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Vector multiplications

  1. Jun 24, 2011 #1
    Vector multiplications....

    It's pretty wierd....but i don't understand that why the dot product of two vector quantities is a scalar quantity and the cross product is a vector quantity....what do these multiplications actually mean..........
     
  2. jcsd
  3. Jun 24, 2011 #2
    Re: Vector multiplications....

    Good question! Let me share some thoughts. These will be a bit hand-wavy, but I hope good food for thought.

    Say we wanted to invent a product -- call it "[itex]\star[/itex]" -- of two vectors [itex]a[/itex] and [itex]b[/itex]. We'll say
    [tex]a \star b = C[/tex]
    where we don't yet know what kind of thing [itex]C[/itex] is -- a scalar, a vector, or something else entirely.

    How would we want the product to behave? We know vectors have a magnitude and a direction... how does each of these figure into the picture? It seems to me we'd want a product with the following ingredients:
    • proportional to the product of the magnitudes. In other words, if we scaled up the magnitudes with scalars [itex]\alpha[/itex] and [itex]\beta[/itex], we'd want the product to scale up by the same amount: [itex](\alpha a) \star (\beta b) = (\alpha \beta) (a \star b)[/itex].
    • also proportional to some function of the angle [itex]\theta[/itex] between them. (We want to account for their relative directions.)
    • Since our product might not be a scalar, we might need to keep track of its direction. Let's call that [itex]\hat{D}[/itex]; it's a function of the directions of [itex]a[/itex] and [itex]b[/itex], but not their magnitudes.

    So whatever our product is, it should look something like this:
    [tex]a \star b = |a| |b| f(\theta) \hat{D}[/tex]
    A few choices in particular turn out to be especially useful.

    First, we often want to measure "sameness of direction" for two vectors. A natural choice here is
    [tex]f(\theta) = \cos(\theta)[/tex]
    After all, this function is maximum when [itex]a[/itex] and [itex]b[/itex] point in the same direction ([itex]\theta = 0[/itex]), minimum when they're in the opposite direction ([itex]\theta = \pi[/itex]), and takes on intermediate values for intermediate angles. The resulting product is called the "inner product" or "dot product", '[itex]\cdot[/itex]'. So we have:
    [tex]a \cdot b = |a| |b| \cos(\theta) \hat{D}_{(\cdot)} [/tex]

    What about the direction, [itex]\hat{D}_{(\cdot)}[/itex]? Think of it this way: suppose we grabbed the pair of vectors, rotated them, and turned them around any way we liked -- just so long as we're careful not to stretch their lengths, or change the angle between them! Would that affect their "sameness of direction"? Not at all. So, this "dot product" doesn't have a direction -- it's invariant under rotations; it must be a scalar. We have [itex]\hat{D}_{(\cdot)} = 1[/itex], and we can write the usual dot product expression,
    [tex]a \cdot b = |a| |b| \cos(\theta) [/tex]

    The dot product is extremely useful. It lets you make projections onto other vectors, test for orthogonality -- in fact, it's even used to define what we mean by distances and angles!

    There is one more very useful product, which I hope to get some time to post later on tonight. What I'm really trying to emphasize here is the geometric meaning of these products: we use them because they give us useful information about the relationship of two vectors.
     
  4. Jun 24, 2011 #3
    Re: Vector multiplications....

    The other kind of product we often want to make is an area based on the two vectors. Remember, the area of a rectangle is the product of its length and width. If we think of the length and the width as orthogonal vectors, we have a good example of the kind of "area product" I have in mind.

    To see what happens for arbitrary angles, consider the following picture.
    a_wedge_b.png
    If we put a copy of [itex]a[/itex] at the head of [itex]b[/itex], and vice versa, we make a parallelogram (shown in green). The area of this parallelogram is a useful kind of product to have between [itex]a[/itex] and [itex]b[/itex].

