For a 3D scene in the world coordinates, its View Reference Point $\mathrm{VRP}$ is at $(5,2,1)$, and a viewer is looking towards point $A=(1,1,1)$. Construct a transform matrix which will map  world  coordinate  points  to  a  right-handed  $(UVN)$  viewing  space,  so  that  VRP  is    the  origin,  the  line  joining  $\mathrm{VRP}$ to  $A$  is  the  positive  $N$  axis,  and  the  view-up  Vector  is $(0,0,1)$.     I have calculated the translation matrix:  $$ T= \begin{pmatrix} 1 &amp; 0 &amp; 0 &amp; -5 \\ 0 &amp; 1 &amp; 0 &amp; 2 \\ 0 &amp; 0 &amp; 1 &amp; -1 \\ 0 &amp; 0 &amp; 0 &amp; 1 \\ \end{pmatrix} $$  I have calculated the rotation matrix:  $$ R= \begin{pmatrix} -0.6 &amp; -0.8 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 1 &amp; 0 \\ -0.8 &amp; 0.6 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 1 \\ \end{pmatrix} $$  But what would the final transformation matrix $M$ be? My lecture slides tell me it would be rotation matrix multiplied by translation matrix, with the last column multiplied by $-1$ except for the $1$. This would equal:  $$ M = R \times T =  \begin{pmatrix} -0.6 &amp; -0.8 &amp; 0 &amp; -1.4 \\ 0 &amp; 0 &amp; 1 &amp; 1 \\ -0.8 &amp; 0.6 &amp; 0 &amp; -5.2 \\ 0 &amp; 0 &amp; 0 &amp; 1 \\ \end{pmatrix} $$  But this answer here:   just adds them together somehow. Which one is correct?
For a 3D scene in the world coordinates, its View Reference Point $\mathrm{VRP}$ is at $(5,-2,1)$, and a viewer is looking towards point $A=(1,1,1)$. Construct a transform matrix which will map world coordinate points to a right-handed $(UVN)$ viewing space, so that $\mathrm{VRP}$ is the origin, the line joining $\mathrm{VRP}$ to $A$ is the positive $N$ axis, and the view-up Vector is $(0,0,1)$.   To my knowledge, the final answer should be a single transform matrix. To calculate the final matrix, I first find the matrix which translates the coordinate system to the to the origin in homogeneous matrix form:  Translation matrix:  $$ T=         \begin{bmatrix}         1 &amp;0 &amp;0 &amp;-5 \\         0 &amp;1 &amp;0&amp; 2 \\         0 &amp;0 &amp;1 &amp;-1 \\         0 &amp;0 &amp;0 &amp;1 \\         \end{bmatrix} $$  Using the unit vectors of the coordinate axes, the resulting rotation matrix is:  $$ R=         \begin{bmatrix}         -0.6 &amp;-0.8 &amp;0 &amp;0 \\         0 &amp;0 &amp;0 &amp;1 \\         -0.8 &amp;0.6 &amp;0 &amp;0 \\         0 &amp;0 &amp;0 &amp;1 \\         \end{bmatrix} $$ Then the final matrix (the final answer for full marks) would be the rotation matrix multiplied by the translation matrix:  $$ M=         \begin{bmatrix}         -0.6 &amp;-0.8 &amp;0 &amp;1.4 \\         0 &amp;0 &amp;0 &amp;1 \\         -0.8 &amp;0.6 &amp;0 &amp;5.2 \\         0 &amp;0 &amp;0 &amp;1 \\         \end{bmatrix} $$ Is my method and final answer correct?    Following the answer to this question, I have calculated the translation matrix:  $$ T= \begin{pmatrix} 1 &amp; 0 &amp; 0 &amp; -5 \\ 0 &amp; 1 &amp; 0 &amp; 2 \\ 0 &amp; 0 &amp; 1 &amp; -1 \\ 0 &amp; 0 &amp; 0 &amp; 1 \\ \end{pmatrix} $$  I have calculated the rotation matrix:  $$ R= \begin{pmatrix} -0.6 &amp; -0.8 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 1 &amp; 0 \\ -0.8 &amp; 0.6 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 1 \\ \end{pmatrix} $$  But what would the final transformation matrix $M$ be? My lecture slides tell me it would be rotation matrix multiplied by translation matrix, with the last column multiplied by $-1$ except for the $1$. This would equal:  $$ M = R \times T =  \begin{pmatrix} -0.6 &amp; -0.8 &amp; 0 &amp; -1.4 \\ 0 &amp; 0 &amp; 1 &amp; 1 \\ -0.8 &amp; 0.6 &amp; 0 &amp; -5.2 \\ 0 &amp; 0 &amp; 0 &amp; 1 \\ \end{pmatrix} $$  But the answer here just adds them together somehow. Which one is correct?
