Is there a vector that satisfies the property of the number 1 for cross product?

hoopsmax25
Messages
13
Reaction score
0

Homework Statement



Show that there is no vector ⃗e that has the property of the number 1 for cross product, namely
that ⃗e × ⃗x = ⃗x for all ⃗x.

Homework Equations



I'm sort of stuck on how to show this.

The Attempt at a Solution


I set e=(e1,e2,e3) and x=(x1,x2,x3) and used cross product to multiply it out but got stuck there.
 
Physics news on Phys.org
cant you do this as a proof by contradiction using the definition of cross product?

a x b = c (c is then perpendicular to a) and (c is perpendicular to b)

so now you assume that e exists and then what do you get?
 
yeah i understand that. i think that's where i am stuck at. i don't know what the next step would be.
 
hoopsmax25 said:
yeah i understand that. i think that's where i am stuck at. i don't know what the next step would be.

well if e exists then e x x = x right which means that x is perpendicular to e and x perpendicular to x bt can x be perpendicular to itself?
 
Oh ok, so obviously it cannot. So is there a way to show that the contradiction by writing it out?
 
One might also note that if \vec{x} is any non-zero vector in the same direction as \vec{e}, then \vec{e}\times\vec{x}= \vec{0}\ne \vec{x}.
 
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...
Back
Top