How to Find the X Component of a Vector | Decomposing Vectors Homework

  • Thread starter Mr Davis 97
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In summary, when finding the x component of a vector that is pointing in a direction specified in degrees, there are two ways to do it. One way is to tack on a negative sign and calculate -(magnitude)cos(angle). The other way is to not tack on a negative sign and calculate (magnitude)cos(180-angle). Either method is acceptable, as long as you are aware of the direction and use the appropriate angle in the cosine function.
  • #1
Mr Davis 97
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Homework Statement


The magnitude of the vector is 6 m, and points 35 degrees north of west. Find the x component of the vector.

Homework Equations



x = |R|cos(theta)

The Attempt at a Solution



I am confused about what degrees to input into the cosine function. I know that the vector is pointing 35 degrees north of west, which means that the x component will be negative, but there are two ways I could do it am I am not sure which one I should do. First, I could tack on a negative and calculate -(6 m)cos35 = -4.9 m. Another way is not tack on a negative and calculate (6 m)cos145 = -4.9 m. Which method is better, and which one should I use on a regular basis?
 
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  • #2
Both ways are fine, as long as you don't try to memorize a rule and blindly apply it. Personally, I like the first approach, and just tack on the negative sign when I know there should be one.
 
  • #3
Mr Davis 97 said:

Homework Statement


The magnitude of the vector is 6 m, and points 35 degrees north of west. Find the x component of the vector.

Homework Equations



x = |R|cos(theta)

The Attempt at a Solution



I am confused about what degrees to input into the cosine function. I know that the vector is pointing 35 degrees north of west, which means that the x component will be negative, but there are two ways I could do it am I am not sure which one I should do. First, I could tack on a negative and calculate -(6 m)cos35 = -4.9 m. Another way is not tack on a negative and calculate (6 m)cos145 = -4.9 m. Which method is better, and which one should I use on a regular basis?
Well, you could learn the angles for the cardinal points of the compass, but why do that when you can just make something up?

East = 0°
North = 90°
West = 180°
South = 270°
 
  • #4
SteamKing said:
Well, you could learn the angles for the cardinal points of the compass, but why do that when you can just make something up?

East = 0°
North = 90°
West = 180°
South = 270°

Huh?
 
  • #5
Mr Davis 97 said:
Huh?
You've never seen the following diagram (or a similar one)?:
ucad.gif
It should have been used when you studied trigonometry.

After all,
cos (0°) = 1.0
sin (0°) = 0.0

cos (90°) = 0.0
sin (90°) = 1.0

cos (180°) = -1.0
sin (180°) = 0.0

cos (270°) = 0.0
sin (270°)= -1.0

etc., etc.
 

FAQ: How to Find the X Component of a Vector | Decomposing Vectors Homework

What is the X component of a vector?

The X component of a vector is the horizontal length or magnitude of the vector. It is the projection of the vector onto the X-axis.

How do you find the X component of a vector?

To find the X component of a vector, you can use the formula X = |V| * cos(θ), where |V| is the magnitude of the vector and θ is the angle between the vector and the X-axis.

Can you decompose a vector into its X and Y components?

Yes, a vector can be decomposed into its X and Y components using trigonometry. The X component is the horizontal projection of the vector and the Y component is the vertical projection.

Why is it important to find the X component of a vector?

Finding the X component of a vector is important in many applications, such as physics and engineering. It allows us to analyze the motion of an object in a specific direction and calculate the forces acting on it.

What is the difference between the X component and the Y component of a vector?

The X component of a vector is the horizontal projection of the vector, while the Y component is the vertical projection. They are perpendicular to each other and together make up the magnitude and direction of the vector.

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