cross product question

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trickslapper

Junior Member
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Sep 17, 2010
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Ok, i'm really confused on how a cross product works with 2d vectors. I've read that a cross product is not defined for two dimensions and that makes sense to me. But i was given a question on my homework that gave me two 2d vectors. The questions were what is the magnitude of the vector and what is the direction of the vector?

This is for a robotics class that has a lot of students that have not even taken precal II so i really don't think the professor is pulling something tricky on us. Is there a way for me to get the magnitude (and i don't have an angle for the vectors either so i can't use abs(v) * abs(u) * sin(theta) = magnitude)

any ideas ?
 


trickslapper said:
i was given a question on my homework that gave me two 2d vectors.

The questions [are] what is the magnitude of the vector and what is the direction of the vector?

Perhaps they meant to write "each vector" instead of "the vector".

If they actually want the magnitude and direction for a single vector, then perhaps they're talking about the vector which results from adding the two given vectors.
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What lead you to consider that this exercise requires a cross product?

If we have the components of a 2d vector, then that's sufficient to find its magnitude and direction.

 
The question is "Find the magnitude and direction of u x v". I think i have to use the fact that if i have unit vectors and i multiply their magnitudes i will get cos(theta) and solve for theta.
 
The cross product is certainly defined on R2, just as it is defined on every other plane. Given any two linearly independent vectors, there is a single plane which is spanned by them, so R2 is simply a special case spanned by the two standard coordinate vectors (1,0) and (0,1). Just "pretend" vectors (a,b) and (c,d) are actually (a,b,0), and (c,d,0).

For that reason, given a set of linearly independent vectors {u,v} you can look at the cross product as a function from P^2->R^3 instead of R^3 -> R^3, where P=<u,v> is the plane spanned by your two vectors. I know Wikipedia says it is defined on vectors in 3-space, and it is probably good to think of it that way for your application. I'm just making the point that there is no difference really what plane the vectors reside in as they are isomorphic to each other.

The direction of the vector is orthogonal to the vectors that are being multiplied and hence cannot lie in the plane they generate. The direction of the vector is given by its components which you can normalize if desired. Make sure you take the product in the right order if the application demands to know what direction some thing may move (as you might know, there are two orthogonal vectors, one pointing in the opposite direction of the other). The magnitude is its length, you square each component, add them, then take the square root.
 
Ok, that makes sense... but if i am working in two dimensions and the cross product has to be orthogonal to the other two vectors where does the cross product exist..you know like in an xy graph ?
 
The direction of the cross product of two linearly independent vectors in R^2 will be either "out of the page" or "into the page", not given by theta or anything. However the Mangnitude of the cross product can be calculated as u x v = |u|*|v|*|sin(theta)| where theta is the angle bewteen them.
 
Ok i was thinking that it would be out of the page or into the page..and since i don't have theta i just solve for the cross product and pretend that the third entry in each vector is zero right ?

thanks i guess i was just overthinking it
 
You can get theta with the dot product. u (dot) v = |u|*|v|* cos(theta)
 


trickslapper said:
i guess i was just overthinking it

Or, you were maybe mislead by a poorly-worded exercise.

<a, b> and <a, b, 0> are not the same thing.
Click to expand...

Is this robotics course an on-line course? If so, from what type of school?

 
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These days robotics are being taught at many "certification" schools - at a technical level (where most of the students are not even required to take pre-calc). So you get this product .....
 
Calculating cross product of two vectors manually is really a big task. Inorder to make it easy we built a -------- which helps to calculate cross product of two vectors in seconds.
 
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