Y1_UniMaths
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Can someone help me with 9 iii please. I’ve worked out the enlargement SF as root2 but not sure how to work out the rotation thanks.
https://imgur.com/a/uUoOMbk
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Notnek
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(Original post by Y12_FurtherMaths)
Can someone help me with 9 iii please. I’ve worked out the enlargement SF as root2 but not sure how to work out the rotation thanks.
https://imgur.com/a/
Do you know the general form of a rotation matrix?
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Y1_UniMaths
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(Original post by Notnek)
Do you know the general form of a rotation matrix?
Yes.
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Notnek
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(Original post by Y12_FurtherMaths)
Yes.
So now you know the enlargement matrix (based on the scale factor) and you have the rotation matrix, you can multiply them and equate them to the given matrix to find \theta. Post your working if you get stuck.
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Y1_UniMaths
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(Original post by Notnek)
So now you know the enlargement matrix (based on the scale factor) and you have the rotation matrix, you can multiply them and equate them to the given matrix. Post your working if you get stuck.
Ok cheers I’ll try that
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RDKGames
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(Original post by Y12_FurtherMaths)
Can someone help me with 9 iii please. I’ve worked out the enlargement SF as root2 but not sure how to work out the rotation thanks.
https://imgur.com/a/uUoOMbk
You can just factor out \sqrt{2} and you will be left with a matrix of rotation.

You can either use the general rotation matrix (which is probably in your formula booklet), or you can draw a small diagram to see that the horizontal basis vector \begin{pmatrix} 1 \\ 0 \end{pmatrix} gets mapped to \begin{pmatrix} \dfrac{1}{\sqrt{2}} \\[0.1cm] \dfrac{1}{\sqrt{2}} \end{pmatrix}. Draw it on, and determine what angle this vector makes with \begin{pmatrix} 1 \\ 0 \end{pmatrix}. (It should be obvious...)
Last edited by RDKGames; 2 years ago
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Notnek
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(Original post by Y12_FurtherMaths)
Ok cheers I’ll try that
Alternatively you can just sketch how the unit vectors are changed by this matrix and that will give you the angle of rotation quickly.
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Notnek
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(Original post by RDKGames)
You can just factor out \sqrt{2} and you will be left with a matrix of rotation.

You can either use the general rotation matrix (which is probably in your formula booklet), or you can draw a small diagram to see that the horizontal basis vector \begin{pmatrix} 1 \\ 0 \end{pmatrix} gets mapped to \begin{pmatrix} \dfrac{1}{\sqrt{2}} \\ \dfrac{1}{\sqrt{2}} \end{pmatrix}. Draw it on, and determine what angle this vector makes with \begin{pmatrix} 1 \\ 0 \end{pmatrix}. (It should be obvious...)
It's probably simpler to consider the original matrix instead i.e. \begin{pmatrix} 1 \\ 0 \end{pmatrix} \rightarrow \begin{pmatrix} 1 \\ 1 \end{pmatrix}.

EDIT: Actually there's not really any difference
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Y1_UniMaths
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(Original post by Notnek)
Alternatively you can just sketch how the unit vectors are changed by this matrix and that will give you the angle of rotation quickly.
How would I do that? I used your method and got 45 degrees anti-clockwise which I’m pretty sure is correct so thanks for that.
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RDKGames
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(Original post by Notnek)
It's probably simpler to consider the original matrix instead i.e. \begin{pmatrix} 1 \\ 0 \end{pmatrix} \rightarrow \begin{pmatrix} 1 \\ 1 \end{pmatrix}.
Indeed, it's just my own little head insisting on decomposing the main matrix into a product of individual transformations :flute:
Just nicer numbers with the original, I guess.
Last edited by RDKGames; 2 years ago
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Notnek
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(Original post by Y12_FurtherMaths)
How would I do that? I used your method and got 45 degrees anti-clockwise which I’m pretty sure is correct so thanks for that.
Yes that's right.

You should know that the columns of a 2x2 matrix are the images of the unit vectors under the matrix so the unit vector \begin{pmatrix} 1 \\ 0 \end{pmatrix} has become \begin{pmatrix} 1 \\ 1 \end{pmatrix} and the unit vector \begin{pmatrix} 0 \\ 1 \end{pmatrix} becomes \begin{pmatrix} -1 \\ 1 \end{pmatrix}.

If you draw a quick sketch showing how these unit vectors change, you'll instantly see the angle of rotation. Also, it's not too hard to see the scale factor from the diagram.
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Y1_UniMaths
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(Original post by Notnek)
Yes that's right.

You should know that the columns of a 2x2 matrix are the images of the unit vectors under the matrix so the unit vector \begin{pmatrix} 1 \\ 0 \end{pmatrix} has become \begin{pmatrix} 1 \\ 1 \end{pmatrix} and the unit vector \begin{pmatrix} 0 \\ 1 \end{pmatrix} becomes \begin{pmatrix} -1 \\ 1 \end{pmatrix}.

If you draw a quick sketch showing how these unit vectors change, you'll instantly see the angle of rotation. Also, it's not too hard to see the scale factor from the diagram.
Ahh I see how that works with the angle of rotation thanks. How can you tell the scale factor though? (from either diagram)? Without working out the lengths of each line
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Notnek
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(Original post by Y12_FurtherMaths)
Ahh I see how that works with the angle of rotation thanks. How can you tell the scale factor though? (from either diagram)? Without working out the lengths of each line
If you were to rotate \begin{pmatrix} 1 \\ 0 \end{pmatrix} 45 degrees it would still have length 1. But \begin{pmatrix} 1 \\ 1 \end{pmatrix} has length \sqrt{2} so the SF must be \sqrt{2}.
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RDKGames
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(Original post by Y12_FurtherMaths)
Ahh I see how that works with the angle of rotation thanks. How can you tell the scale factor though? (from either diagram)? Without working out the lengths of each line
(Original post by Notnek)
If you were to rotate \begin{pmatrix} 1 \\ 0 \end{pmatrix} 45 degrees it would still have length 1. But \begin{pmatrix} 1 \\ 1 \end{pmatrix} has length \sqrt{2} so the SF must be \sqrt{2}.
To add, it is a fairly simple and obvious case with this example. But generally, you would just work out the length of the new vector via everyone's favourite GCSE theorem, and the angle would just be worked out by considering tan of an angle (which doesn't sound specific, because arctanning for the angle isnt always gonna give you the full anticlockwise rotation angle, but this would be part of the process generally)

Inspection can rarely give away the exact values, and this is just one of the nice examples.
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Y1_UniMaths
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(Original post by Notnek)
If you were to rotate \begin{pmatrix} 1 \\ 0 \end{pmatrix} 45 degrees it would still have length 1. But \begin{pmatrix} 1 \\ 1 \end{pmatrix} has length \sqrt{2} so the SF must be \sqrt{2}.
Oh I see thank you for your help
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Y1_UniMaths
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(Original post by RDKGames)
You can just factor out \sqrt{2} and you will be left with a matrix of rotation.

You can either use the general rotation matrix (which is probably in your formula booklet), or you can draw a small diagram to see that the horizontal basis vector \begin{pmatrix} 1 \\ 0 \end{pmatrix} gets mapped to \begin{pmatrix} \dfrac{1}{\sqrt{2}} \\[0.1cm] \dfrac{1}{\sqrt{2}} \end{pmatrix}. Draw it on, and determine what angle this vector makes with \begin{pmatrix} 1 \\ 0 \end{pmatrix}. (It should be obvious...)
This makes sense. Thanks for your input I appreciate it!
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