Glide plane, composite isometries

Maths and statistics discussion, revision, exam and homework help.

This thread is sponsored by:
Announcements Posted on
Important: please read these guidelines before posting about exams on The Student Room 28-04-2013
Sign in to Reply
  1. SubAtomic's Avatar
    • Exalted and Worshipped Member
    • Posts: 1,343
    Glide plane, composite isometries
    Hi, loving the isometries at the mo I must say Hopefully it is me being simple again.

    Here's the question

    Let q be the transformation that reflects the plane in the x-axis and let r be the transformation that rotates the plane clockwise about the origin through a quarter of a turn. Algebraically these have the forms

    \displaystyle \boxed{ \begin{array} \ q : \mathbb{R}^2 \longrightarrow \mathbb{R}^2 \\ (x,y)\mapsto (x,-y)\end{array} \ \mbox{and} \begin{array} \ r : \mathbb{R}^2 \longrightarrow \mathbb{R}^2 \\ (x,y) \mapsto(y,-x) \end{array}}

    Determine the composite isometries r \circ q and  q \circ r , and interpret them geometrically.

    Now I can work out how \displaystyle r(q(x,y)) \implies r(x,-y)=(-y,-x)

    But I cannot understand how q(r(x,y)) \implies q(y,-x)=(y,x)

    Don't understand how this bit works \displaystyle q(y,-x)=(y,x)

    not clicking
  2. ghostwalker's Avatar
    • Outcast of Imrryr
    • Location: CA13
    Re: Glide plane, composite isometries
    Perhaps using different symbols might help:

    q((a,b)) = (a,-b)

    And just replace a with x, and b with -y.
  3. jack.hadamard's Avatar
    • Benevolent Member
    • Posts: 736
    Re: Glide plane, composite isometries
    Well, if I understand it correctly, then

    q(r(x,y)) = q(y, -x) = (y, -(-x))

    is just a sign flip of the ``y"-coordinate.
  4. SubAtomic's Avatar
    • Exalted and Worshipped Member
    • Posts: 1,343
    Re: Glide plane, composite isometries
    Hmm, but I thought that by doing a similar thing like I did with the first thing I would end up with q(y,-x)=(-x,-y), daft I know, but it doesn't click.
  5. ghostwalker's Avatar
    • Outcast of Imrryr
    • Location: CA13
    Re: Glide plane, composite isometries
    (Original post by SubAtomic)
    doing a similar thing like I did with the first thing
    You'd have to elaborate on what you mean by that.

    Another way to look at it is:

    q simply changes the second coordinate to the negative of itself.

    r flips the order of the two coordinates and then forms the negative of the new second coordinate.
  6. nuodai's Avatar
    • PS Helper
    • TSR Legend
    Re: Glide plane, composite isometries
    I think you're suffering from the irritatingly ambiguous terminology that people (including myself) use when referring to the x-axis and y-axis, whereby they still call them the 'x-axis' and 'y-axis' when (x,y) is used to denote a point. So when you follow the image of a point (x,y) through a succession of transformations, the 'x-axis' refers always to the 1st coordinate and not to the letter x, and the 'y-axis' refers always to the 2nd coordinate and not to the letter y.

    So for instance q(y,x) = (y,-x) and not (-y,x), even though in the definition of q it is the component labelled 'y' which changes sign. Likewise, q(x+y,x-y) = (x+y, y-x), and q(1,2) = (1,-2), and so on.
    Last edited by nuodai; 21-07-2012 at 16:10.
  7. SubAtomic's Avatar
    • Exalted and Worshipped Member
    • Posts: 1,343
    Re: Glide plane, composite isometries
    (Original post by ghostwalker)
    ...

    (Original post by nuodai)
    ...
    Yep, that cleared it up, think my book is also a bit ambiguous at times, tells me to make sure I always do something a certain way then the very next page it completely blows the 'always do something a certain way' proposition out the water. Lol and not lol at the same time.

    Could visualise it just couldn't work out what was going on with the algebra.

    Cheers.
  8. nuodai's Avatar
    • PS Helper
    • TSR Legend
    Re: Glide plane, composite isometries
    (Original post by SubAtomic)
    Yep, that cleared it up, think my book is also a bit ambiguous at times, tells me to make sure I always do something a certain way then the very next page it completely blows the 'always do something a certain way' proposition out the water. Lol and not lol at the same time.

    Could visualise it just couldn't work out what was going on with the algebra.
    I can see why you'd be confused: you're taught that if f is a function then f(a) is what you get when you substitute a in place of each occurrence of x in the expression for f(x).

    The key here is that q is defined geometrically: that is, it's defined in terms of the basis vectors of the space (i.e. the coordinates themselves) rather than the labelling of the coordinates. You can get around this by considering it as a transformation of vectors instead. If \mathbf{i},\mathbf{j} are unit vectors in the direction of the x- and y-axes, respectively, then you can say that q(\mathbf{i})=\mathbf{i} and q(\mathbf{j})=-\mathbf{j}. Then since (x,y) = x\mathbf{i} + y\mathbf{j} it becomes clearer what happens, namely q(x\mathbf{i} + y\mathbf{j}) = x\mathbf{i} - y\mathbf{j}. Likewise r(\mathbf{i})=\mathbf{j} and r(\mathbf{j})=-\mathbf{i}, and so on.
    Last edited by nuodai; 21-07-2012 at 16:33.
  9. SubAtomic's Avatar
    • Exalted and Worshipped Member
    • Posts: 1,343
    Re: Glide plane, composite isometries
    (Original post by nuodai)
    I can see why you'd be confused: you're taught that if f is a function then f(a) is what you get when you substitute a in place of each occurrence of x in the expression for f(x).

    The key here is that q is defined geometrically: that is, it's defined in terms of the basis vectors of the space (i.e. the coordinates themselves) rather than the labelling of the coordinates. You can get around this by considering it as a transformation of vectors instead. If \mathbf{i},\mathbf{j} are unit vectors in the direction of the x- and y-axes, respectively, then you can say that q(\mathbf{i})=\mathbf{i} and q(\mathbf{j})=-\mathbf{j}. Then since (x,y) = x\mathbf{i} + y\mathbf{j} it becomes clearer what happens, namely q(x\mathbf{i} + y\mathbf{j}) = x\mathbf{i} - y\mathbf{j}. Likewise r(\mathbf{i})=\mathbf{j} and r(\mathbf{j})=-\mathbf{i}, and so on.

    Thanks, much appreciated for that different take on things. This is one of only two questions in the book on composite isometries, takes some getting used to that's for sure.

    All the best:cool:
Sign in to Reply
Share this discussion:  
Article updates
Moderators

We have a brilliant team of more than 60 volunteers looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

Reputation gems:
The Reputation gems seen here indicate how well reputed the user is, red gem indicate negative reputation and green indicates a good rep.
Post rating score:
These scores show if a post has been positively or negatively rated by our members.