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[Ecrofirt]

how can you determine if the following three points lie on a straight line:
A(5, 1, 3) B(7, 9 , -1) C(1, -15. 11)

I read through the section in the book, and it didn't discuss anything like this
This is the very beginning of a chapter on vectors, and the only thing the section discusses is that there *is* a z plane, and it shows the distance formula

 

[TheQueen'sOwn]

Wow, I should know how to do this... damn you Grade 12 Discrete math.... I've forgotten already .

EDIT:
Find the equation of the plane that contains all of those points. Then the equation of a line on that plane (using two of the points). Filling in (2/3rds) of the 3rd point coordinates should yield the correct value for x, y or z??? I dunno. Guessing here.

 

[truffleshuffle83]

ill try and help since you were generous with the codes.

1. parameterize the first two points
(5-7,1-9,3--1)=(-2,-8,4)
let Po=(5,1,3)
so
x=5-2t
y=1-8t
x=3+4t

symmetric equations result in
(x-5)/-2=(y-1)/-8=(z-3)/4

using the last set of numbers

(1-5)/-2=(-15-1)/-8=(11-3)/4 ======= 2
your points all lie on the same line


calc three up in the hizzy bitches. if this is wrong i take no responsibility

 

[Shompola]

easy.. check if the vector AB and BC are parallel.

 

[Animal]

and what about the vector AC?
you'd have to check for all possible combinations from what i remember. The parallel vectors could be made of any of the points.

 

[Shompola]

AC is redunant... What important IS that the vectors you check share a one point with each other, in my case point B.

But if you want you can check if AB and AC are parellel. Same thing basicly. As said the important thing is that the two vectors share a single point.

 

[truffleshuffle83]

why you guys arguing?? i already solved it :lol

 

[Gorgie]

Create two vectors, AB, and BC. (I'll be using these, but one can use CB and BA)

[7,9,-1]-[5,1,3] = [2,8,-4]

[1,-15,11]-[7,9,-1] = [-6,-24,12]

If one of these vectors can be made into the other vector by multiplying a scalar value, then the vectors are paraell, and in this case, on the same straight line, since we know the head of one vector touchs the tail of the next vector.

Alternativaly, if you can't see a simple scalar to multiply one by to get the other, remember that if two vectors are parallel, the normal vector will be 0 vector. To find normal vector, cross multiply.

i j k
2 8 -4
-6 -24 12

={8*12-4*24} - (24-24) +(-48 + 48) = [0,0,0]

So for these 3 points, they do exist in a straight line.

 

[Shompola]

why you guys arguing?? i already solved it :lol

Sorry mate. I knew you did it correctly. I just simple posted an alternative, a linear algebra alternative wich Gorgie showed.