Need help solving for unknown components

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Homework Help Overview

The discussion revolves around determining the possible values of the variable \( a \) in the context of vector addition and magnitudes. The vectors are defined as \( \mathbf{A} = (2x - y) \) m and \( \mathbf{B} = (x + ay) \) m, with the condition that the magnitude of their sum equals 5 m.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the notation and definitions of the vectors, questioning the roles of \( x \) and \( y \). Some suggest using the dot product to find the cosine of the angle between the vectors, while others propose adding the vectors directly to find the resultant magnitude. There is also discussion about the potential quadratic equation that may arise from the problem.

Discussion Status

Participants are actively engaging with the problem, clarifying notation and exploring different methods to approach the solution. Some have provided guidance on how to manipulate the vectors and set up equations, but there is no explicit consensus on a single method or interpretation yet.

Contextual Notes

There is some ambiguity regarding the definitions of \( x \) and \( y \), and whether they correspond to unit vectors in a Cartesian coordinate system. Additionally, the problem's constraints regarding the magnitude of the vectors and the variable \( a \) are under consideration.

blazeuofa
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Homework Statement


Vector A = (2x-y)m and Vector B= (x+ay)m (a is alpha, m is meters)

If |A+B|=5m, determine the possible values of a.

Homework Equations



The magnitude of A= sqrt 5

The Attempt at a Solution



|A^2 + B^2|=5^2
|(2x-y)^2 + (x+ay)^2|= 5^2
|(2x-y)^2 + (x+ay)^2|= 25

I cannot figure out how to solve this problem. Please help!
 
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Vector A = (2x-y)m and Vector B= (x+ay)m (a is alpha, m is meters)

This statement is not clear. Here what is x and y? Are they equivalent i and j?
If that is the case, find cos(theta) between A and B using dot product. Use this cos(theta) in R^2 = A^2 + B^2 +2AB*cos(theta) to get values of alpha.
 
I'm a little confused by your notation, do you mean:

A = <2, -1>; B = <1, a>?

If so add the vectors then take the magnitude and solve for a it should get you a quadratic, I think.
 
Feldoh said:
I'm a little confused by your notation, do you mean:

A = <2, -1>; B = <1, a>?

If so add the vectors then take the magnitude and solve for a it should get you a quadratic, I think.

yes that's what i mean. What would be the result of adding those? That's where I am stuck
 
blazeuofa said:
yes that's what i mean. What would be the result of adding those? That's where I am stuck
Then follow the post 2.Or
R^2 = A^2 + B^2 +2A.B and solve for alpha
 
Last edited:
what do you get when you square B?
 
A+B = <3, (a-1)>

[tex]|A+B| = \sqrt{9 + (a-1)^2} = 5[/tex]

Just solve for a...
 
1 + a^2
 

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