Proof u cross u = 0 ?

[goku900]

I needed to prove that u x u = 0 in three ways 1) by definition 2) by determinate form and 3) by the definition that u x v = -(v x u )

I really don't know how to start the third one. My textbook says its the anticommunitive property but there is no proof for it. I'm trying to find it online also.

 

[Romsek]

I needed to prove that u x u = 0 in three ways 1) by definition 2) by determinate form and 3) by the definition that u x v = -(v x u )

I really don't know how to start the third one. My textbook says its the anticommunitive property but there is no proof for it. I'm trying to find it online also.

are you given that (u x v) = -(v x u) or are they asking you to prove this?

if you are given it just plug u and u in and the fact that u must be zero is pretty obvious.

if you have to prove it just work the numbers for 3 dimensions and that should be good enough. I doubt they are asking you to prove this in the general case of n dimensions.

Intuitively you can see this using the right hand rule. If you sweep from u to v your thumb is pointing in some direction. If you sweep v to u it's pretty clear your thumb will be pointing in the opposite direction.

 

[goku900]

are you given that (u x v) = -(v x u) or are they asking you to prove this?

if you are given it just plug u and u in and the fact that u must be zero is pretty obvious.

if you have to prove it just work the numbers for 3 dimensions and that should be good enough. I doubt they are asking you to prove this in the general case of n dimensions.

Intuitively you can see this using the right hand rule. If you sweep from u to v your thumb is pointing in some direction. If you sweep v to u it's pretty clear your thumb will be pointing in the opposite direction.

I have to prove that u x u = 0 with that last formula. Ill try what u said

How do you mean plug u and u in. Am I doing this arbitrarily? Isn't this the same as using the determinate form then. I don't understand how u and u plug in

 

[Romsek]

I have to prove that u x u = 0 with that last formula. Ill try what u said

How do you mean plug u and u in. Am I doing this arbitrarily? Isn't this the same as using the determinate form then. I don't understand how u and u plug in

you have a formula (u x v) = -(v x u)

well let u=u and let v=u as well

so you end up with (u x u) = -(u x u)

what does that tell you about (u x u) ?

 

[goku900]

you have a formula (u x v) = -(v x u)

well let u=u and let v=u as well

so you end up with (u x u) = -(u x u)

what does that tell you about (u x u) ?

I guess the only number that satisfies that would be 0. Ok so from there though could you plug in real numbers for the vector u ?

 

[Romsek]

I guess the only number that satisfies that would be 0. Ok so from there though could you plug in real numbers for the vector u ?

u is totally arbitrary so you can make u whatever u like. u x u will still be zero.

 

[HallsofIvy]

However, putting specific numbers in for u, say, u= <1, 2, 3>, and showing that u x u= 0 would only prove it for that one vector. You need to use "general numbers" or variables. If u= <a, b, c> what is u x u?

You say prove it "by definition" but what definition are you using? You also say "by determinant form" but I have seen the "determinant form", $\displaystyle \vec{u}\times\vec{v}= \left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ u_x & u_y & u_z \\ v_x & v_y & v_z\end{array}\right|$ as the definition. I have also seen $\displaystyle <u_x, u_y, y_z>\times <v_x, v_y, v_z>= (u_yv_z- u_zv_y)\vec{i}+ (u_zv_x- u_xv_z)\vec{j}+ (u_xv_y- u_yv_x)\vec{k}$ as the definition as well as $\displaystyle \vec{u}\times\vec{v}= |\vec{u}||\vec{v}|sin(\theta)$ where $\displaystyle \theta$ is the angle between the two vectors. As the angle between u and itself is 0, that last is particularly easy for this problem.

 

[Romsek]

However, putting specific numbers in for u, say, u= <1, 2, 3>, and showing that u x u= 0 would only prove it for that one vector. You need to use "general numbers" or variables. If u= <a, b, c> what is u x u?

You say prove it "by definition" but what definition are you using? You also say "by determinant form" but I have seen the "determinant form", $\displaystyle \vec{u}\times\vec{v}= \left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ u_x & u_y & u_z \\ v_x & v_y & v_z\end{array}\right|$ as the definition. I have also seen $\displaystyle <u_x, u_y, y_z>\times <v_x, v_y, v_z>= (u_yv_z- u_zv_y)\vec{i}+ (u_zv_x- u_xv_z)\vec{j}+ (u_xv_y- u_yv_x)\vec{k}$ as the definition as well as $\displaystyle \vec{u}\times\vec{v}= |\vec{u}||\vec{v}|sin(\theta)$ where $\displaystyle \theta$ is the angle between the two vectors. As the angle between u and itself is 0, that last is particularly easy for this problem.

Just cause we all love math. Index notation shows u x u = 0 almost instantly.

$\displaystyle \{u\otimes u\}_i=\epsilon _{\text{ijk}} u_j u_k$

$\displaystyle u_j u_k$ is clearly symmetric

and $\displaystyle \epsilon _{\text{ijk}}$ is of course anti-symmetric

The contraction of a symmetric and anti-symmetric tensor is 0.