The question is:   Given two vectors $\vec v, \vec u \ne 0$ and $\alpha$ scalar.  is there a vector $\vec w$ such that $\vec u \times \vec w = \vec v$ and $\vec u \cdot \vec w = \alpha $?  In cases that there isn't such a vector $\vec w$, explain why.  In cases that there exists $\vec w$, find it.  My Work:  In order to get some idea on how to do the proof, I tried to take a simple example with coordinations:  $\vec u = (1,1,1)$, $\vec w = (1,0,1)$, then $\alpha = 2$, and $\vec v = \begin{vmatrix}i&amp;j&amp;k\\1&amp;1&amp;1\\1&amp;0&amp;1\end{vmatrix}=(1,0,-1)$ now the question is if I change $\alpha$ or $\vec v$ or $\vec u$, will I be able to find a $\vec w$? So let's try $\alpha = 3$, now $\vec w$ can be $(1,1,1)$ or $(2,0,1$) ... (for the dot product).  but in the cross product, I can notice that $\vec w$ cannot be equal to $\vec u$ otherwise we get $\vec v=0$.  so I need $\vec v = (1,0,-1)=\begin{vmatrix}i&amp;j&amp;k\\1&amp;1&amp;1\\w_1&amp;w_2&amp;w_3\end{vmatrix}=(w_3-w_2,w_1-w_3,w_2-w_1)$ So I need to find $w_1+w_2+w_3=3$.  But I really just didn't continue here because even if I found it, I don't see myself reaching the proof or atleast getting an idea of how to generalize things, and thought that I'm in somewhat wrong direction or I'm not seeing something important. After that I tried to use $\vec u, \vec v$ to express $\vec w$ (Just because that's the only way to generalize a vector $\vec w$).  So if I choose lets say: $\vec w = \vec u - \vec v$ I would get $\vec v = \vec u \times (\vec u - \vec v)=\vec u\times \vec u-\vec u \times\vec v = -\vec u \times \vec v$. and again I couldn't find out how to make it equal to $\vec v$, and on top of that I would need to check that it works with the scalar $\alpha$ and the dot product, so I'm a little lost.   End of my work  I would really appreciate any hints or feedback about my two approaches and if they will bring me somewhere, and any pushes in the right direction and help will be awesome, thanks in advance to everyone.
Let $\vec{k}$, $\vec{v}$, and $\vec{u}$ be vectors, where $\vec{u}$ is unknown and $\vec{k}$ and $\vec{v}$ are known vectors. Given:     $\vec{u}\cdot\vec{k}=c$  $\vec{u} \times \vec{k}= \vec{v}$   From this relations, how can i determine the vector $u$? I tried to construct orthogonal coordinate system from $(\vec{k}, \vec{v},\vec{k}\times \vec{v})$ but i couldn't proceed from there. Any idea?
Ubuntu 14.04.5/16.04 and newer on AMD graphics I have 14.04.5/16.04 or newer and an AMD GPU. I'm not getting the performance I want, and some programs say they need things like OpenCL.  I've heard about fglrx, but if I try to install it on my machine, it breaks and I have no GUI.  Is there anything I can do to get my AMD card working to its full potential? Let $X, Y$ be an exponentially distributed random variables with parameters $a, b$. Then $X$ has pdf:    $$f_X(x) =\begin{cases} a e^{-a x},&amp; x\geq 0\\ 0,&amp; \text{otherwise}.\end{cases}$$ Suppose $X$ and $Y$ independent. Show that $$\mathbb P(X&gt;Y) = \frac{b}{a+b}.$$   Now I thought the following: $$f(x,y) = f_X(x)\ f_Y(y) = abe^{-ax -by},\qquad\text{for } x,y &gt; 0.$$ And then $$\mathbb P(X&gt;Y) = \int_0^\infty \int_0^x a b e^{-ax -by}\,dydx$$ However, if I solve this (manually or using Wolframalpha), I can't seem to end up with $\frac{b}{a+b}$. Any ideas? Let $trcA=0$.why $A=M+N$ where $M$ and $N$ are nilpotent matrices? How to find Hamiltonian from this simple Lagrangian? (tricky) $$L~=~ \frac{1}{2} \dot{q}  \sin^2{q} $$  Is it zero or not defined? Encode URL in JavaScript? I have the following problem illustrated by a simple example: If I cite the paper of an author T. Coven in my document, the outcome is 'Coven (2004b)'. If I cite the appendix to this paper, the outcome is 'Coven (2004a)'. Now the appendix pops up much less frequently than the original paper, therefore I would like to reverse the order (exchange 'a' and 'b' in the citations). Is this possible?  