Prove: u × (v + w) = (u × v) + (u × w)

  • Thread starter Thread starter mk_29
  • Start date Start date
mk_29
Messages
2
Reaction score
0
For all u, v,w, we have v × u = −(u × v) and (u + v) × w = (u × w) + (v × w).

Apply these facts to prove, without using co-ordinates, that for all vectors u, v,w, we have

u × (v + w) = (u × v) + (u × w).

I need help, do not know were to start.

Thaks
 
Physics news on Phys.org
Start with ux(v+w)=-(v+w)xu. Do you see why you can do that? It's your first property.
 
Dick said:
Start with ux(v+w)=-(v+w)xu. Do you see why you can do that? It's your first property.

Oh, ok i see what you have done. Know it should easy for me 2 prove this.

Thanks for the help
 
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