# Vector product in 2 dimensions [closed]

If I have a vector A=4i+3j and B=5i-2j, how can I find the vector product AxB? I know that given the angle, its C=AB sin theta, but how can I solve this without the angle?

-

## closed as off topic by pho, Mark Eichenlaub, Colin K, David Z♦Feb 5 '11 at 22:00

Questions on Physics Stack Exchange are expected to relate to physics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here. If this question can be reworded to fit the rules in the help center, please edit the question.

Seems more like a question for math.stackexchange.com –  MBN Feb 5 '11 at 16:26
This is not a physics question. –  pho Feb 5 '11 at 18:31
for simple questions like these, it's easiest to use wikipedia. en.wikipedia.org/wiki/Cross_product –  Mark Eichenlaub Feb 5 '11 at 19:47

Luboš's answer is correct (and efficient), but, for someone new to vector algebra, perhaps confusing. The easy way to explain this is to do the cross product for each component of the vectors, rather than the whole vector at once.

So, for example, using your definition for the cross product, take only the x-component of the vectors and plug them in. You'll get (4)(5)sin(0)=0. A little thought would tell you that you can skip this for vector components with the same unit vector, since sine of the angle between them will always give zero.

Taking one more example, grab the x-component of A and the y-component of B. You get (4)(-2)sin(90)=-8. Remember when doing this that you must use the counterclockwise angle between the vectors, starting from the first vector, or you won't get the minus sign you should for the jXi component (unless you use the right-hand rule to determine whether it's in the positive or negative k direction, in which case you can use the smaller angle between the vectors).

The second way you could approach this is to convert your A and B vectors into magnitude and angle expressions. |A| = sqrt(4^2 + 3^2)=5, and angle is given by the inverse tangent of 3/4. This, by the way, is a terrible way to do this sort of problem, as it creates a lot of unnecessary work.

Having said that, the matrix method that Luboš explains is by far the best method, as it's efficient and is the easiest to accomplish, especially with 3D vectors. If you don't know how that method works, you can grab the following Powerpoint presentation for a good explanation: http://www.whrhs.org/21331011516827430/lib/21331011516827430/Cross_Product.ppt. If you don't have a program that can open that file, feel free to google "matrix method for cross product," although many explanations, while correct, are less easy to follow than the one linked.

-
Dear @Mitchell, let me admit that I have no idea what you're talking about in the whole comment. You may multiply random components and fail to add them or subtract them but the result won't be the vector product, will it? Because Tim wrote pretty explicitly what the number should be, it's clear that none of your alternative formulae is correct. Moreover, concerning the last paragraph, I haven't used any matrices at all, especially because I have already run into problems with misunderstandings while using matrices in one of the previous answers (about coupled oscillators). ;-) –  Luboš Motl Feb 5 '11 at 16:46
@Luboš If it doesn't make sense to you, just ignore it. I'm reasonably certain that the methods outlined do work, but if the OP is able to successfully follow any of the answers to this now-closed question, I'd just as soon he use the more elegant methods you and @sb1 outlined. –  Mitchell Feb 6 '11 at 12:18

Vector Product of 2 vectors $ai + bj + ck$ and $Ai + Bj + Ck$ is given by the determinant $\left| \begin{array}{ccc} \ i & j & k \\ a & b & c \\ A & B & C \end{array} \right|$

In your example it is $\left| \begin{array}{ccc} \ i & j & k \\ 4 & 3 & 0 \\ 5 & -2 & 0 \end{array} \right|$ $=$ $-23k$

-

By the vector product of 2-dimensional vectors $(a_1,a_2)$ and $(b_1,b_2)$, you probably mean the third component of the vector product of $(a_1,a_2,0)$ and $(b_1,b_2,0)$ which is clearly $a_1 b_2 - a_2 b_1$ (the first two components of the product vanish), as indicated by your comment that the magnitude of the vector product is $\sin(\theta)$ just like for 3-dimensional vectors.

So the vector product of $4i+3j$ and $5i-2j$ in your notation is $4\times (-2) -3\times 5=-23$. If you expressed the vectors as complex numbers $z_1=4+3i$ and $z_2=5-2i$, you could get the vector product as $\mbox{Re}(i z_1 z_2^*)$.

-