Solving p and q in a Vector Equation

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Homework Help Overview

The discussion revolves around solving for the constants p and q in a vector equation related to mechanics, specifically involving vector components and their equality.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster considers trial and error for finding p and q, while another participant suggests using the equality of vector components to form linear equations.

Discussion Status

Participants are exploring the relationship between vector components and how to set up equations based on their equality. There is an indication of a productive direction as one participant provides a method to derive the equations needed to solve for p and q.

Contextual Notes

The original poster expresses uncertainty about the method to approach the problem, indicating a possible lack of familiarity with solving linear equations derived from vector components.

Xaif
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Hello, I have a quick question that I think seems stupid but I can't find an answer for it. It's about vectors in a mechanics module. Anyhow, this is the question:

Determine p and q:

p(4i + 3j) + q(12i + 5j) = 68i + 33j, where p and q are contants.

I can only think to do this by trial and error to get p = 3.5 and q = 4.5. Is there a proper method to work out something like this?
 
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Yes there is.

If the components (i and j in this case) of two vectors are equal, then the two vectors are equal. So, in your equation, you need the components on one side to be equal to the components on the other side. This should give you two linear equations, which I am going to assume you can solve.
 
[tex]p(4\mathbf{i} + 3\mathbf{j}) + q(12\mathbf{i} + 5\mathbf{j}) = 68\mathbf{i} + 33\mathbf{j}[/tex]

[tex](4p + 12q)\mathbf{i} + (3p + 5q)\mathbf{j} = 68\mathbf{i} + 33\mathbf{j}[/tex]

[tex]4p + 12q = 68\ \mbox{ and }\ 3p + 5q = 33[/tex]

[tex]\left (\begin{array}{cc}4 & 12\\ 3 & 5\end{array}\right )\left (\begin{array}{c}p\\ q\end{array}\right ) = \left (\begin{array}{c}68\\ 33\end{array}\right )[/tex]

[tex]\left (\begin{array}{c}p\\ q\end{array}\right ) = \left (\begin{array}{cc}4 & 12\\ 3 & 5\end{array}\right )^{-1}\left (\begin{array}{c}68\\ 33\end{array}\right )[/tex]
 
Ah, thankyou. It was quite simple really :blushing:
 

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