Need help solving for unknown components

  • Thread starter Thread starter blazeuofa
  • Start date Start date
  • Tags Tags
    Components
AI Thread Summary
The discussion revolves around solving for the unknown component "a" in the vectors A = (2x-y)m and B = (x+ay)m, given that the magnitude of A+B equals 5m. Participants clarify the notation and suggest using the dot product to find the cosine of the angle between the vectors. They propose adding the vectors to find A+B = <3, (a-1)> and then calculating the magnitude. The equation |A+B| = √(9 + (a-1)²) = 5 leads to a quadratic equation to solve for "a." The conversation emphasizes the need for clarity in notation and the steps to derive the solution effectively.
blazeuofa
Messages
14
Reaction score
0

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!
 
Physics news on Phys.org
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)>

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

Just solve for a...
 
1 + a^2
 
Back
Top