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# Advanced Math/Vectors, cross product and span

Question
QUESTION: I am having difficulty understand what to do for this question. Do i just need to take the cross product right away between vector u and vector x and then reaarange the xs, ys and zs in the equation into a form that matches that of w? Thanks for any help. It is greatly appreciated.

Given vector u = <1,1,-1> and vector w = <-3,-4,-7>, find all vectors vector x = <x,y,z> such that (vector u)cross product(vector x)= vector w.

ANSWER: Pretty much. You know that taking the cross-product gives a new vector (it happens to be perpendicular to the 2 vectors but that doesn't seem to be important in this problem). So lets say you get a vector x = (x1, x2, x3). Then

u x x = w

will consist of 3 eqns in 3 unknowns and you can solve for x1, x2, x3 by successive elimination. An interesting aspect of this eqn is that if the cross product is represented by a matrix multiplication

Ux =w

then the matrix U is skew-symmetric and has a determinant of zero (for 3x3 matrix). This doesn't mean it there is no solution but ut does mean you can't solve it using Cramer's rule (which involves dividing by the determinant) but it may (does?) mean that one of the components of v will be a free variable (not determined). That may be why the question is phrased find "all" vectors.

---------- FOLLOW-UP ----------

QUESTION: I have obtained the cross product to be <x_3+x_2, -x_1-x_3, x_2-x_1>
I have out this equal to w:
<x_3+x_2, -x_1-x_3, x_2-x_1> = <-3,-4,-7>

But now I seem to be having a bit of difficulty solving the system of equations. I am having difficulty isolating it for a specific variable that would then allow me to find the values of the other variables. Also, when it says to write the answer in this form: (a,b,c)+<d,e,f>t , where do I obtain the (a,b,c) from? I know that the values found for the vector x will be the <d,e,f> part.
Thanks for any help.

This is a trickier problem than it first seems. The vector v in the eqn u x v = w is not unique (i'm changing the name of the vector x, used before, to v for clarity). As I alluded to before, there are a family of vectors that can satisfy this eqn. For instance, we can also write u x v = u x (v +au) = (u x v) + a(u x u) = w, where a is a constant, because u x u = 0, i.e., parallel lines have a zero cross-product as can be seen by the definition u x u = |u||u|sin(0°) = 0. Before I go too far, you need to check that u and w are perpendicular, i.e., u･w = 0, otherwise there is no solution!

This boils down to needing to choose an angle at which the 2 vectors meet. As an aside, the cross-product eqn represents the relationship of a force times a moment arm to give a torque, FxR = T, where R is the point and direction along F that the force F is applied. The simplest configuration is when the force is applied at a right angle, for which sin(angle) = 1 and |T| = |F||R|.

For the present case, this means that u and v should be perpendicular. It is tempting to just use the dot product between u and v, set it = 0 and see what happens. There is another way, though. SInce the cross-product makes w perpendicular to u and v, it follows that w x u will be parallel to v. This gives a direction vector for v,

w x u = -11i + 10j -1k

according to my calculations, where i, j and k are the usual unit vectors along x, y ,z  (this should be perpendicular to u and w of course).

So now we need a point for v to go through. Letting this point be given by the intersection of u and w (from which v emanates as well),

u(a,b,c) = w(a,b,c) -> (a,b,c) + t(1,1,-1) = (a,b,c) + s(-3,-4,-7), where t and s are parameters. The only way I can reconcile this is using the symmetric eqns is to set t = s and get a = b= c = 0.

I'm curious what study/course materials you are using and whether it covers the properties of cross-products. Can you send me something?

Randy

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#### randy patton

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college mathematics, applied math, advanced calculus, complex analysis, linear and abstract algebra, probability theory, signal processing, undergraduate physics, physical oceanography

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