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Problem with dot product of vectors

  1. Jun 27, 2008 #1
    It began in class with this problem.

    Find the dot product of 2 vectors:
    v1 a vector with components <4, 8>
    v2 a vector of length 1 angle pi/4

    So, i have 2 ways of doing it.
    1) v1.v2 = v1x.v2x + v1y.v2y
    2) v1.v2 = |v1||v2|cos(theta)
    And they should come out the same but.

    1) v1.v2 = 4cos(pi/4) + 8sin(pi/4) = 6√(2)
    1) v1.v2 = √(42 + 82)cos(pi/4) = 2√(10)

    and 6√(2) ≠ 2√(10), so i wondered if i did something wrong. Then it tried different numbers and found out these two ways of computing the dot product of two vectors are never the same for an angle pi/4 except for vectors with components <a, 0> or <0, a> where a is any real number.

    I wonder if i computed anything wrong causing different results or is is that they are not the same?
  2. jcsd
  3. Jun 27, 2008 #2


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    Hi cshum00!

    (have a pi: π and a theta: θ and a root: √ :smile:)

    θ is the angle between the two vectors.

    If v2 is at π/4 to the x-axis, it is at (1/√2,1/√2), but the angle between it and (4,8) is not π/4. :smile:
  4. Jun 27, 2008 #3
    Re: Problem with dor product of vectors

    I see so it means that, v1x.v2x + v1y.v2y = |v1||v2|cosθ is true only if v1 or v2 is on the x-axis and therefore the angle making between them is θ.

    Thanks, it clarifies at lot.
  5. Jun 27, 2008 #4


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    No, θ in that formula has nothing to do with the x-axis.

    θ is the angle between whatever two vectors you're dot-producting. :smile:
  6. Jun 27, 2008 #5
    |v1||v2|cosθ this formula doesn't but
    v1x.v2x + v1y.v2y this formula does

    so, if the angle of v2 is θ and
    v1x.v2x + v1y.v2y = |v1||v2|cosθ
    then v1 have to be in the x-axis

    although now you cleared me that in the original problem, the angle in v2 was not θ but with the x-axis

    thanks again
  7. Jun 28, 2008 #6


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    NO, v1 and v2 can be any two vectors, [itex]\theta[/itex] the angle between them. It is only if v1 lies on the (positive) x-axis that the angle between then is the angle v2 makes with the positive x-axis.

  8. Jun 28, 2008 #7
    Yes, you are right. I am making so much confusion with the angle given for v2 which is the one between v2 and the x-axis and not θ
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