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R = sA + tB vector equation

  1. Feb 24, 2008 #1
    1. The problem statement, all variables and given/known data
    Let A and B be the two points with position vectors A and B. Show that the line passing through thezse points may be represented by the vector equation:

    R = sA + tB

    2. Relevant equations
    R = Ro + tV
    where Ro is a point on the line and t is some scalar, and V is a vector pointing in the direction of the line.

    3. The attempt at a solution

    I have tried writing R = A + t(A - B)
    and R = B + s(A - B), and manipulating the equations, however, i dont find the solution they are looking for. Could someone please help me get started on this? thank you
  2. jcsd
  3. Feb 24, 2008 #2


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    It would help a lot if you said what s and t are! One thing I note is that they cannot be independent parameters- two parameters would give the equation of a plane. I notice that if s= 0 and t= 1, then R= B. And if s= 1 and t= 0, R= A. In order that this give a line the equation relating s and t must be linear. Now, what linear function, s= at+ b, gives s= 0 when t= 1 and s= 1 when t= 0? I think there is a good change that when you get the answer you will say, "Oh, of course!"
  4. Feb 24, 2008 #3
    s + t = 1 is the relationship i suppose, however, I cannot use this to show the desired relationship. Is it a matter of manipulating the equations i came up with, or should I try something else? Also, thank you for the help
  5. Feb 24, 2008 #4
    Okay, so you know R = A + t(A - B) is the equation of the line, right? Play around with this and see if you can't get it to look something like something*A + something*B.
  6. Feb 24, 2008 #5
    R = B + sA - sB = B(1-s)+ sA = Bt + sA.
    Indeed, thanks a lot for all the help!
  7. Feb 25, 2008 #6


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    Would the "manipulation" going from s+ t= 1 to s= 1- t be too difficult? Replace s by 1- t in your equation and see what happens.
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