Vectors: Collinear


by Macleef
Tags: collinear, vectors
Macleef
Macleef is offline
#1
Feb25-08, 08:30 PM
P: 30
1. The problem statement, all variables and given/known data

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)

2. Relevant equations

[tex]\frac{x_{1}}{x_{2}}[/tex] = [tex]\frac{y_{1}}{y_{2}}[/tex]

[tex]\frac{x_{1}}{x_{2}}[/tex] = [tex]\frac{y_{1}}{y_{2}}[/tex] = [tex]\frac{z_{1}}{z_{2}}[/tex]

3. 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?
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HallsofIvy
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#2
Feb25-08, 08:32 PM
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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)?
sutupidmath
sutupidmath is offline
#3
Feb25-08, 08:34 PM
P: 1,635
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!

sutupidmath
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#4
Feb25-08, 08:35 PM
P: 1,635

Vectors: Collinear


Well, Halls is faster!!!!
Macleef
Macleef is offline
#5
Feb25-08, 08:40 PM
P: 30
Quote Quote by HallsofIvy View Post
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. . .
sutupidmath
sutupidmath is offline
#6
Feb25-08, 11:00 PM
P: 1,635
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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