Vectors: Collinear Demo - P,Q,R & D,E,F

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Homework Help Overview

The discussion revolves around demonstrating the collinearity of points defined by vectors in two-dimensional and three-dimensional space. The points in question are P(15, 10), Q(6, 4), R(-12, -8) for part (a) and D(33, -5, 20), E(6, 4, -16), F(9, 3, -12) for part (b).

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss forming vectors between the given points and checking for proportionality to establish collinearity. There are questions about the relevance of the original equations provided and the correct interpretation of vector relationships.

Discussion Status

Some participants have offered guidance on how to approach the problem by suggesting the formation of vectors and checking their proportionality. There is an ongoing exploration of the correct method to demonstrate collinearity, with various interpretations being discussed.

Contextual Notes

There are mentions of confusion regarding the definitions of variables in the equations provided, and some participants express uncertainty about the next steps after finding the vectors.

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Homework Statement



Using vectors, demonstrate that these points are collinear.

a) P(15 , 10) , Q(6 , 4) , R(-12 , -8)

b) D(33, -5, 20) , E(6, 4, -16) , F(9, 3, -12)

Homework Equations



\frac{x_{1}}{x_{2}} = \frac{y_{1}}{y_{2}}

\frac{x_{1}}{x_{2}} = \frac{y_{1}}{y_{2}} = \frac{z_{1}}{z_{2}}

The Attempt at a Solution



a)
Vector PQ = (-9 , -6)
Vector QR = (-18 , -12)
Vector RP = (27 , -18)

(-9 / 27 / -18) = (-6 / -18 / -12)

Therefore, not collinear.

b)
Vector DE = (-27, 9, -36)
Vector EF = (3, -1, 4)
Vector FD = (24, -8, 32)

(-27 / 3 / 24) = (9 / - 1 / -8) = (-36 / 4 / 32)
Am I correct or did I do something wrong? If I did, can you please point it out and tell me on how to fix it?
 
Last edited:
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Since you don't say what x1, x2, y1, and y2 have to do with this problem you "relevant equation" really doesn't make sense.

The problem said "Usining vectors". Okay, what is the vector from P(15, 10) to Q(6,4)? What is the vector from P(15, 10) to R(-12, -8)?
 
well what u need to do is form two vectors like PQ as a vector, and QR, also as a vector. Where P is the starting point wheras Q is the end of the vector for the first one, and similarly for the second. I assume you are working on a cartesian system of coordinates. so now if you manage to show something similar to

PQ=k*QR, where k is a constant, than i guess you also have managed to show that those four points are collinear, since they all lie in a line!
 
Well, Halls is faster!
 
HallsofIvy said:
Since you don't say what x1, x2, y1, and y2 have to do with this problem you "relevant equation" really doesn't make sense.

The problem said "Usining vectors". Okay, what is the vector from P(15, 10) to Q(6,4)? What is the vector from P(15, 10) to R(-12, -8)?

I don't follow. . .

The question does mention what are x1, x2, y1, y2, (if relevant) z1 and z2. . .

And from my attempted solution, I did find the vectors. . .but I don't know what to do after finding the vectors. . .
 
why don't u just follow the suggestions! for the first part you have
PQ(-9,-6)
QR(-18,-12) ,now you see that their coordinates are proportional, that is

-9/-18=-6/-12=1/2.

Just do the same thing with the other!
Also you can proceede with PR like halls suggested, you will get the same thing.
 

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