Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How to check if r1(t) = r2(t)

  1. Dec 3, 2011 #1
    1. The problem statement, all variables and given/known data
    Show that r1(t) and r2(t) define the same line, where
    [itex]r_{1}[/itex](t) = <3,-1,4> + t<8,12,-6>
    [itex]r_{2}[/itex](t) = <11,11,-2> + t<4,6,-3>

    2. Relevant equations

    3. The attempt at a solution
    I set [itex]r_{1}[/itex](t) = [itex]r_{2}[/itex](t) and got the value of t which is 2.
    then I plugged that t value into [itex]r_{1}[/itex](t) and [itex]r_{2}[/itex](t) which both of them came out to be <19,23,-8>. Is this how to do it?
  2. jcsd
  3. Dec 3, 2011 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    In general, you should use a different variable for the parameter in the two expressions; i.e.
    r1(t)=3,-1,4> + t<8,12,-6>

    r2(s)=<11,11,-2> + s<4,6,-3>​

    All that you have shown is that the two lines intersect at <19,23,-8> .

    See if you can find a linear relationship between s & t that makes the two lines equivalent.
  4. Dec 4, 2011 #3


    User Avatar
    Science Advisor

    [itex]r_{1}[/itex](t) = <3,-1,4> + t<8,12,-6>
    [itex]r_{2}[/itex](t) = <11,11,-2> + t<4,6,-3>

    As SammyS suggested, use another letter, say, s (in honor of SammyS, of course!) as parameter for the second equation. Then, where the lines intersect, we must have
    x= 3+ 8t= 11+ 4s
    y= -1+ 12t= 11+ 6s
    z= 4- 6t= -2- 3t.

    You can solve the first equation for, say, s as a function of t. Replace s in the other two equations with that and see what happens.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook