# Vector cross product anti-commutative property

• greg_rack
In summary, the magnitude of the cross product between two vectors, ##\vec{a}## and ##\vec{b}##, is equal to the product of the magnitudes of the two vectors and the sine of the angle between them. However, the direction of the cross product can differ depending on the order in which the vectors are multiplied, as well as the direction of the angle between them. This is due to the right hand rule for cross products, which dictates the direction of the resulting vector. Therefore, while the magnitudes may be equal, the direction of the cross product can be different.

#### greg_rack

Gold Member
Homework Statement
Demonstrate that:
$$\vec{a}\times\vec{b}=-\vec{b}\times\vec{a}$$
Relevant Equations
none
That may sound really silly, and that may be due to my lack of understanding of the operations itself, but:
if ##|\vec{a}\times\vec{b}|=|\vec{a}|\cdot|\vec{b}|sin\theta##, being ##\theta## the angle between the two vectors, how could ##\vec{b}\times\vec{a}## be different? Wouldn't it be just the same magnitude(of course) and direction as those of the first case?

Are you familiar with the right hand rule for cross products?

kuruman said:
Are you familiar with the right hand rule for cross products?
Yes, I am... the original thought behind the thread is: "oh ok, right-hand rule tells that... but why is it analytically and mathematically true?".
Shifting the problem towards unit vectors, online I could only find that ##\hat{j}\times\hat{i}=-\hat{k}## and so on... but no proof for that.

greg_rack said:
That may sound really silly, and that may be due to my lack of understanding of the operations itself, but:
if ##|\vec{a}\times\vec{b}|=|\vec{a}|\cdot|\vec{b}|sin\theta##, being ##\theta## the angle between the two vectors, how could ##\vec{b}\times\vec{a}## be different? Wouldn't it be just the same magnitude(of course) and direction as those of the first case?

You have that ##\mathbf{u} = \mathbf{a} \times \mathbf{b} = |\mathbf{a}||\mathbf{b}| \sin{(\theta)} \mathbf{n}_1## and ##\mathbf{v} = \mathbf{b} \times \mathbf{a} = |\mathbf{a}||\mathbf{b}| \sin{(\theta)} \mathbf{n}_2##. The unit vectors ##\mathbf{n}_1## and ##\mathbf{n}_2## point in opposite directions, ##\mathbf{n}_1 = -\mathbf{n}_2## as per the right hand rule. Then ##|\mathbf{u}| = |\mathbf{v}| = |\mathbf{a}||\mathbf{b}| \sin{\theta}##, however ##\mathbf{u} = -\mathbf{v}##.

For a right handed orthonormal basis set ##\{ \mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3 \}##, you have that by definition$$\mathbf{e}_i \times \mathbf{e}_j = \sum_{k=1}^3 \varepsilon_{ijk} \mathbf{e}_k$$for any ##i,j \in \{1,2,3 \}##. The ##\varepsilon_{ijk}## is the Levi-Civita symbol, which might look complicated, but it's just a neater way of dealing with the permutations,

Anyway, take your two vectors ##\mathbf{a} = a_1 \mathbf{e}_1 + a_2 \mathbf{e}_2 + a_3 \mathbf{e}_3## and ##\mathbf{b} = b_1 \mathbf{e}_1 + b_2 \mathbf{e}_2 + b_3 \mathbf{e}_3##, then can you work out$$(a_1 \mathbf{e}_1 + a_2 \mathbf{e}_2 + a_3 \mathbf{e}_3) \times (b_1 \mathbf{e}_1 + b_2 \mathbf{e}_2 + b_3 \mathbf{e}_3)$$ and also$$(b_1 \mathbf{e}_1 + b_2 \mathbf{e}_2 + b_3 \mathbf{e}_3) \times (a_1 \mathbf{e}_1 + a_2 \mathbf{e}_2 + a_3 \mathbf{e}_3)$$using that the cross product is distributive over addition? By the way, it's a lot quicker if you use the so-called Einstein summation notation, but I haven't used that here for simplicity. Also, you can think in terms of ##\mathbf{e}_1 \equiv \mathbf{\hat{x}}##, ##\mathbf{e}_2 \equiv \mathbf{\hat{y}}##, etc. if you prefer

