Basic vector operations, using cross and dot product

• greg_rack
In summary, you need to use the determinant form of the cross product to solve for the x, y, and z components of a 3D vector.

greg_rack

Gold Member
Homework Statement
##(\textbf{r}\times \dot{\textbf{r}})\times \textbf{r}=\dot{\textbf{r}}(\textbf{r}\cdot \textbf{r})-\textbf{r}(\textbf{r}\cdot\dot{\textbf{r}})##

Where ##\textbf{r}## is a vector
Relevant Equations
Vector cross and dot product
Hi guys,
I am losing my mind over this passage...
I cannot understand how to get from the first expression with the cross products to the second ##\dot{\textbf{r}}(\textbf{r}\cdot \textbf{r})-\textbf{r}(\textbf{r}\cdot\dot{\textbf{r}})##

Delta2
greg_rack said:
Homework Statement:: ##(\textbf{r}\times \dot{\textbf{r}})\times \textbf{r}=\dot{\textbf{r}}(\textbf{r}\cdot \textbf{r})-\textbf{r}(\textbf{r}\cdot\dot{\textbf{r}})##
Where ##\textbf{r}## is a vector
Relevant Equations:: Vector cross and dot product

Hi guys,
I am losing my mind over this passage...
I cannot understand how to get from the first expression with the cross products to the second ##\dot{\textbf{r}}(\textbf{r}\cdot \textbf{r})-\textbf{r}(\textbf{r}\cdot\dot{\textbf{r}})##
Often the best way is to use the determinant form of the cross product and equate both sides component by component.

Astronuc and Delta2
Thread moved to Calculus & Beyond section.

etotheipi said:
My post #3 was deleted?
Yes, as being a violation of forum rules for providing too much help. You should have received a warning PM about the deletion.

And I'm deleting it again.
On helping with questions: Any and all assistance given to homework assignments or textbook style exercises should be given only after the questioner has shown some effort in solving the problem. If no attempt is made then the questioner should be asked to provide one before any assistance is given. Complete solutions can be provided to a questioner after the questioner has arrived at a correct solution. If the questioner has not produced a correct solution, complete solutions are not permitted, whether or not an attempt has been made.
If you continue providing full answers, expect consequences.

SammyS
Are you familiar with the BAC-CAB rule?

PeroK said:
Often the best way is to use the determinant form of the cross product and equate both sides component by component.
By calculating the cross product with the determinant form, for 3D vectors, how do I reach a solution?
In other terms, how do I apply this definition

In order to get to the desired form?

PS: I have never taken a linear algebra course, so I have no familiarity at all with matrices and such notations... probably that's why I'm getting stuck.

greg_rack said:
By calculating the cross product with the determinant form, for 3D vectors, how do I reach a solution?
In other terms, how do I apply this definition
View attachment 283135
In order to get to the desired form?

PS: I have never taken a linear algebra course, so I have no familiarity at all with matrices and such notations... probably that's why I'm getting stuck.
You do that process again to get ##(\vec a \times \vec b) \times \vec c##

You only need algebra and a bit of cleverness.

Astronuc
greg_rack said:
I am losing my mind over this passage...
Don't lose your mind! When you meet these more abstract formulas, first prove it by example (just to show yourself it is in fact true), then progressively get more abstract.

In your post #7, when it says ##a_z## or ##b_y## those are the components of that vector in that direction. Example: Let ##\vec{A} = <3,4,5>## then ##A_x= 3, A_y = 4, A_z = 5## so, example: ##A_xA_y = 3*4 = 12##

greg_rack
romsofia said:
Don't lose your mind! When you meet these more abstract formulas, first prove it by example (just to show yourself it is in fact true), then progressively get more abstract.

In your post #7, when it says ##a_z## or ##b_y## those are the components of that vector in that direction. Example: Let ##\vec{A} = <3,4,5>## then ##A_x= 3, A_y = 4, A_z = 5## so, example: ##A_xA_y = 3*4 = 12##
Thanks a lot for the advice! :)

greg_rack said:
By calculating the cross product with the determinant form, for 3D vectors, how do I reach a solution?
In other terms, how do I apply this definition
View attachment 283135
In order to get to the desired form?

PS: I have never taken a linear algebra course, so I have no familiarity at all with matrices and such notations... probably that's why I'm getting stuck.
You just need to work step by step. It may help as well to work component by components, i.e. show that the x-component on the lefthand side is equal to the x-component on the righthand side. Rewrite the lefthand side as ##\vec a \times \vec r## where ##\vec a = \vec r \times \dot {\vec r}##. Use the definition to compute the components of ##\vec a## in terms of the components of ##\vec r## and ##\dot {\vec r}##.

The x-component of the lefthand side is
$$(\vec a \times \vec r)_x = a_y r_z - a_z r_y.$$ Plug in the expressions you found for ##a_y## and ##a_z##.

Now compute the x-component of the righthand side in terms of the components of ##\vec r## and and ##\dot {\vec r}##. It should be straightforward to manipulate what you got the lefthand side into the form you got for the righthand side.

Repeat the process for the y- and z- components.

greg_rack
Reducing it to components is the standard proof, e.g. see https://en.m.wikipedia.org/wiki/Triple_product#Vector_triple_product.
But maybe there's a neater way.
##(\vec v\times\vec w)## is normal to each of that pair, so is normal to the plane containing them. ##\vec u\times(\vec v\times\vec w)## is normal to that product, so lies in the v w plane, so ##=a\vec v + b\vec w## for some a, b.
Taking the dot product with u we get ##0=a\vec u.\vec v+b\vec u.\vec w##.
Hence ##\vec u\times(\vec v\times\vec w)=\lambda((\vec u.\vec v)\vec w-(\vec u.\vec w)\vec v)##. It remains to show ##\lambda=1##, but there I am stuck.

vela
haruspex said:
It remains to show ##\lambda=1##, but there I am stuck.
I found ##\lambda = -1##. I let ##\vec w = \vec w_\parallel + \vec w_\perp## where the components are parallel and perpendicular to ##\vec v##. Then I considered ##\vec w_\perp \times (\vec v \times \vec w)##.

vela said:
I found ##\lambda = -1##. I let ##\vec w = \vec w_\parallel + \vec w_\perp## where the components are parallel and perpendicular to ##\vec v##. Then I considered ##\vec w_\perp \times (\vec v \times \vec w)##.
Ah yes, I didn't check which sign I needed.
But in principle ##\lambda## as I arrived at it is a function of the three vectors, so showing it to be -1 in a special case is not general enough.
Also, if the proof depends on the properties of the scalar triple product and those are proved by resort to components in an orthonormal basis then we're no better off.

Did the creator of this thread solve his problem?

Given ##(\textbf{r}\times \dot{\textbf{r}})\times \textbf{r}=\dot{\textbf{r}}(\textbf{r}\cdot \textbf{r})-\textbf{r}(\textbf{r}\cdot\dot{\textbf{r}})##, look at the term on the left. Remember that cross product is antisymmetric.

@greg_rack, can you let us know whether the responses given so far were enough for you to figure out this problem?

Thanks!

PeterDonis said:
@greg_rack, can you let us know whether the responses given so far were enough for you to figure out this problem?

Thanks!
Yes, sure! I've been able to carry on the math.
At the beginning I only got daunted by the number of terms deriving from the two cross products(not for a lack of cleverness, as @PeroK pointed out), and thought I was on the wrong path, having never dealt with linear algebra this way.
Just needed to insist, but it still got me 3/4 pages full of steps :)

Hint. Sometimes brute force is most simple!

Just use the standard rules to calculate dot and cross products of vectors using Cartesian coordinates (Cartesian right-handed basis) to show that for any three vectors ##\vec{a}##, ##\vec{b}##, and ##\vec{c}##
$$\vec{a} \times (\vec{b} \times \vec{c})=\vec{b} (\vec{a} \cdot \vec{c}) - \vec{c} (\vec{a} \cdot \vec{b}).$$

PeroK
greg_rack said:
Yes, sure! I've been able to carry on the math.
At the beginning I only got daunted by the number of terms deriving from the two cross products(not for a lack of cleverness, as @PeroK pointed out), and thought I was on the wrong path, having never dealt with linear algebra this way.
Just needed to insist, but it still got me 3/4 pages full of steps :)
Until you've found the clever step, you may not be seeing what you're missing!

Another pretty elegant way is to use the definition of the Levi-Civita tensor components in terms of a right-handed Cartesian basis,
$$\epsilon_{jkl}=\vec{e}_j \cdot (\vec{e}_k \times \vec{e}_l),$$
from which you can derive the formula (Einstein summation convention applied)
$$\epsilon_{jkl} \epsilon_{jmn}=\delta_{km} \delta_{ln}-\delta_{kn} \delta_{lm},$$
which is equivalent to the formula for arbitrary three vectors (the "bac-cab formula") in #18.

All you need to know is the obvious formula
$$\vec{a} \cdot \vec{b}=\sum_{j} (\vec{a} \cdot \vec{e}_j) (\vec{e}_j \cdot \vec{b})$$
for arbitrary vectors ##\vec{a}## and ##\vec{b}## and that for the Cartesian basis vectors ##\vec{e}_j \cdot \vec{e}_k=\delta_{jk}##.

PeroK