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How to find which of three points are on a line?

  • Thread starter rocapp
  • Start date
  • #1
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Homework Statement


Which of these three points falls on the line?

l: r(t)=(i+2j)+t(6i+j-5k)

P(1,2,0); Q(-5,1,5); R(-4,2,5)

I have the answer, but I don't understand why P and Q fall on the line but R does not. Is it because the i and j magnitudes are different for R?


Homework Equations




The Attempt at a Solution



P:
(1+6t) = 1
t=0

2=2+t
t=0

t=0

Q:
-5=1+6t
t=-1

1=2+t
t=-1

t=0

R:
-4=1+6t
t=-5/6

2=t+2
t=0

t=0
 

Answers and Replies

  • #2
329
34
First, you are right that P and Q fall on the line and R does not. And as you see, R is not on the line because using those points you get 3 incompatible equations in t.

Whether a point falls on a line depends on all 3 magnitudes, i, j and k, and the trivial answer to your question is that most combinations will not work i.e. most 3 dimensional points are not on any given line.

However, working backwards from the equation we can get some insight. For example r(0) = i + 2j. So the point that fits there is P = (1,2,0). When you look at r(-1) = -5i + j - 5k, you see that Q = (-5,1,-5) is a point on the line.

But your incompatible equation for R tells you that there is no possible t for which r(t) = R. It's not really deeper than that.
 
  • #3
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Thanks! I think I understand, but what about the t=0 in R? Does this value not count since there is already an incompatible value in its components?
 
  • #4
Ray Vickson
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Homework Statement


Which of these three points falls on the line?

l: r(t)=(i+2j)+t(6i+j-5k)

P(1,2,0); Q(-5,1,5); R(-4,2,5)

I have the answer, but I don't understand why P and Q fall on the line but R does not. Is it because the i and j magnitudes are different for R?


Homework Equations




The Attempt at a Solution



P:
(1+6t) = 1
t=0

2=2+t
t=0

t=0

Q:
-5=1+6t
t=-1

1=2+t
t=-1

t=0

R:
-4=1+6t
t=-5/6

2=t+2
t=0

t=0
The line has parametric equations x = 1 + 6t, y = 2 + t, z = -5t. A point (a,b,c) lies on the line if the equations a = 1+6t, b = 2+t, c = -5t are compatible; that is, we must get the same t from all three equations.
 
  • #5
95
0
Thanks a bunch! That's what I needed.
 

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