How Do You Calculate the Magnitude of a Constant Vector in Different Dimensions?

tomfrank
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Homework Statement



Calculate ||1,1,1||in R3
Calculate ||1,1,1,1|| in R4.
Calculate ||1,1,...,1|| in Rn.

Homework Equations


All I have in this problem is that, Where do I start?


The Attempt at a Solution

 
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What does ||1,1,1|| mean? This isn't a leading question; I want to know what that symbol is supposed to represent.
 
I'm guessing || means the magnitude, or the scalar length of the vector.

[Moderator's note: solution deleted]
 
Last edited by a moderator:
I would guess that too, but I'm trying to get the poster to write it correctly for starters. You wouldn't normally write a vector as 1,1,1.
 
Last edited:
it means the magnitude, i believe.
 
So how do you calculate the magnitude of a vector in R3, R4, Rn?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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