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Homework Help: Vector Equations

  1. Feb 6, 2008 #1
    1. The problem statement, all variables and given/known data
    find a vector equation for the straight line given by y=4x+3

    2. Relevant equations

    3. The attempt at a solution

    i think i need to change this into cartiesen then to parametrics but have no idea how to change that equation into cartisean
  2. jcsd
  3. Feb 6, 2008 #2


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    Welcome to PF rose,

    The equation is already in Cartesian form, all you need to do now is parameterise it.
  4. Feb 7, 2008 #3
    Ok, thanks for pointing that out but as a person who has never been taught any of this before any chance of an example of how to do that? I've done it before when it has been for exampl r1=i+2j+2K+ landa(i-j+3k) yes i know how to get that into parametrics but not if its just y=4x+3
  5. Feb 7, 2008 #4


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    Okay, you have done this when you have 3 dimensions and I think you are saying you can find the vector equation of a line through 2 given points: the line through (1, 2, 2) and (2, 1, 5) has "direction vector" (2-1)i+ (1-2)j+ (5-2)k= i- j+ 3k so we can write the vector equation as r= (i+ 2j+ 2k)+ lambda(i- j+ 3k).

    One problem you have is that the single equation y= 4x+ 3 is that a single equation in 3 dimensions describes a plane, not a line. So you must mean "in the xy-plane" which means that z= 0.

    Now, one way to do that is to find two points! if x= 0, y= 3 so one point the line passes through is (0, 3, 0). If x= 1, y= 7 so another point is (1, 7, 0). Use exactly the same method as above to find the vector equation of the line through those two points. You should see that the component of the k vector is 0: the line is always in the xy-plane, of course.

    One thing that may be confusing you is that the is no one vector equation. The parameter, lambda, has no "geometrical" meaning and can be chosen almost arbitrarily. When we have all the other coordinates written as functions of the other (here y is a function of x) we can always use that one coordinate as the parameter. That is, if y= f(x), we can write x= lambda, y= f(lambda) and have the vector equation r= lambda i+ f(lambda) j.
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