How Does the Cross Product Relate to Rank 1 Tensors?

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Homework Statement


I don't know how to prove it.
Let us consider two arbitrary vectors
⃗ A and ⃗B.
Let us define the vector product of them as
⃗C = ⃗A × ⃗B
Show that the vector ⃗C belongs to the Rank 1 tensor. In other words,
prove that
C′i = λij Cj
where
Ci ≡ ϵij k Aj Bk
C′i ≡ ϵijk A′j B′k

Homework Equations


The Attempt at a Solution


I just tried, but I don't know about it.
 
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zorrorojo said:

The Attempt at a Solution


I just tried, but I don't know about it.

Hi zorrorojo, welcome to PF!:smile:

You'll need to be more specific than "I just tried, but I don't know about it" in order to get assistance here.

What exactly did you try? Show your attempt.
 
pls answer these qoestion:
1)expand the following
aibj ϵijk=
ϵjkl (du,l/dx,k)
 

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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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