MHB Finding distance between vectors

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To find the distance between vectors A and B given in cylindrical coordinates, first convert these vectors to Cartesian coordinates. The distance between the two points represented by the vectors can then be calculated using the standard distance formula. It is important to clarify that vectors themselves do not possess a fixed position; rather, the distance refers to the separation between the points represented by the vectors. This approach allows for a clear understanding of the spatial relationship between the two vectors. Converting to Cartesian coordinates simplifies the calculation of the distance.
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Given $\vec{A}=2.15a\rho+6.25a\phi+3az$ and $\vec{B}=0.11a\rho+3.35a\phi+2az$. Determine the distance from $\vec{A}$ to $\vec{B}$.


what to do first?
 
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This question doesn't make a whole lot of sense. I am assuming you are dealing with a cylindrical polar orthonormal system, which is fine. But there is no such thing as finding "distances between vectors", because vectors do NOT have any position! All they have is direction and magnitude. That is why they can be picked up and moved anywhere you like.
 
maybe the problem is asking about a vector from A to B.
How can I do that?
 
You can convert cylindrical coordinates to Cartesian coordinates and use the standard formula for distance.
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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