What do the numbers in a man page mean? So, for example, when I type man ls I see LS(1). But if I type man apachectl I see APACHECTL(8) and if I type man cd I end up with cd(n).  I'm wondering what the significance of the numbers in the parentheses are, if they have any. Show that if $|f(z)| \leq M |z|^n$ then $f$ is a polynomial max degree n I can't prove this statement, can anybody show me how to prove it?    $$f:\mathbb{C}\rightarrow \mathbb{C} \in \mathcal{O}(\mathbb{C}), \exists n\in \mathbb{N}, R &gt;0 , M&gt;0 : |f(z)| \le M|z|^{n} \ \ \forall |z|&gt;R \Rightarrow \deg(f)\le n $$  To show is that if there exists such an $M$, that then $f$ is a polynomial of max degree $n$. I started like this:   $$f(z) = \sum_{n=0}^{\infty} a_n (z-z_0)^n$$ So if I put this into the inequality: $$|f(z)| = \left| \sum_{n=0}^{\infty} a_n (z-z_0)^n \right| \le M |z|^n .$$ What is the difference between 'can', 'could', 'may' and 'might'? I'm a native English speaker and I've been doing some research into English grammar for a programme I'm working on. However, on looking into modal verbs, I've only just come to appreciate how subtle and complicated they are.  What is the difference between these sentences? What possible different response might one have to each?        I can go to the cinema tonight.   I could go to the cinema tonight.   I may go to the cinema tonight.   I might go to the cinema tonight.      Granted, the first of these looks more like a statement of ability or possibility than the three that follow. But in what ways does it differ from the others, since all suggest ability and possibility? And crucially, what distinguishes the other three.  Perhaps it's a matter of tone, of context, of the speaker... The question may appear quite simple but I haven't found a satisfactory answer elsewhere.    looked quite useful in that it covered levels of confidence and certainty but it was talking about the potential state of something rather than the possibility for the person who is speaking. Are there any differences in that sense?   looked at these verbs used in question form (e.g., differences between 'Can I...?' and 'May I...?') but I'm not so interested in that for now.  Any thoughts appreciated. std::auto_ptr to std::unique_ptr How to repair a broken .EXE file association "is a field and used like a type" error Open source computer algebra systems for general relativity I would like to use an open source computer algebra system (CAS) for research in general relativity. Kindly suggest a good choice between Ipython and Sage. I am more interested in the theoretical aspect rather than numerical computations. Finding a ray that intersects a polygon as many times as possible bash: ./program: cannot execute binary file: Exec format error Which $n$ can be expressed as $x^2 − y^2$? Which $n$ can be expressed as $x^2 − y^2$?  Attempt: Any number which is odd or or which is a multiple of 4. Boot stuck at logo screen after upgrading to 14.10 Is the term "errored out" a grammatically correct phrase, or just a colloquial one? How long can I store soaked beans before cooking? Fullscreen YouTube video crash Let $a\in \mathbb{Z}$ be such that $a=b^2+c^2$ where, $b,c \in \mathbb{Z}-\{0\}$. Then $a$ cannot be written as    $pd^2$ where $d \in \mathbb{Z}$ and $p$ is prime with $p \equiv 1~(mod~4)$ $pd^2$ where $d \in \mathbb{Z}$ and $p$ is prime with $p \equiv 3~(mod~4)$ $pqd^2$ where $d \in \mathbb{Z}$ and $p,q$ are primes with $p \equiv 1~(mod~4)$ and $q \equiv 3~(mod~4)$ $pqd^2$ where $d \in \mathbb{Z}$ and $p,q$ are primes with $p,q \equiv 3~(mod~4)$   The similar question with some part missing is asked , but I want to discuss additional options too.  Here, I got 1) is false Since $13=3^2+2^2$ For other options What should I do How do I split a string so I can access item x? Is it just for me, or has the privilege to review new posts and late answers changed for everyone?  I recently started being more active in reviewing, but the rep requirement just shot up and I can not review anymore? The rep requirement has shot up to 500. Is this just for me, as a 'process", or has it been changed site wide? A world where nothing can be hidden. 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