Solving the Triple Scalar Product: Finding the Value of a(dot)(a(cross)b)

CaityAnn
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Homework Statement


What is the value of a(dot)(a(cross)b) ? Why?
I am supposed to find an actual value.
Sorry I don't know code, these variables are all vectors. A is dotted with vectors a and b which are cross product.

Homework Equations


I know this can be written as a determinate of all three variables but I do not see how this gives me a single answer.


The Attempt at a Solution

 
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axb is perpendicular to a and b. The relevant part of this is that it is perpendicular to a, so axb is some vector c perpendicular to a. So what is a.c when c is perpendicular to a?
 
Try this, write out the cross product by hand.

Once you have the cross product, take the scalar product.

(There is a trick to this question, see if you can find it).
 
^ ^physicist, you're evil! Yes, once CaityAnn has gone through that pain, perhaps she will learn that it is better to think!
 
It may be evil, but I find if you work through it this way, after doing it a couple of times you start seeing the trick...
 
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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