What is the correct way to calculate unit vectors for given vector components?

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Homework Help Overview

The discussion revolves around calculating unit vectors from given vector components in a two-dimensional context. Participants are examining the relationship between the components and the hypotenuse, which represents the magnitude of the vector.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using similar triangles to derive unit vectors from the components. There is mention of setting up a relationship involving the lengths of the components and the hypotenuse. Some participants express uncertainty about the correctness of their calculations and question the notation used for unit vectors.

Discussion Status

There is an ongoing exploration of the calculations involved in determining unit vectors. Some participants suggest that the calculations may not be necessary, while others affirm the correctness of the values obtained. Multiple interpretations of the problem are being considered, particularly regarding the need for calculations versus the given components.

Contextual Notes

Participants note that the x and y components are explicitly provided in the problem, leading to questions about whether additional calculations are required. There is a concern about receiving homework credit based on the interpretations of the unit vector calculations.

H2Owned
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Homework Statement


http://img22.imageshack.us/img22/1254/10031833.jpg


Homework Equations





The Attempt at a Solution


My attempt was based on the idea that both triangles formed by the vector components (x component, y component, and the hypotenuse which is the magnitude of the vector itself) would be similar.

I took the length of each of the sides, -3.20 for x and 2.10 for y and individually solved for the unit vector by setting up this relationship:
(length of component)/(length of hypotenuse)=(length unit component)/1

The system rejected both of my answers. For x i got -0.836 and for y i got 0.549.
 
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H2Owned said:

Homework Statement


http://img22.imageshack.us/img22/1254/10031833.jpg


Homework Equations





The Attempt at a Solution


My attempt was based on the idea that both triangles formed by the vector components (x component, y component, and the hypotenuse which is the magnitude of the vector itself) would be similar.

I took the length of each of the sides, -3.20 for x and 2.10 for y and individually solved for the unit vector by setting up this relationship:
(length of component)/(length of hypotenuse)=(length unit component)/1

The system rejected both of my answers. For x i got -0.836 and for y i got 0.549.

Your answer looks fine to me. I calculated the same value for the two components. I like that similar triangle method to find unit vectors. In vector classes, you usually just think about it this way(which involves the exact same math): you have a vector A with magnitude A at angle C. You want the same direction(the same angle C) at a magnitude of 1, so you divide by A(the hypotenuse). So yeah, it's the same calculation: each component divided by the hypotenuse.
 
Last edited by a moderator:
Maybe it has to do something with unit vector notation. is there a specific notation?Or maybe we both got the answer wrong. Either way I am not getting my homework credit right now
 
please help me figure this out, or at least give me your opinions.
 
What are you calculating here? Are not the components given in the problem?
The x component is -3.20 and the y component is +2.10, according to the problem. Am I missing something here?
 
nasu said:
What are you calculating here? Are not the components given in the problem?
The x component is -3.20 and the y component is +2.10, according to the problem. Am I missing something here?

it turns out that there isn't any calculation necessary, and your answer is correct. i thought the question was asking for a unit vector with the same angle as the given vector.
A Unit vector is a vector with a magnitude of 1, that's where all of my calculations came from - which apparently were unnecessary.
 

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