2
$\begingroup$

I recently realised that asking a question and answering our own question is allowed here, so here is a question I've seen commonly on many sites:

"How does one multiply two vectors?"

This is very open-ended (but basic) question, but here's my answer to it (below in the answers section).

$\endgroup$
  • $\begingroup$ There is also the Hadamard Product, see [here][1]. [1]: math.stackexchange.com/questions/20412/… $\endgroup$ – al-Hwarizmi Jun 8 '13 at 8:08
  • 1
    $\begingroup$ I have a professor who likes to call the geometric product, "the real product" of vectors (real as in genuine, not as in the real numbers). $\endgroup$ – user137731 Jul 18 '14 at 18:13
3
$\begingroup$

The dot product and cross product both appear as components of the tensor product of two vectors, $v^\mu w^\nu$, which gives a rank-2 tensor. The dot product is the contraction/trace $v^\mu w_\mu$, which is useful due to its invariance properties, and the cross product appears as the tensor $v^\mu w^\nu - v^\nu w^\mu$ (which in three dimensions allows for an ugly misrepresentation in terms of axes of rotation, which allows this tensor to be written as a vector, even though it doesn't transform as one).

An alternative formulation of this is in the geometric algebra notation, where the cross product is written as the "wedge product", the dot product is still the inner product and the tensor product is called the "geometric product" and is the sum of the two.


ARCHIVED ANSWER (Jun 2013) follows, I no longer endorse the below contents --

  1. Dot product (Scalar Product)

    The dot product, you could say, very hand-wavily measures both the overall size of 2 vectors and how parallel they are.

    The dot product is related to the magnitudes and angles of the two vectors by: $$\vec a\cdot\vec b=||\vec a||\mbox{ }||\vec b||\cos\theta$$

    So, if the two vectors are orthogonal, their dot product is 0. If they are parallel, their dot product is the product of the two magnitudes. The latter case always happens in scalars. So, in, this sense, the dot product actually is a generalisesation of the normal ordinary scalars' product for scalars in $\mathbb C$

    Of course, it is better to use the dot product when measuring orthogonality, only. In fact, often, orthogonality (and not perpendicularity) is defined in terms of the dot product being equal to 0.

    Also, note that for complex vectors, $$\Re(\vec a\cdot\vec b)=||\vec a||\mbox{ }||\vec b||\cos\theta$$

    Generally, the dot product is calculated by: $${\mathbf{a}}\cdot{\mathbf{b}} = \sum {{a_i}\overline {{b_i}} }$$

  2. Cross Product (Vector Product)

    The cross product of two vectors in $\mathbb R^3$ is a vector orthogonal to these two vectors and has a magnitude of $$||\vec a\times\vec b||=||\vec a||\mbox{ }||\vec b||\sin\theta$$

    It can be calculated as: $${\mathbf{a}} \times {\mathbf{b}} = \left| {\begin{array}{ccccccccccccccc} {{{\hat e}_1}}&{{{\hat e}_2}}&{{{\hat e}_3}} \\ \leftarrow &{{{\mathbf{a}}^T}}& \to \\ \leftarrow &{{{\mathbf{b}}^T}}& \to \end{array}} \right|$$

    (not really a determinant -- just a mnemonic, etc. etc.)

    Thus, the magnitude of the cross products describes their "orthogonal-ness" and their overall "size". It is 0 whenever the two vectors are parallel. Oh, and of course, these definitions become relatively very complicated in more than 3 dimensions. In more than 3-dimensions, one has to use: $$\vec a\times\vec b=/(\vec a\wedge\vec b)$$

    Here, $\wedge$ is the exterior product and $/$ is a duality between the cross products and the exterior (wedge $\wedge$) products. I once showed that the following generalisation is possible: $$/\left( {{{{\mathbf{\hat e}}}_m} \wedge {{{\mathbf{\hat e}}}_n}} \right) = {\left( { - 1} \right)^{m + n + 1}}\mathop \bigwedge \limits_{k \ne m,n}^{} {{{\mathbf{\hat e}}}_k}$$

    ...in any dimension...

  3. Exterior Product (Wedge Product)

    The Exterior Product of 2 vectors is the bivector spanned by them.

    Of course, there are many more products, such as the tensor product (The outer product is a special case for vectors and the Kronecker Product is for matrices), the natural product, the Clifford Product etc. Actually, the natural product was defined by me in http://ccsenet.org/journal/index.php/jmr/article/view/18102 in a hope to obtain a geometric interpretation of matrices, though it works only for singular matrices.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.