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Homework Help: Solving parallel vectors, quicker method

  1. Sep 2, 2010 #1
    Ok, so I'm new to vectors, and I've been reading up on them and I think I understand how paralell vectors work, or at least instinctively.
    a || xa, where x is a constant.

    So I started doing some questions from my textbook, and I could manage, but I'm not sure if what I'm doing is the most efficient way to solve the problems.

    For example:
    Find s if 3i+4j -s(i-2j) is parallel to 2i + j

    What I've done is:
    3i+4j -s(i-2j)
    => 3i+4j -si + 2js = x(2i+j)
    => i(3-s) + j(4+2s) = 2xi+jx
    => 3-s = 2x, and
    4+2s = x
    => 3-s = 2(4+2s)
    => 3-s = 8 + 4s
    => 3-8 = 4s + s
    => -5 = 5s
    => s = -1

    When it comes to math I've never believed in there being a "correct" way to do stuff, but in this case, is there a "better" way to do these sorts of problems?
     
    Last edited: Sep 3, 2010
  2. jcsd
  3. Sep 2, 2010 #2

    ehild

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    Homework Helper

    Your solution is correct, and there is no "better way" in Maths. There can be shorter or simpler ways, but your method is simple enough. You can make it more complicated if you like:smile:

    ehild
     
  4. Sep 2, 2010 #3
    yay.
    And yeah like I said, I agree, there is no "better" way, solving a problem quickly just compensates for what could have been enjoyed through the frustration of anwering the question.
    Oh and, I don't wan't to create a new post, so, I might as well ask here, how do you measure the bearing of a vector?
     
  5. Sep 3, 2010 #4

    ehild

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    Homework Helper

    Sorry, I don't know what the bearing of a vector is. I never heard it. Could you please explain? Or it is better to start a new thread.

    ehild
     
  6. Sep 3, 2010 #5
    The short answer is with a protractor, plotter, or some other device.

    I assume you meant calculate; in which case you find the slope of the vector, from that you can use trig to get the angle.
     
  7. Sep 3, 2010 #6
    Yeah, ok, I'd figured out what it meant yesterday, thanks anyway.
     
  8. Sep 3, 2010 #7
    There is another method, called the cross product, it's probably lurking a few pages ahead in your textbook! If not, Google. If two vectors a and b are parallel, their cross product, a x b = 0. It's defined for vectors in 3 dimensions, but you can still use it in your 2 dimensional example by just setting the third component of each vector equal to 0.

    ((3 - s)i + (4 + 2 s)j) x (2i + 1j) = (-5 -5s)k

    which is 0 when s = -1, as you found. When you see the fiddly formula for calculating the components, you might wonder whether it's quicker. But you'll get use to it with practice, and it's certainly quick if you have a computer handy. E.g. in Mathematica:

    Code (Text):

    In[1]:= Cross[{3 - s, 4 + 2 s, 0}, {2, 1, 0}] == 0

    Out[1]= {0, 0, -5 - 5 s} == 0
     
     
  9. Sep 3, 2010 #8
    You can find the angle between two vectors using what's called the dot product: a . b = |a| |b| cos(a,b), where |a| is the length of a, and cos(a,b) the cosine of the angle between them. a . b can be calculated by multiplying corresponding components and adding them together to give a number: (a1 i + a2 j + a3 k) . (b1 i + b2 j + b3 k) = a1 b1 + a2 b2 + a3 b3. So the angle is

    Cos-1((a . b)/(|a| |b|)).

    If you want the angle between a vector and one of the coordinate axes, dot it (that is, take its dot product) with the appropriate basis vector, e.g.

    Cos-1((a . i)/(|a|)),

    the length of i being 1.
     
  10. Sep 3, 2010 #9
    Yeah, the dot product stuff is in the next book. I'm gonna start that in a week or two. Can't wait =)
     
    Last edited: Sep 3, 2010
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