I am using Biblatex. Here is the bib-file of the above example (named: lib.bib):  @article{co:2004, author = {T. Coven}, title = {Title A}, journaltitle = {The X Journal}, date = {2004} } @article{coapp:2004, author = {T. Coven}, title = {Appendix to Title A}, journaltitle = {The X Journal}, date = {2004} }   MWE:  \documentclass[a4paper, 12pt, headsepline, headings=small,]{scrreprt} \usepackage[backend=bibtex8,     style=authoryear-icomp,     dashed=false,     autocite=footnote,     maxcitenames=3,     mincitenames=1,     maxbibnames=100,     firstinits=true,     sorting=nty     ]{biblatex} \bibliography{lib}  \begin{document} \cite{co:2004} and \cite{coapp:2004} \end{document} Why is.vector on a data-frame doesn't return TRUE? I tried to search for similar questions, as I always do, but was not able to find one. I just created a new  on Data.Stack Exchange with my existing Google OpenID and it seems to be not associated with my regular Stack Exchange account. Is this by design? and if so, why? pgfplots fillbetween with multiple curves I have a TikZ picture like this:  \begin{tikzpicture}   \begin{axis}[     axis lines=middle,     xmin=-0.5,xmax=3.5,ymin=-0.5,ymax=3.5   ]   \addplot[very thick,blue, samples=300, domain=0:3, name path=A] {sqrt(x)};    \addplot[very thick,yellow!70!red, samples=300, domain=0:3, name path=B] {9-3*x};    \addplot[very thick,red!90!teal, samples=300, domain=0:3, name path=C] {3};    \addplot[very thick,green!70!magenta, samples=300, domain=0:3, name path=D] coordinates{(1/9,-6) (1/9,6)};   \end{axis} \end{tikzpicture}     I’d like to fill the area enclosed by these four curves.  Similar questions have been answered using pgfplots’s fillbetween package, but it only seems to support filling between two curves. What are my options? I am new to Linux and Ubuntu. When I don't know how to do something with this OS I search on the web and Ask Ubuntu and always find the answer.  But sometimes it's not a great answer nor complete tutorial. I follow the instructions anyway, but I don't know what all the commands mean.  So my question is:   When I follow a tutorial, am I 100% safe? If not, then how can I tell I am in danger? Does using links to other pages dilute the linking page's link juice/page rank? There are lot's of questions here about passing link juice to other pages, including .  I've read several of them over the months, but I always come away wondering the same question.  Does linking to another page, dilute or burn some of the link juice the linking page has already earned?  For example, if my current page has for example, a link juice of 10, and I link to another page, it passes a certain amount of juice (I'll use 90% as an example) of the linking page's juice, giving the page being linked to a .9 boost in it's own link juice.  However, does that mean the linking page's page's remaining juice is now less than 10?  While some of the questions would imply this to be the case, it seems counter intuitive as no one would ever want to link to any page outside of their own content for fear of losing page rank. How to make a disk image and restore from it later? I'm a new Linux user. I've reinstalled my Wubi from scratch at least ten times the last few weeks because while getting the system up and running (drivers, , etc.) I've broken something (X, grub, unknowns) and I can't get it back to work. Especially for a newbie like me, it's easier (and much faster) to just reinstall the whole shebang than try to troubleshoot several layers of failed &quot;fixing&quot; attempts. Coming from Windows, I expect that there is some &quot;disk image&quot; utility that I can run to make a snapshot of my Linux install (and of the boot partition!!) before I meddle with stuff. Then, after I've foobar'ed my machine, I would somehow restore my machine back to that working snapshot. What's the Linux equivalent of Windows disk imagers like  or ?  Note: I found a similar question: Changing capitalization of filenames in Git I have some HTML menus, which I show completely when a user clicks on the head of these menus. I would like to hide these elements when the user clicks outside the menus' area.  Is something like this possible with jQuery?  $("#menuscontainer").clickOutsideThisElement(function() {     // Hide the menus }); “Fire” a weapon before firearms existed? Did the verb “fire a weapon” exist before the actual introduction of firearms on battlefields?  