What is the Norm of a Vector and How is it Used in Linear Algebra?

LeakyFrog
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Hello, I am studying for an exam in Linear Algebra. My teacher gave us an outline of things that we need to know and one of them is this:

Find the norm of a vector v in n-dimensional space. Use it to find a unit vector in the same direction as v.

I was just hoping someone might be able to explain what exactly that means. I understand what a unit vector is but I'm a little hazy on the rest. Such as what exactly is a normal vector to another vector? Is that simply a vector that is perpendicular to the other? Any help would be awesome.
 
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Not normal, norm. They are two different things. In n-dimensional space, the most common norm is the square root of the inner product, or dot product. It gives a concept of length to a vector. How would you use a vector's length to find a unit vector in the same direction?
 
Dude thank you soooo much for clearing that up. Makes perfect sense now. And I learned what norm means. F' Yeah!
 
I was kind of worried! Reviewing for a test in linear algebra and did not know what the "norm" of vector was? But I can see confusing "norm" and "normal".
 
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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