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Why sine is used for cross product and cosine for dot product?

  1. Jun 8, 2012 #1
    While we calculate cross product of two vectors let A and B
    we write ABsinθ.

    And while we calculate dot product of them we write ABcosθ.

    Why particularly we use sinθ for cross product and cosθ for dot product.Is there any physical reason why we choose sine for cross product and cosine for dot product or is it convention?


    *(how can I write vector variables?i don't know.that's why i wrote vector variables without vector (→) sign.)
     
    Last edited: Jun 8, 2012
  2. jcsd
  3. Jun 8, 2012 #2

    HallsofIvy

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    The primary purpose of "cross product" is to calculate areas. If you have two vectors, [itex]\vec{u}[/itex] and [itex]\vec{v}[/itex], the area of the parallelogram having those two vectors as two sides is, of course, "base times height". The "base" is the length of one of the vector, [itex]|\vec{u}|[/itex], say. To Find the "height", draw a line from the tip of [itex]\vec{v}[/itex] perpendicular to [itex]\vec{u}[/itex]. That gives a right triangle having [itex]|\vec{v}|[/itex] as hypotenuse. The "height" is the length of the "opposite side" which is given by [itex]|\vec{v}|sin(\theta)[/itex] where [itex]\theta[/itex] is the angle between the vectors. That is, the area of the parallelogram formed by vectors [itex]\vec{u}[/itex] and [itex]\vec{v}[/itex] is [itex]|\vec{u}||\vec{v}|sin(\theta)[/itex].

    A primary use of the dot product, on the other hand, is to find the projection of one vector on the other. Again, draw a line from the tip of [itex]\vec{u}[/itex] perpedicular to [itex]\vec{v}[/itex]. The projection of [itex]\vec{u}[/itex] onto [itex]\vec{v}[/itex] is now the near side of the right triangle produced. Its length is [itex]|u|cos(\theta)[/itex]. To find the actual vector projection, multiply that length by a unit vector in the direction of [itex]\vec{v}[/itex] which is [itex]\vec{v}/|\vec{v}[/itex]. That is, the vector projection of [itex]\vec{u}[/itex] on [itex]\vec{v}[/itex] is given by [itex]|\vec{u}|cos(\theta)\left[\vec{v}/|\vec{v}|\right][/itex] which we can simplify by multiplying both numerator and denominator by [itex]|\vec{v}|[/itex]:
    [tex]\frac{|\vec{u}||\vec{u}|cos(\theta)}{|\vec{v}|^2} \vec{v}[/tex]
    The numerator of that fraction is [itex]\vec{u}\cdot\vec{v}[/itex].

    I have done the vectors and other math symbols by using "LaTeX" which is implemented on this board To see the code used, click on the formula or click on "quote". There are several tutorials on the internet.
     
  4. Jun 8, 2012 #3
    I think the cos in the dot product and the sin in the cross product are used because they give simple formulas. If you wanted to calculate a dot product that used sin instead, you wouldn't get a nice and simple formula for calculating it like x1*x2+y1*y2+z1*z2, as it is when you use cos. With the cross product, you get something much nastier if you want the length of the vector be related to cos instead of sin. With sin you get a nice and simple formula. Then there are various uses figured out for them, such as the cross product in various physical laws etc.
     
  5. Jun 8, 2012 #4
    This just has to do with rotations of vectors. Consider a 2D vector space spanned by [itex]e_1, e_2[/itex]. If a vector [itex]a[/itex] forms an angle [itex]\theta[/itex] with [itex]e_1[/itex], then its components are

    [itex]a = e_1 \cos \theta + e_2 \sin \theta[/itex]

    The geometry of Euclidean space dictates that the vector will have this form--rotations naturally involve cosines and sines. The cosine expresses that the less this vector [itex]a[/itex] is rotated, the more it has in common with [itex]e_1[/itex], for instance.
     
  6. Jun 8, 2012 #5

    mathman

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    The basic reason is that θ is the angle between the vectors involved, so the formulas for dot and cross product are the correct expressions in terms of this angle. There is no alternative.
     
