What is the Matrix Notation for a Rotation About the Origin in ℝ2?

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In summary, Simon Bridge provides a summary of the content. He explains that a rotation around the origin in ℝ2 will drive the point P = (4,3) to the point ρ(P) = (3,4). He then asks for help with finding the angle of the rotation and its matrix notation. He solves for the angle using the trigonometric functions and expresses the result in terms of g(x) = Ax + v. Finally, he provides a helpful tip on how to work out the trigonometric functions given a set of constants.
  • #1
phyzz
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Homework Statement



A rotation ρ about the origin in ℝ2 drives the point P = (4,3) to the point ρ(P) = (3,4).
Find the angle of rotation as well as its matrix notation.

Homework Equations



Ok so I made a sketch and I realized I needed to find θ = θ1 - θ2 where θ1 and θ2 equal arctan(4/3) and arctan(3/4) respectively.

I guess it's correct but then how would I work out the sin and cos of that angle θ without a calculator?

SO. I worked out the cosine angle between the vectors 0P and 0ρ(P) and got cosθ = 24/2√25

But now I don't know how to work out the sine!

My goal is to replace everything into the ℝ2 rotation matrix (row 1; row 2):
(cosα -sinα; sinα cosα)

and express as g(x) = Ax + v

Thanks a lot!
 
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  • #2
A point at ##P=(a,o)## is at the end of the hypotenuse of a right-angled triangle with adjacent side ##a## and opposite side ##o## ... so the hypotenuse has length ##h=\sqrt{a^2+o^2}## and you can find sine cosine and tangent by SOH CAH TOA.
 
  • #3
hi phyzz! :smile:
phyzz said:
I guess it's correct but then how would I work out the sin and cos of that angle θ without a calculator?

sec2 = 1 + tan2

cos2 = 1/sec2

sin = cos*tan :wink:
 
  • #4
How do you know that this is a right angle though?
 

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  • #5
tiny-tim said:
hi phyzz! :smile:sec2 = 1 + tan2

cos2 = 1/sec2

sin = cos*tan :wink:

Ohhhhh my bad I didn't see it at first. I just used sin^2 + cos^2 = 1
and the cos of the angle I found between the two vectors.

Thanks!
 
  • #6
... in fact, P=(3,4) and Q=(4,3) each make a special kind of triangle.
The rotation angle is the difference between complimentary angles in this triangle - if you recall your special triangles, you don't need the trig.

Of course, since you know the start and finish, you could just expand the matrix equation into a pair of simultanious equations and solve them?
 
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  • #7
Simon Bridge said:
... in fact, P=(3,4) and Q=(4,3) each make a special kind of triangle.
The rotation angle is the difference between complimentary angles in this triangle - if you recall your special triangles, you don't need the trig.

Of course, since you know the start and finish, you could just expand the matrix equation into a pair of simultanious equations and solve them?

Thank you Simon Bridge!

In other news, when I work out (in this case a is a constant) [tex] \left \| T_{\gamma}(t) \right \| = \left \| \frac{1}{\sqrt{1 + a^{2}}} (-asin(t),\sqrt{1+a^{2}}cos(t),-sin(t)) \right \|[/tex]

I know I can just do [tex]\left \|(-asin(t),\sqrt{1+a^{2}}cos(t),-sin(t)) \right \|[/tex] and then multiply by the constant [tex]\frac{1}{\sqrt{1 + a^{2}}} [/tex]at the very end.

My question:

Given that [tex] N_{\gamma}(t) = \frac{1}{\sqrt{1 + a^{2}}} (-acos(t),-\sqrt{1+a^{2}}sin(t),-cos(t)) [/tex]

and I want to work out [tex] T_{\gamma}(t) \times N_{\gamma}(t) [/tex]

can I just do [tex]\frac{1}{\sqrt{1 + a^{2}}} \frac{1}{\sqrt{1 + a^{2}}}[/tex] multiplied by the result of [tex](-asin(t),\sqrt{1+a^{2}}cos(t),-sin(t))\times (-acos(t),-\sqrt{1+a^{2}}sin(t),-cos(t))[/tex]? Thanks so much again!
 
  • #8
Don't see the connection of all that with your question... in ##T_\gamma(t)\times N_\gamma(t)## - is that ##\times## supposed to be a cross product?

FYI: putting a backslash in front of the function name turns ##sin\theta## into ##\sin\theta##
 
  • #9
Yes, cross product

Thanks for the tip!
 
  • #10
If ##\vec{u}=a(x,y,z)## and ##\vec{v}=b(p,q,r)## then is ##\vec{u}\times\vec{v}=ab## ... is that your question?
Does a cross product usually produce a scalar?
 
  • #11
What I meant was if we have constants in front of two vectors we want to cross product, are we allowed to multiply the constants together and perform the cross product to the vector bits?
Sorry for the vagueness!
 
  • #12
Yes, the "associative" law holds for the cross product: [itex](a\vec{u}\times (b\vec{v})= (ab)(\vec{u}\times\vec{v})[/tex]
 

FAQ: What is the Matrix Notation for a Rotation About the Origin in ℝ2?

What is R2 rotation about (0,0)?

R2 rotation about (0,0) is a type of geometric transformation that involves rotating an object or coordinate system on a two-dimensional plane around the origin point (0,0). This means that all points on the object or coordinate system will move in a circular motion around the origin point.

How is R2 rotation about (0,0) different from other types of rotations?

R2 rotation about (0,0) is different from other types of rotations, such as R1 rotation or R3 rotation, because it only occurs on a two-dimensional plane. This means that the object or coordinate system will only rotate around the origin point on a flat surface, rather than in three-dimensional space.

What is the purpose of R2 rotation about (0,0)?

The purpose of R2 rotation about (0,0) is to change the orientation of an object or coordinate system on a two-dimensional plane. This transformation can be used in various applications, such as computer graphics, robotics, and engineering, to manipulate and analyze different shapes and patterns.

How is R2 rotation about (0,0) represented mathematically?

In mathematics, R2 rotation about (0,0) is typically represented using a rotation matrix, which is a square matrix that describes the transformation of points on a two-dimensional plane. The rotation matrix is composed of sine and cosine values, which determine the amount and direction of rotation.

What are some real-world examples of R2 rotation about (0,0)?

R2 rotation about (0,0) can be seen in various real-world applications, such as rotating a map to match the direction of a compass, rotating the wheels of a car to change its direction, or rotating a satellite dish to align with a specific satellite in space. It is also commonly used in computer graphics to create 2D animations and games.

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