Dot product between cross products

quietrain
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Homework Statement


show (axb) . (cxd) =
|a.c b.c|
|a.d b.d|

The Attempt at a Solution



i have no idea. i don't know if the lines at the side are modulus lines or matrix brackets

but it seems that it has something to do with distribution law.

but i can't start proving if i don't even understand what the right hand side of the equation is ><

thanks for help!
 
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Right side is a determinant of this matrix. It's not so hard to prove, you could in example use Levi-Civita symbol to write your cross product, it goes in 3 steps afterwards.
 
irycio said:
Right side is a determinant of this matrix. It's not so hard to prove, you could in example use Levi-Civita symbol to write your cross product, it goes in 3 steps afterwards.

oh i have done it. but i didn't use that symbol? thanks anyway!
 
Well, you don't have to use it, that's just my favourite way to deal with cross product stuff :).
 
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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