  7. Jun 8, 2012 #6

    mathwonk

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    as others have said, one involves projecting on the line parallel to a given one (hence cosine), and the other involves projecting on the line perpendicular to a given one (sine).
     
  8. Jun 13, 2012 #7
    For example if you have to vectors [itex]\vec{A}=(f,g,h)[/itex] and [itex]\vec{B}=(u,v,w)[/itex]. Their lengths are then [itex]A= \sqrt{f^2+g^2+h^2}[/itex] and [itex]B= \sqrt{u^2+v^2+w^2}[/itex]. The length of the vector connecting their endpoints is [itex]C=\sqrt{(f-u)^2+(g-v)^2+(h-w)^2}=\sqrt{f^2-2fu+u^2+g^2-2gv+v^2+h^2-2hw+w^2}[/itex].
    From the cosine theorem we have [itex]C^2=A^2+B^2-2ABcos(\alpha)[/itex], where [itex]\alpha[/itex] is the angle between the two vectors.
    From it we get that [itex]ABcos(\alpha)=\frac{A^2+B^2-C^2}{2}=\frac{f^2+g^2+h^2+u^2+v^2+w^2-f^2+2fu-u^2-g^2+2gv-v^2-h^2+2hw-w^2}{2}=\frac{2fu+2gv+2hw}{2}=fu+gv+hw[/itex].
    A lot of that stuff cancelled and now we got a nice formula for calculating [itex]ABcos(\alpha)[/itex] just from the components of the vectors.
    However, lets say we want [itex]ABsin(\alpha)[/itex] in terms of the components of the vectors.
    Then [itex]AB\sqrt{1-sin^2(\alpha)}=fu+gv+hw[/itex]
    [itex]A^2B^2(1-sin^2(\alpha))=(fu+gv+hw)^2[/itex]
    [itex]A^2B^2sin^2(\alpha)=A^2B^2-(fu+gv+hw)^2[/itex]
    [itex]ABsin(\alpha)=\sqrt{A^2B^2-(fu+gv+hw)^2}=\sqrt{(f^2+g^2+h^2)(u^2+v^2+w^2)-(fu+gv+hw)^2}=[/itex]
    [itex]=\sqrt{f^2v^2+f^2w^2+g^2u^2+g^2w^2+h^2u^2+h^2v^2-2fugv-2fuhw-2gvhw}[/itex]
    I can't think of a way of simplifying that any further. Now that isn't a very nice formula, and probably not useful at all to memorize this, easier to start with the cos one and work from there if you ever need to derive ABsin(a) from the vector components.
    Sorry couldn't get the right notation for the vector lengths.
     
  9. Jun 16, 2012 #8
    Let me offer you a basic physical justification.

    First dot product:
    If the vectors are pointing in the same direction the dot product gives you the full product of their magnitudes |A||B| since cos(0) =1. If the vectors are not pointing in the same direction the dot product gives you less. When they are pointing at 90 degree angles to each other it gives you zero. In other words the dot product is the actual product but reduced as the angle between the vectors increases. Imagine pulling a wagon by its handle. You are trying to pull it horizontally, but the handle is at some angle. The steeper the angle the less of your pulling effort works horizontally. If the angle is at 90 degrees (straight up) you will not be able to pull the wagon horizontally no matter how hard you pull. You will be able to pull it the easiest if the handle is horizontal (angle 0). This is an example of use of dot product.

    Cross product:
    Here if the vectors are at 90 degree angles to each other you get the full product of their magnitudes |A||B|. As the angle *decreases* you get less, and at 0 degrees you get 0 (just opposite of dot product). Imagine pulling on a big wrench in order to twist off a tight bolt. Your best bet is to pull in a direction that is at a right angle to the wrench. If you pull at a smaller angle, less of your effort will be torquing the bolt. If you pull at a 0 degree angle (along the length of the wrench) none of your effort is working to untwist bolt. This is an application where cross product is used.
     
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