Multivariable Calculus: Points in a Straight Line

  • #1
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Determine whether the points lie in a straight line:
1) A (2,4,2), B (3,7,-2), and C (1,3,3)
2) D (0,-5,5), E (1,-2,4), and F (3,4,2)

I'm not sure what method I need to use to show that they are or are not in a straight line. I know that the three points in a are not in a line but those in group b are in a line.
 
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  • #2
hilbert2
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Suppose we have three points in 3d space, and their position vectors are ##\vec{a},\vec{b}## and ##\vec{c}##. If the three points are on the same line, the difference vectors ##\vec{b}-\vec{a}## and ##\vec{c}-\vec{b}## have same (or opposite) direction and their cross product ##(\vec{b}-\vec{a})\times (\vec{c}-\vec{b})## is zero.
 
  • #3
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I get the cross product thing, but this is from a section be for we were introduced to dot and cross products, so I can't use them.
 
  • #4
Ray Vickson
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Determine whether the points lie in a straight line:
1) A (2,4,2), B (3,7,-2), and C (1,3,3)
2) D (0,-5,5), E (1,-2,4), and F (3,4,2)

I'm not sure what method I need to use to show that they are or are not in a straight line. I know that the three points in a are not in a line but those in group b are in a line.

If A,B,C are in a straight line, C-A must be a numerical (scalar) multiple of B-A. This works in any number of dimensions, not just three.
 
  • #5
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If A,B,C are in a straight line, C-A must be a numerical (scalar) multiple of B-A. This works in any number of dimensions, not just three.

Should I use the distance formula to get the scalar then?
 
  • #6
Ray Vickson
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Should I use the distance formula to get the scalar then?

Instead of asking, just start writing out the conditions that one (vector) difference is a scalar multiple of the other, and figure out if (a) this can possibly be true; and (2) if true, the value of the scalar.
 

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