Vector cubed

1. Sep 17, 2007

Ok so if I have (a_x + a_y + a_z)^2

I get a_x^2 + a_y^2 + a_z^2 as the answer because the cross terms go to zero. For example a_x . a_y = 0 etc...

What if I have (a_x + a_y + a_z)^3 ?

Will the answer be a_x^3 + a_y^3 + a_z^3 , or will it be a_x^3 + a_y^3 + a_z^3 + a_x^2 a_y + a_x^2 a_z + a_y^2 a_x + a_y^2 a_z + a_z^2 a_x + a_z^2 a_y ?

My motivation for thinking it might be the first one is that a term like a_x^2 a_y could be expanded into one like a_x . a_x . a_y and then would go to zero if we take the dot product a_x a_y . On the other hand the factor a_x . a_x is a scalar and a scalar a_x^2 times a vector a_y will remain as a_x^2 a_y . On the.. third.. hand if we take the supposed commutative properties of a dot product then (a_x . a_x) . a_z should be the same as a_x . (a_x . a_z) which I seem to argue is not the case. We cannot argue that a_x . (a_x . a_z) != 0 but we can argue that (a_x . a_x) . a_z = a_x . (a_x . a_z) = 0 ... So I guess the first anwer is the right one.

Confirmation or help in the right direction would be much appreciated. Thanks.

2. Sep 17, 2007

neutrino

I don't think you can actually "square" or "cube" a vector. What you can say is that the dot or scalar product of a vector with itself is equal to square of the modulus of the vector, i.e.,

$$\vec{r}.\vec{r} = \left|\vec{r}\right|^2$$​

But $$\vec{r}.\left(\vec{r}.\vec{r}\right)$$ is meaningless, since the term within the parentheses is a scalar, and scalar products are defined only between vectors.

The multiplication defined for vectors is different from the multiplication defined for real numbers.

3. Sep 17, 2007

Math Jeans

Ya. Basically, a dot product of two vectors is a scalar. If you try to take another dot product with a dot product, you end up with a vector dotted with a scalar which is meaningless unless you are merely multiplying the vector by the factor of the scalar in which you would be using two different types of multiplication from squaring to cubing.

4. Sep 17, 2007

How would you interpret $$(\mathbf{a}_x + \mathbf{a}_y + \mathbf{a}_z)^3$$, where $$\mathbf{a}_l = |a_l| \hat{i}$$, ?

5. Sep 17, 2007

neutrino

Sum of three vectors raised to the power 3? That's not defined mathematically.

6. Sep 17, 2007

So if I have $$\mathbf{a}^3$$, where a is a vector, the only thing it could mean is $$|\mathbf{a}|^3$$?

Last edited: Sep 17, 2007
7. Sep 17, 2007

neutrino

It can't be...it's not a meaningful expression.

Maybe you came across this

$$\left(\mathbf{a}.\hat{i} + \mathbf{a}.\hat{j} + \mathbf{a}.\hat{k}\right)^3$$​

or something similar?

EDIT: I just read your edited post. Maybe what you read was just the magnitude cubed. In what context did you come across this term?

Last edited: Sep 17, 2007
8. Sep 17, 2007

dynamicsolo

Did you actually encounter this in your reading or in a problem or are you just curious?

There are only two mathematically defined ways of combining three vectors. That's because the multiplication operations are binary, so you have to select two of the three for the first operation. That first operation has to produce a vector in order to perform the second binary operation with the third remaining vector. So you only get these possibilities:

A · (B X C) , the so-called scalar triple product or

A X (B X C) , the so-called vector triple product.

I can't recall having ever seen an A^3 notation involving a vector. If you tried to apply either of these to the situation A=B=C, you'd get zero since A X A = 0.

9. Sep 17, 2007

dynamicsolo

I think the consensus is that the writer must mean the magnitude cubed. Was the name of the vector notated in boldface or not? If the vector name was just in plain face, it *would* mean the magnitude cubed and is the only way this would make sense. (If it was in boldface, the writer probably just slipped up...)

10. Sep 17, 2007

Quite possibly. I'm basically doing a calculation and checking the result is the same as what somebody else got. We've both been lazy with notation and now I'm confused. Along the way we had somehting like $$\int d^3 \mathbf{b} f(\mathbf{a},\mathbf{b})$$. After integration we both got terms with a^3 in them, and neither of us have indicated what that means exactly... Mind you at one point all the a terms where written as a^2 and then after simplification we had these terms with a^3. I know for sure that a^2 means |a|^2 ... so I guess a^3 must mean |a|^3 .

Last edited: Sep 17, 2007
11. Sep 17, 2007

Some physics textbooks denote the volume element in a volume integral as $$d^3\vec{r}$$, but that usually represents the usual dxdydz (in the Cartesian system, of course.)