Proving the Lagrange Triple Vector Identity for Orthogonal Coordinates

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Homework Statement


Prove

a \times (b \times c) = (a * c)b - (a*b)c

For orthagonal coordinates, a,b,c

Homework Equations



Cross Product and Dot Product

The Attempt at a Solution



I thought about expanding both sides out and proving they are equal, but I just realized that the left side of the theorem would give me a vector and the right side would give me a scalar. Perhaps I don't understand the theorem perfectly. Can someone explain the notion about this theorem and how I would go about proving it?
 
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I should clarify a bit

a,b,c are for cartesian orthagonal coordinates (i,j,k vectors are all normal to each other)
 
If a,b and c are vectors the right side IS a vector. (a.c)b-(a.c)c is scalar*vector minus scalar*vector. It's a vector. And you can just multiply it out.
 
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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