The cross product of two vectors produces a third vector that is perpendicular to both original vectors due to its mathematical definition, which is rooted in the properties of quaternions and vector algebra. This perpendicularity is demonstrated through the determinant of a matrix formed by the vectors, where the determinant becomes zero when the two rows are identical, confirming orthogonality. Additionally, the magnitude of the cross product is defined as the product of the magnitudes of the vectors and the sine of the angle between them, reinforcing its directional properties. The right-hand rule is used to determine the direction of the resulting vector, addressing any ambiguity. Understanding the cross product is essential in physics and engineering as it describes effects, such as moments, that are not represented by the original vectors alone.
#1
mess1n
24
0
Yo guys, I was wondering:
Why does the cross product of two vectors produce a third vector which is perpendicular to both the original vectors?
I've come across it in math and it's one of those things I've accepted... but I'm not quite sure why it happens.
As you may know, the definition of u x v, is the determinant of the following matrix:
\bmatrix<br />
\hat{x} & \hat{y} & \hat{z} \\<br />
u_{1} & u_{2} & u_{3} \\<br />
v_{1} & v_{2} & v_{3}\endbmatrix
That is *not* the definition of the cross product. That is just an incredibly bad (but very widely used) abuse of notation. Defining something in terms of abuse of notation is just a bad idea.
The cross product was originally defined to reflect the quaternion product, in which i2=j2=k2=ijk=-1. This leads to the anti-commutative expressions ij=k, ji=-k, jk=i, kj=-i, ki=j, ik=-j. Gibbs and Heaviside thought quaternions were overly cumbersome and a bit artificial (we live in a three dimensional world). The two independently developed vector analysis, stealing useful chunks of Hamilton's quaternions along the way.
Another way to look at the cross product is that the cross product of two vectors u and v is defined as having magnitude equal to uvsinθ, where u and v are the magnitudes of the two vectors and θ is the angle between the two vectors. The cross product vector points in a direction normal to both u and vby definition. Which way it points is a bit ambiguous; there are two choices. The right-hand rule removes this ambiguity.
I like the definition in terms of the Levi-Civita symbol. In every expression below, there's a sum over all indices that appear exactly twice, but I will not write any summation sigmas for those. (Einstein's summation convention).
(x\times y)_i=\varepsilon_{ijk}x_jy_k
This definition makes it extremely easy to prove the orthogonality:
x_i(x\times y)_i=\varepsilon_{ijk}x_ix_jy_k=0[/itex]<br />
<br />
The last step is based on this extremely useful result (excercise): If S_{ij}=S_{ji} and A_{ij}=-A_{ji}, then S_{ij}A_{ij}=0.
Why does the cross product of two vectors produce a third vector which is perpendicular to both the original vectors?
I've come across it in math and it's one of those things I've accepted... but I'm not quite sure why it happens.
It is not a question of why. It is just a matter of definition.
#7
elibj123
237
2
D H said:
That is *not* the definition of the cross product. That is just an incredibly bad (but very widely used) abuse of notation. Defining something in terms of abuse of notation is just a bad idea.
The cross product was originally defined to reflect the quaternion product, in which i2=j2=k2=ijk=-1. This leads to the anti-commutative expressions ij=k, ji=-k, jk=i, kj=-i, ki=j, ik=-j. Gibbs and Heaviside thought quaternions were overly cumbersome and a bit artificial (we live in a three dimensional world). The two independently developed vector analysis, stealing useful chunks of Hamilton's quaternions along the way.
Another way to look at the cross product is that the cross product of two vectors u and v is defined as having magnitude equal to uvsinθ, where u and v are the magnitudes of the two vectors and θ is the angle between the two vectors. The cross product vector points in a direction normal to both u and vby definition. Which way it points is a bit ambiguous; there are two choices. The right-hand rule removes this ambiguity.
I think the notation will forgive me for being so abusive, I certainly didn't want to harm its feelings.
It is not a question of why. It is just a matter of definition.
That is one definition. It is never safe to assume that everyone learns the same definition when there are several equivalent definitions of the same thing. Another equivalent definition is <a, b, c> X <x, y, z>= <bz- cy, cx- az, ay- bx>, which is the same as elibj123's without the "abusive" notation.
From that, you can calculate that
<a, b, c>.<bz- cy, cx- az, ay- bx>= abz- acy+ bcx- abz+ acy- bcx= (abz-abz)+ (acy-acy)+ (bcx- bcx)= 0
I don't think the determinant notation is abusive. It's just necessary to clearly state that we're extending the definition of "determinant" to apply to matrices that have vectors on one of the rows (or one of the columns). We don't even have to change the formula that defines the determinant
\det A=\sum_P s(P) A_{1,P1}\cdots A_{n,Pn}
The sum is over all permutations P, and s(P) is the sign of the permutation P (i.e. s(P)=1 if P is even and s(P)=-1 if P is odd). The reason why things work out so nicely is that the definition never multiplies two matrix elements from the same row or two from the same column.
I highly recommend learning to use the Levi-Civita symbol though. It makes a lot of proofs involving cross products extremely easy.
#11
Studiot
5,440
10
Code:
I think you should be worrying about a law suit!
Wow, I'm new here and already I've discovered a maths professor with a sense of humour.
I might even get to like this place.
Mess1n, this is a good question you have posted in the mathematics section of a physics forum.
The answers given have been mathematical, rather than physical so I will try to provide a physical explanation or motivation.
In mathematics there are many vector algebras, not all of which possesses a cross product.
The vector algebra used by physicists and engineers use a particular vector algebra, because it is useful to have such an operation.
It is useful to have one because it describes effects, observable in the physical world, which are in some way additional to the original physical effect or quantity under consideration.
For example
Consider a physical vector quantity, force, and the statement
"A force can have no effect at right angles to its line of action"
This is true insofar as the horizontal component can exert no effect (force) in the vertical direction.
But what about moment?
This is another effect that a force can exert, but which is certainly not a force.
By saying that moment is not a force we are saying that it cannot be found in the list of all possible force vectors that can be generated by linear combination of the horizontal and vertical components. Mathematically, moment is not in the vector space of forces.
However the moment effect does exist and we can represent it by a 'vector' generated by a
'cross product' of distance and the generating force.
It should be noted that this 'vector' is fundamentally different from the generating force with its own properties and effects.
So the motivation is to chose a mathematics (vector algebra) which correctly represents the full picture.
