like MATLAB, Python (with libraries such as NumPy), or online vector calculators. These tools can perform the necessary arithmetic and trigonometric calculations to find both the magnitude and direction of a vector.

How do I verify my calculations for magnitude and direction?

To verify your calculations, you can double-check each step. For magnitude, ensure

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