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Prove this Vector Operation

  1. Apr 28, 2009 #1
    1. The problem statement, all variables and given/known data
    if vectors [tex]\vec{a}[/tex] and [tex]\vec{b}[/tex] have opposite directions, how to show that |[tex]\vec{a}[/tex]| + |[tex]\vec{b}[/tex]| = |[tex]\vec{a}[/tex] - [tex]\vec{b}[/tex]|?


    2. Relevant equations
    quadratic equation, definition of absolute value


    3. The attempt at a solution
    [tex]|\vec{a}-\vec{b}| = \sqrt{\vec{a}^{2}-2\vec{a}\bullet\vec{b}+\vec{b}^{2}}[/tex]
    and then I got
    [tex]|\vec{a}-\vec{b}| = \sqrt{\vec{a}^{2}-2|\vec{a}||\vec{b}|cos\Phi+\vec{b}^{2}}[/tex]
    So then cosine of the angle is equal to -1, and I don't know how to go from there.
     
    Last edited: Apr 28, 2009
  2. jcsd
  3. Apr 28, 2009 #2
    Looks like a notation issue.

    [tex]
    \begin{align}
    \vec{a}^2 &= a^2\notag\\

    |\vec{a}|&=a\notag
    \end{align}
    [/tex]

    Everything under the radical is a scalar.
     
  4. Apr 29, 2009 #3

    Mark44

    Staff: Mentor

    This makes no sense. In the first equation you have the square of a vector, which implies multiplication of some kind. The only kinds of multiplication that are available are 1) multiplication by a scalar, 2) the dot product, and 3) (for vectors in R^3) the cross product.

    In the second equation, you claim that the magnitude of a vector is equal to the vector itself.
     
  5. Apr 29, 2009 #4

    Mark44

    Staff: Mentor

    You're given that the vectors have the opposite directions, which means that a = -kb for some positive scalar k. Also, since a.b = |a||b|cos(theta), and theta = pi, you have a.b = -|a||b|.

    Put both of these ideas together, and the result you want should fall out pretty readily.
     
  6. Apr 29, 2009 #5
    Again, I think it's a notation issue. You'll often see shorthands like this in the literature. Since you're using boldface:
    [tex]
    \begin{align}
    \mathbf{a}^2 &= \mathbf{a} \cdot \mathbf{a} = a^2 = |\mathbf{a}|^2
    \intertext{and}
    |\mathbf{a}| &= a.
    \end{align}
    [/tex]

    I can see where there might be problems, if instead we did something like
    [tex]
    \mathbf{a}^2 = \mathbf{aa} = \mathbf{a} \otimes \mathbf{a}
    [/tex]
    and instead ended up with a dyadic (an outer product).

    Notice that the arrows are missing on the RHS of each equation in my previous post.
     
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