One final question in algebra and geometry

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Homework Help Overview

The discussion revolves around the relationship between two vectors, u and v, which are stated to be linearly independent. The original poster is tasked with finding a scalar s such that the vectors (1-s)u - 2/3v and 3u + sv are parallel.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to equate the two vector expressions to find s, questioning whether the vectors can be considered parallel given their linear independence. Other participants clarify that parallel vectors do not need to be equal and suggest using a scalar relationship to express this.

Discussion Status

Participants are exploring the implications of the linear independence of u and v, with one suggesting a system of equations derived from the vector expressions. There is acknowledgment that the equations may yield multiple solutions for s, but the same value of s must satisfy both equations.

Contextual Notes

There is a mention of the problem being unfamiliar to the participants, which may affect their approach and understanding. The discussion also highlights the non-linear nature of one of the equations involved.

byronsakic
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the vectors u and v are linearly independent. find s, if vectors (1-s)u - 2/3v and 3u +sv are parallel.
i got them equal to each other...but then if you draw them, both of them would be in the exact place which makes them collinear...so would you say it is parallel because vectors u and v are linearly independent?
this is my solution.
geo1.jpg
 
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You can't have two different values of s (one for u and one for v).

Also, two parallel vectors need not be equal.

If A is parallel to B (A,B vectors), then A = kB (k is a scalar).
 
Last edited:
any further suggestions?

like this is a question we've never seen before so i don't know how to approach the question, using a scalar a = kb with the s would give me k and s right. but i have to solve for s.
 
Last edited:
You know that {(1-s)u- (2/3)v}= k{3u+ sv} AND that u and v are independent.

{(1-s)- 3k}u- {(2/3)+sk}v= 0. Since u and v are indepent, you must have

1-s- 3k= 0 and (2/3)+ sk= 0, two equation in two unknowns. Since one of the equations is non-linear, there may be two correct solutions. However, the same value of s applies to both the u and v equations.
 
thanks a lot :D
 

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