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greg_rack
greg_rack said:
... if ##|\vec{a}\times\vec{b}|=|\vec{a}|\cdot|\vec{b}|sin\theta##, being ##\theta## the angle between the two vectors, how could ##\vec{b}\times\vec{a}## be different? Wouldn't it be just the same magnitude(of course) and direction as those of the first case?
Hi. The equation ##|\vec{a}\times\vec{b}|=|\vec{a}|\cdot|\vec{b}|sin\theta## only tells you the product's magnitude. It has no information about the product's direction. A vector's magnitude is always positive. Part of the confusion may be that ##sin\theta## can be positive or negative, so the right -hand side of the equation can be positive or negative. I suppose we should really write the equation as ##|\vec{a}\times\vec{b}|=||\vec{a}|\cdot|\vec{b}|sin\theta|## so the right-hand side is always positive. But it's rather clumsy.

Assuming you are working at an introductory level, then think in terms of simple components. Note that if ##\vec{c}=\vec{a}\times\vec{b}## then
##{c_x} = {a_y}{b_z} − {a_z}{b_y}##
and similar expressions for ##{c_y}## and ##{c_z}##.
That should enable you to do a proof easily.

greg_rack and PeroK
greg_rack said:
Homework Statement:: Demonstrate that:
$$\vec{a}\times\vec{b}=-\vec{b}\times\vec{a}$$
Relevant Equations:: none
... being ##\theta## the angle between the two vectors, ...
How exactly do you define this angle? Angles have signs. Suppose you define the angle between two vectors as the angular displacement going from the first vector in the cross product towards the second in the clockwise direction and not as the smaller angle between the vectors. Also suppose that the "smaller" angle between the two vectors is going counterclockwise from ##\vec a## to ##\vec b## (this choice doesn't matter.) When you write ##|\vec a \times \vec b|=|\vec a | | \vec b|\sin \theta## then ##|\vec a | | \vec b|\sin \theta## is positive. Now consider ##\vec b \times \vec a##. To be consistent with the procedure, you start at ##\vec b ## and go counterclockwise to ##\vec a##. The sine of the angle is now ##\sin(2\pi-\theta)=-\sin\theta.## Note that negative sign here indicates direction in that the cross product is perpendicular to the plane defined by the two vectors and there are two antiparallel perpendicular directions.

So when you say,
greg_rack said:
if ##|\vec{a}\times\vec{b}|=|\vec{a}|\cdot|\vec{b}|sin\theta##, being ##\theta## the angle between the two vectors, how could ##\vec{b}\times\vec{a}## be different?
you have already thrown out the difference between the two ways of taking the cross product. In short, always using the small angle as the "angle between the two vectors" works only when you want to calculate the magnitude. Just because two cross products have the same magnitude does not mean that they have the direction.

greg_rack
@kuruman I'm not sure I exactly follow the argumentation above. The angle ##\theta## in ##\mathbf{a} \times \mathbf{b} = |\mathbf{a}||\mathbf{b}|\sin{(\theta)} \mathbf{n}## always satisfies ##\theta \in [0, \pi]##, which means that ##\sin{(\theta)} \geq 0## in all cases, and the coefficient of ##\mathbf{n}## is always a positive number. The direction of the resulting vector is built into ##\mathbf{n}##, and not the angle.

etotheipi said:
The direction of the resulting vector is built into ##\mathbf{n}##, and not the angle.
That is absolutely correct. I was trying to indicate, perhaps clumsily, that if one sticks to a consistent definition of angle, the two definitions of cross product result in the same magnitude but opposite signs. One would not get the same direction for the two cross products as OP seemed to suggest. There are better approaches and I think the quickest is to use the determinant formulation of the cross product and what happens when you interchange two rows.

Steve4Physics and etotheipi
kuruman said:
That is absolutely correct. I was trying to indicate, perhaps clumsily, that if one sticks to a consistent definition of angle, the two definitions of cross product result in the same magnitude but opposite signs. One would not get the same direction for the two cross products as OP seemed to suggest. There are better approaches and I think the quickest is to use the determinant formulation of the cross product and what happens when you interchange two rows.

Ah, okay I see now. Thanks!

kuruman
Thank you very much guys, now I definitely have a deeper comprehension of the cross product.
I got that there are many ways to demonstrate this property, some more mathematically formal and elegant, others clumsier but still useful to get the physics behind(@kuruman, I definitely got the point of being "consistent" with the procedure, in calculating angles).
Thanks again for your help and time @etotheipi @kuruman @Steve4Physics!

kuruman