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Help Understanding Dot Product

  1. Jul 9, 2011 #1
    I'm learning about dot products, and I'm having a bit of trouble grasping why axbx + ayby + azbz = ab*cosΘ. I understand how it works in two dimensions, I think, but three is still fuzzy.

    This is what I came up with for two dimensions. The angle between the vectors is simply the difference between the angle between each vector and the x-axis.

    = ab*cos(Θab)
    = ab*(cosΘacosΘb + sinΘasinΘb)
    = a*cosΘab*cosΘb + a*sinΘab*sinΘb
    = axbx + ayby

    Now I understand intuitively that the dot product rule should work in three dimensions, since you could always orient the axes so that the two vectors lie on the x-y plane, in which case the above applies. But I'd like to see an analytical proof like the one above, except including azbz. I would feel a lot better about this if I could see that.
  2. jcsd
  3. Jul 10, 2011 #2
    If you rotate the vectors a and b in a way such that a is aligned along the z axis, the expression for the scalar product reduces to [itex]a \cdot b=a b_z =a b cos \theta[/itex]. The last passage follows the change of coordinate system (from cartesian to spherical).
  4. Jul 10, 2011 #3


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    Hey there Opus_723.

    Dot products (as well as general inner products of which a dot product is one example of an inner product) provide a description of geometry of the respective vector space.

    The best way to think about this is to think first about the notion of length. This is captured by what is called a norm.

    One norm you should be familiar with is the pythagorean theorem of a triangle. c^2 = a^2 + b^2, where a and b are the sides of a triangle and c is the length of a hypotenuse. The sides a and b are perpendicular to each other and the length of c is given by SQRT(a^2 + b^2).

    Now it turns out that in an inner product space (provided that it meets the axioms) can be written in terms of a norm. We can form the inner product using the definition of the norm, and since the norm is known with euclidean geometry (the pythagorean theorem but in any dimension), we can use this to get an expression of the inner product for euclidean spaces (think right angle geometry that you are used to).

    In geometries that are not "flat" (i.e. curved geometries), we have to use more advanced mathematics, but the idea is the same. If the conditions for an inner product are satisfied, you can find ways of calculating the inner product in that geometry.

    Hermann Grassman wrote a theory about geometric calculus using inner and outer products and if you read modern accounts of geometric calculus, you'll find that these come in at a very abstract level. For linear systems, these are more straightforward, but for general systems (i.e. curved space), you are dealing with differentials and it can get hairy.

    Also with regard to the |a||b|cos(theta) the best way to understand this is to look at the Cauchy-Schwarz inequality. This inequality says that the inner product (or in your case dot product) |<a,b>| = |a . b| <= ||a|| x ||b|| where |x| is the absolute value of x, and ||a|| is the length of vector a. This equality establishes that the cos(theta) term relates to an "angle" and is why we associate geometry with inner products.
  5. Jul 10, 2011 #4
    I use the cosine law for the intuition. Let's say you have a vector a and b. Their difference is a-b. These 3 vectors can form a triangle. Let [itex]/theta[/itex] be the angle between a and b. By the cosine law, we have
    Since the norm of the vector squared is equal to the vector dot itself, we have
    [tex](\mathbf{a}-\mathbf{b})\cdot(\mathbf{a}-\mathbf{b})=\mathbf{a}\cdot \mathbf{a}+\mathbf{b}\cdot \mathbf{b} -2|\mathbf{a}||\mathbf{b}|\cos\theta.[/tex]
    Since the dot product is distributive, we have
    [tex]\mathbf{a}\cdot \mathbf{a}-2\mathbf{a}\cdot\mathbf{b}+\mathbf{b}\cdot \mathbf{b}=\mathbf{a}\cdot \mathbf{a}+\mathbf{b}\cdot\mathbf{b} -2|\mathbf{a}||\mathbf{b}|\cos\theta.[/tex]
    Simplifying yields
    [tex]-2\mathbf{a}\cdot \mathbf{b}=-2|\mathbf{a}||\mathbf{b}|\cos\theta.[/tex]
    Further simplifying yields
    [tex]\mathbf{a}\cdot \mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos\theta.[/tex]
  6. Jul 10, 2011 #5
    dalcde, Thanks, that makes the most sense so far. That satisfies me enough to start using it. As a side note, does anyone know where I could find some history as to how to dot and cross products were originally developed?
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