# Vector Cross Product Question

1. Dec 23, 2012

### fehilz

1. The problem statement, all variables and given/known data

From John Taylor's Classical Mechanics:

Show that definition (1.9) of the cross product is equivalent to the elementary deinition that R x S is perpendicular to both R and S, with magnitude rssinθ and direction given by the right hand rule. [Hint: It is a fact (though quite hard to prove) that the definition (1.9) is independent of your choice of axes. Therefore you can choose axes so that R points along the x axis and S lies in the xy plane.

Definition 1.9 refers to the determinant of the 3x3 matrix with the three unit vectors in the top row, vector R components in the second and vector S components in the third.

2. Relevant equations

(Provided in question)

3. The attempt at a solution

I have been able to find the solution when choice of axis is independent by squaring the magnitude of the cross product and using the dot product definition involving the cosine function to eventually arrive at the magnitude equation provided, which I verified with a Khan Academy video that used the same method.

However, I've been trying to do the question in the manner described the question (apologies in advance for the lack of LaTex, still need to learn to use it):

I defined a vector R with only a non-zero x component and a vector S with only x and non-zero y components. When the cross product of R and S is performed, only a z component is left in the new that I called vector T equal to Rx times Sy. This solves the direction/right hand rule part of the question.

To find the magnitude, Pythagoras' theorem is used. For T, the magnitude is equal to Rx times Sy since that's the only component. For R, the magnitude is equal to Rx since that's the only component. This is where I'm running into a problem, how can I make the magnitude of S times sine equal to Sy?

Sorry if that was unclear and thanks in advance for any help.

Last edited: Dec 23, 2012
2. Dec 23, 2012

### Dick

Use trig. The vector S makes a right triangle with sides Sx and Sy and hypotenuse the length of S. What is Sy in terms of the length of S and the angle θ?

Last edited: Dec 23, 2012
3. Dec 23, 2012

### fehilz

I did try this and thought I had finished the question. sinθ is Sy/S which I can then sub back into the equation. This yields Rx×s×Sy/S. I previously thought that s and S would cancel leaving the expected RxSy but that doesn't work.

EDIT: Just re-read your comment, so the hypotenuse of the right angle triangle formed by a vector is the magnitude of that vector? That makes more sense, can't believe I forgot that from first year. Thanks a heap!!

Last edited: Dec 23, 2012
4. Dec 23, 2012

### Dick

You're welcome. I edited the comment when I realized I wasn't being clear about the difference between the vector S and the length of S.