    The angular part of our product should be zero when the vectors make an angle of [itex]0[/itex] or [itex]\pi[/itex], and maximum for an angle of [itex]\pi/2[/itex]. That sounds a lot like the sine function! If we make this choice, [itex]f(\theta) = \sin(\theta)[/itex], we have the "outer product" or "wedge product",
    [tex]a \wedge b = |a| |b| \sin(\theta) \hat{D}_{(\wedge)}[/tex]

    Now all that's left is the direction. Once again, suppose we grab [itex]a[/itex] and [itex]b[/itex], and spin them around any way we please -- as long as we don't change their lengths or the angle between them. The area will always be in the plane formed by the two vectors, and that plane does change as we rotate them. So, [itex]\hat{D}_{(\wedge)}[/itex] does contain some kind of information about directionality:
    • It specifies a plane -- the plane that contains both [itex]a[/itex] and [itex]b[/itex].
    • It specifies an orientation -- the direction that turns [itex]a[/itex] into [itex]b[/itex].
    (Note that this last property implies that [itex](a \wedge b) = -(b \wedge a)[/itex] -- order matters!)

    Many people are unfamiliar with the idea of an area being a kind of "direction", and don't have a good intuition for how to, e.g., add two areas that are in different planes. In 2D, there's no problem, because all the areas are coplanar (there's only one plane in 2D, after all!). In 3D there's a little trick. You can actually replace the plane with a vector that's perpendicular to it. This vector's magnitude is equal to the parallelogram's area, and its direction is chosen (by convention) as the direction that a right-handed screw would advance when turned from [itex]a[/itex] to [itex]b[/itex].

    This is, of course, nothing but the good old cross product [itex]a \times b[/itex]!

    Like I said: my two posts so far have been kind of handwavy. But I hope you've found them helpful. Let me know if I've made anything unclear, or you have more questions.

    Cheers,
    Chip
     
    Last edited: Jun 24, 2011
  5. Jun 27, 2011 #4
    Re: Vector multiplications....

    lol @ chogg... That was one of the most thorough explanations I've seen! I didn't read them all the way through but they look good from what I skimmed. It was humorous (in a good way) seeing all of that.

    Another comment for the OP: if you just accept the definitions of the dot and cross product, they will reveal their "use" and meaning to you through examples and practice. Of course, understanding the theory provided by chogg won't hurt either!
     
  6. Jul 7, 2011 #5
    Re: Vector multiplications....

    That was an excellent explanation, Chogg.
     
  7. Jul 8, 2011 #6
    Re: Vector multiplications....

    Dot product and cross product are simply definitions of functions that depend on two vector variables. Definitions need no proof, however their potential usefullness should be justified.
    Here are two simple applications of dot product and cross:

    1. Calculation of work done by a force. If force and displacement are parallel, we multiply scalar force and scalar displacement to get work. When they are not, only the component of force parralel to the movement contributes to work. It is easy to prove that work is then the dot product of vector of force and vector of displacement.

    2. Thanks to the cross product, we can define "left" and "right" mathematically. "left" is a cross product of "up" and "forward" and "right" is the opposite of that. (of course only directions are relevant in this case, so we should assume directions are unit vectors)
     
  8. Jul 8, 2011 #7
    Re: Vector multiplications....

    Actually, the cross product requires a pre-existing definition of left and right. Your choice for the handedness of space gets smuggled in to the definition of the cross product. For some choices of basis, your definition of left will give you the answer you want. For other, equally good choices, it will give the opposite answer.

    But your main point -- that we should justify that our definitions are useful -- is right on target. Example uses for the cross product include torque, angular momentum, magnetism, etc.: a truly vast array of physical phenomena! (Although the wedge product I mentioned earlier is generally a more natural and powerful alternative, and I see no reason to use the cross product in any situation.)
     
  9. Aug 5, 2011 #8
    Re: Vector multiplications....

    really nice post chogg thanks for that
     
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