Scalar product used for length?

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I got asked how the scalar product can be used to find the length of a vector? Could someone please explain
 
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Do you mean vector product, also called the dot product?

V°V = |V|^2.

Or do you mean the triple scalar product:

|A°BxC|= volume of the parallelepiped with three sides defined by A,B,C.


If it is the triple scalar product you will need more information to obtain a vectors length - but the dot product provides it immediately.
 
UltrafastPED said:
Do you mean vector product, also called the dot product?

V°V = |V|^2.

Or do you mean the triple scalar product:

|A°BxC|= volume of the parallelepiped with three sides defined by A,B,C.If it is the triple scalar product you will need more information to obtain a vectors length - but the dot product provides it immediately.

To my knowledge, the vector product and the dot product are never given the same meaning. The vector product yields a vector quantity; the dot product yields a scalar quantity.

I'm not sure how one could use the scalar product to find the length of an arbitrary vector -- it seems like a redundant question to me. Do you know how to find the magnitude of a vector?
 
FeDeX_LaTeX said:
To my knowledge, the vector product and the dot product are never given the same meaning. The vector product yields a vector quantity; the dot product yields a scalar quantity.

There are many vector products: http://mathworld.wolfram.com/VectorMultiplication.html

Two of those listed above result in a scalar. I was trying to determine which one he was referring to; the OPs language was ambiguous.

The vector inner product (dot product) is the obvious method.

But the original requestor has disappeared ... so the problem is dead.
 
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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