Gary
GaryVasco
Posts: 3,352 
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Post by Gary on Sept 15, 2016 23:15:36 GMT
Vasco, As with Ex. 6, the last question seemed the trickiest. nh.ch6.ex7.pdf (217.69 KB) (Revised Sept 21, 2016) Gary |
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Post by Admin on Sept 19, 2016 21:02:58 GMT
Gary
In part (iii) of your solution on line 3 you seem to be assuming that $\textbf{A}\cdot \textbf{P}$ is zero, but $\textbf{P}$ is here an arbitrary quaternion and so the dot product with $\textbf{A}$ is not generally zero. Also near the end of part (iii) you write "Since $\mathbb{A}$ can be any vector orthogonal to $\Pi_{A}$,...", but $\mathbb{A}$ is fixed - orthogonal to $\Pi_{A}$ and passing through the origin.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Sept 19, 2016 23:35:54 GMT
Vasco,
I've been reading over your answers to Ex. 7. You see the relevance of the previous work more clearly than I do, so you are getting your answers more easily. I will have to take some consolation in a few benefits of working by a more roundabout way to see other connections.
In (iii) (c), I believe the use of $\mathbb{P}$ for a vector orthogonal to $\Pi_{A}$ creates a potential confusion with the $\mathbb{P}$ in the first sentence. The second sentence introduces $\mathbf{P}$, which one would expect to be the pure form of $\mathbb{P}$. This $\mathbf{P}$ is orthogonal to A, so it must lie in the plane $\Pi_{A}$.
I realize that the question allows $\mathbb{P}$ to be an arbitrary quaternion, but I think it would be helpful if the question and answers alerted one to these significant changes in $\mathbb{P}$. I rewrote (iii)(c) with this in mind, using $\mathbb{B}$ instead of $\mathbb{P}$.
Also in (iii)(c), is $\mathbb{P} = \alpha\mathbb{A}$ sufficient to describe all the vectors that are orthogonal to $\Pi_{A}$?. What about orthogonal vectors that are translated some distance from $\mathbb{A}$ and therefore don't point toward the origin. Don't we have to take the translation into account? When I add a quaternion in the plane to $\alpha\mathbb{A}$, the result is $-\alpha\mathbb{A} + \mathbb{Q}$, which seems satisfactory.
I see now why you included the orthogonals in (iv). I have included a plot with both the reflections and the quaternion rotation of a random quaternion. I was a bit amazed to see that it worked. Just to see what would happen, I used the Mathematica package Quaternions to do the multiplication, leaving all the v and K values at 0 and converting the I and J values from the real and imaginary parts of the points and back again to plot the result.
I replaced the exercises posted on Sept 15.
Gary
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Post by Admin on Sept 20, 2016 7:15:17 GMT
Vasco, I've been reading over your answers to Ex. 7. You see the relevance of the previous work more clearly than I do, so you are getting your answers more easily. I will have to take some consolation in a few benefits of working by a more roundabout way to see other connections. Gary It seems to me from reading the book, that one of Needham's guiding principles when writing the book was to build as much as possible on previous results, not only recent ones but also those from previous chapters and so I always spend a lot of time looking for them. This is not only true for the text but also the exercises. As a result of this, if my draft paper-and-pencil answer seems long and complicated I tend to always look for another solution - it doesn't always work though. I agree about the confusion here. I was confused at first by P=APA in (ii) and then later in (iii) P'=APA. When I am doing the maths on paper I often use other symbols to help me remember and then go back to Needham's notation when I publish. I think in this case I wrote $\mathbb{P}_A$ for those values of P that lie entirely in the plane $\Pi_A$. I understand what you are saying here but it IS correct to describe all vectors orthogonal to $\Pi_{A}$ as $\alpha\mathbb{A}$, by definition. You are right that the position vector of any point in space can be written as the sum of a vector in the plane $\Pi_{A}$ and a vector parallel to $A$ (hence your $-\alpha\mathbb{A} + \mathbb{Q}$), but that is taken care of in my proof at the end of part (ii)("deduce that (49)...) by combining (b) and (c) - your Q remains fixed, that's (b) and $\alpha\mathbb{A}$ is reversed , that's (c), and so all points in space are reflected in the plane. I'll now go and have a look at your solution. Vasco |
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Post by Admin on Sept 20, 2016 12:01:02 GMT
Gary
Your solution looks good to me except for part (v) where the last 3 or 4 lines may contain a typo or two. I don't understand $\psi/2=-\psi/2=\pi$ which doesn't make sense and also you have changed $-\cos(\psi/2)$ to $\cos(-\psi/2)$ when it should be $\cos(\pi-\psi/2)$. Also, what has happened to $\phi$ introduced at the beginning of part (v)?
It seems to me that $A\cdot B=\cos(\psi/2)$ and $A\times B=V\sin(\psi/2)$.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Sept 20, 2016 14:02:25 GMT
Vasco,
I'll take another look at that.
Gary
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Sept 20, 2016 23:13:42 GMT
Vasco, I've been reading over your answers to Ex. 7. You see the relevance of the previous work more clearly than I do, so you are getting your answers more easily. I will have to take some consolation in a few benefits of working by a more roundabout way to see other connections. Gary It seems to me from reading the book, that one of Needham's guiding principles when writing the book was to build as much as possible on previous results, not only recent ones but also those from previous chapters and so I always spend a lot of time looking for them. This is not only true for the text but also the exercises. As a result of this, if my draft paper-and-pencil answer seems long and complicated I tend to always look for another solution - it doesn't always work though. Vasco, I have prefixed my two responses with *. *My habits are improving in that regard, but not yet where I would like them to be. It is also a question of understanding the meaning of the forms, which takes some practice. I agree about the confusion here. I was confused at first by P=APA in (ii) and then later in (iii) P'=APA. When I am doing the maths on paper I often use other symbols to help me remember and then go back to Needham's notation when I publish. I think in this case I wrote $\mathbb{P}_A$ for those values of P that lie entirely in the plane $\Pi_A$. I understand what you are saying here but it IS correct to describe all vectors orthogonal to $\Pi_{A}$ as $\alpha\mathbb{A}$, by definition. You are right that the position vector of any point in space can be written as the sum of a vector in the plane $\Pi_{A}$ and a vector parallel to $A$ (hence your $-\alpha\mathbb{A} + \mathbb{Q}$), but that is taken care of in my proof at the end of part (ii)("deduce that (49)...) by combining (b) and (c) - your Q remains fixed, that's (b) and $\alpha\mathbb{A}$ is reversed , that's (c), and so all points in space are reflected in the plane. *Yes, I see it now. I'll now go and have a look at your solution. Vasco |
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