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Vectors: Linear Combinations and Parallelism

  1. Jan 15, 2012 #1
    1. The problem statement, all variables and given/known data
    The question states: "find all scalars c, if any exist such that given statement is true."
    It also suggests trying to do without pencil and paper.

    a) vector [c2, c3, c4] is parallel to [1,-2,4] with same direction.

    b) vector [13, -15] is a linear combination of vectors [1,5] and [3,c]

    c) vector [-1,c] is a linear combination of vectors [-3,5] and [6,-11]


    3. The attempt at a solution

    a) for it to be parallel, it has to be a scalar multiple of [1,-2,4] = x. and [c2, c3, c4] = y. then you can say ax = y. but wouldn't that give different values for all 3 c's? Lets say for example a=2. then 2x = y. then it would be:
    i) c2 = 2*1
    ii) c3 = 2*-2
    iii) c4 = 2*4
    which results in a different value of c in each. can i conclude that there DNE a such c?

    b) the scalar multiple could be any real number for both vectors so c could be any real number?

    c) 1[-3,5] + 1/3[6,-11] = [-1,4/3] so c = 4/3. this was achieved by trial and error. firstly is this even right? second, if its right, is there a algebraic way of doing it. it would have 3 unknowns and 2 equations.

    at a glance seems pretty simple, or i just forgot some of the things i learned last year. any guidance would be appreciated.
     
  2. jcsd
  3. Jan 15, 2012 #2

    Dick

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    To solve any of these systematically you need to solve equations. Take the first one. You've got ax=y. So you've got c^2=a*1, c^3=a*(-2), and c^4=a*4. Just showing that a=2 doesn't work doesn't mean NO number can work. You would need to show that. Hint: the first equation tells you a=c^2. Substitute that for 'a' in the other equations. I wouldn't try to do these without pencil and paper until you know how to do them with pencil and paper.
     
    Last edited: Jan 15, 2012
  4. Jan 15, 2012 #3
    oh i tried doing that with 2x=y, now i see why that would be wrong.
    c3 = -2c2
    c = -2 => not valid since c has to be > 0, since its in the 'same direction'. moving on.

    c4= 4c2
    c = 2 => which holds true so then a = 4.

    I know you need to solve equations, but for questions such as b and c, i get one more unknown then the number of equation. For b, wouldn't more than one value of c hold true, intuitively? in fact all real values?

    and for c i get :
    -3x + 6y = -1
    5x - 11y = c
    2 equations with 3 unknowns! how do i go about solving that?
     
  5. Jan 15, 2012 #4

    Dick

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    For the first one, c=(-2) does work. The thing that's supposed to be positive is a, not c. What is a if c=(-2)??

    For the other two, sure you have more unknowns than equations. So you likely have an infinite number of solutions. But work it out. Eliminate one of the variables that's not c and look at the remaining equation. For b) you'll find a value of c that does NOT work.
     
    Last edited: Jan 15, 2012
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