For what values of a and b are these two vectors collinear (linear algebra)?

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Homework Help Overview

The discussion revolves around determining the values of variables a and b for which two vectors, t and v, are collinear in the context of linear algebra. The vectors are defined as t = (3a - 4b + 5, 5b - 2a - 8) and v = (-3, 4).

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the condition for collinearity, suggesting that t = kv for some scalar k. There are attempts to express a and b in terms of k. Some participants express confusion about the steps taken by the teacher and the implications of the derived equations.

Discussion Status

The discussion is active, with participants sharing their interpretations and questioning the steps taken to arrive at the expressions for a and b. Some guidance has been offered regarding the relationship between a and b, but there is no explicit consensus on the final values or the clarity of the solution process.

Contextual Notes

Participants note the complexity of the problem and the potential for multiple interpretations of the results. There is a mention of the dot product and orthogonal spaces as alternative methods for assessing collinearity, which adds to the discussion's depth.

BlueRope
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Homework Statement


For what values of a and b are these two vectors collinear (linear algebra)?

t = (3a -4b +5, 5b -2a -8)
v = (-3, 4)

t and v are vectors


Homework Equations





The Attempt at a Solution



I'd like to know what the steps are. I don't really care about the answer.
 
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t and v are collinear if t = kv for some scalar k. Expand and solve for a and b in terms of k.
 
I was thinking of doing that, until I saw the answer, which was t = 7(a-1)(-3,4) or 7(a-1)vectorz.
 
Basically, he wrote 6a -b -4 = 0

then he isolated b, which gives:

b = 6a -4

then he somehow came with

the answer, which is, like I said:

t = 7(a-1)(-3,4) or 7(a-1)vector z.

I have no idea what the teacher did.
 
Another way of determining vectors are collinear is to see if the dot product of one of the vectors and basis vectors of the other vector's orthogonal space evaluate to 0. In the plane, a vector v's orthogonal space has one dimension and is in fact perp v.

It appears that the equation t * perp v = 0 was solved for b here.
 
I finally understood how he did it. However, I don't understand how it answers the question. He wanted the values of b and a, but we only got the value of k.

If the answer is 7(a-1) vector z then a GENUINE answer would be: b = 0 and a = R or any value, right?
 
I don't understand where 7(a-1) vector z is from here, but letting a = R gives b = 6R - 4. Values a, b such that b = 6a -4 should be enough to answer this question.
 

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