More specifically, does it make sense for a creative work to have archers (or whatever ranged weaponry) be told to “fire!”, when the world they live in has not yet seen firearms?  It seems some kind of an anachronism to me, since before firearms, “fire” would never propel any projectile...  I've seen several movies do it; I can't remember them all, but for instance I verified it in The Lord of the Rings trilogy (left), Kingdom of Heaven (top right), and more recently Frozen (bottom right):    The first two do it several times (with Kingdom of Heaven also using it on ballistae, and The Return of the King even having Aragorn ask Legolas to ), whereas firearms never appear on the battlefields they feature (Kingdom of Heaven is set in the 12th century; as for The Lord of the Rings, apart from one occurrence of a witchcraft-ish bomb, it's archery and medieval siege engines all the way).  Frozen may be debatable, happening probably somewhere in the 18th or 19th century, but since we only see swords, spears and the like, it made me flinch to hear crossbowmen use “fire!”.  I've also noticed other movies avoiding this, using “loose!” instead, such as Troy (left) and Gladiator (right):    This makes me think it would be the right thing to say instead... although I'm still wondering why not use “shoot”, which seems simpler to me. The set of integers $n$ expressible as $n=x^2+xy+y^2$ Let $S$ be the set of integers $n$, such there exist integers $x,y$ with $$n=x^2+xy+y^2$$   Is the implication $$a,b\in S\implies ab\in S$$ true? If yes, how can I prove it?  I worked out $$n\in S\iff 4n\in S$$ and $$n\in S\iff 3n\in S$$ I tried two approaches. The first is to express $$(a^2+ab+b^2)(c^2+cd+d^2)$$ in the form $$f^2+fg+g^2$$ with polynomials $f,g$ with integer coefficients. I however could not find suitable $f$ and $g$. The second approach is based on $$x^2+xy+y^2=\frac{(2x+y)^2+3y^2}{4}$$ If we have $n=x^2+xy+y^2$ , we have $u^2+3v^2=4n$ for some integers $u,v$ with equal parity. The main problem of this approach is to consider the equal parity. Any ideas ? Take most recent (or concatenate) edit comments when editing during grace period Recently, I edited  to update a link to a better resource. I clicked "Save" without entering a description of the edit (edit #1).  Then I realised that the question was asked by a new user who may not be able to easily identify the change that was made. So I edited it again, this time only to add a detailed edit description (edit #2).  Since edit #2 was made during the grace period (within 5 minutes of edit #1), the system has merged them into a single edit, but only the edit comment from edit #1 was recorded in the revision history, whereas edit #2's comment was simply discarded/ignored.  I'd suggest that either the most recent (non-default) edit comment is taken or that the comments are concatenated to provide a more useful edit history What is the best solution for database connection pooling in python? I have developed some custom DAO-like classes to meet some very specialized requirements for my project that is a server-side process that does not run inside any kind of framework.    The solution works great except that every time a new request is made, I open a new connection via MySQLdb.connect.    What is the best "drop in" solution to switch this over to using connection pooling in python?  I am imagining something like the commons DBCP solution for Java.    The process is long running and has many threads that need to make requests, but not all at the same time... specifically they do quite a lot of work before brief bursts of writing out a chunk of their results.  Edited to add: After some more searching I found  which looks decent, but as I'm relatively new to python I guess I just want to make sure I'm not missing a more obvious/more idiomatic/better solution. Is there any way to fix the wavy lines that appear when photographing a striped dress?   It's not just in viewing online: when it was printed at wallet size, the wavy lines showed up. New 16GB of RAM installed yet I see 15.3 on my system. Why? Just switched 4GBs to 16GBs of RAM. However, when I look at how much memory I have, it says 15.3. I'm just wondering why did my memory drop down when I just installed 16 GBs of RAM.