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

[phyzz]

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!

 

[Simon Bridge]

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.

 

[tiny-tim]

hi phyzz!

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

 

[phyzz]

How do you know that this is a right angle though?

 

[phyzz]

hi phyzz! sec² = 1 + tan²

cos² = 1/sec²

sin = cos*tan

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!

 

[Simon Bridge]

... 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?

 

[phyzz]

... 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) $$\left \| T_{\gamma}(t) \right \| = \left \| \frac{1}{\sqrt{1 + a^{2}}} (-asin(t),\sqrt{1+a^{2}}cos(t),-sin(t)) \right \|$$

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

My question:

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

and I want to work out $$T_{\gamma}(t) \times N_{\gamma}(t)$$

can I just do $$\frac{1}{\sqrt{1 + a^{2}}} \frac{1}{\sqrt{1 + a^{2}}}$$ multiplied by the result of $$(-asin(t),\sqrt{1+a^{2}}cos(t),-sin(t))\times (-acos(t),-\sqrt{1+a^{2}}sin(t),-cos(t))$$? Thanks so much again!

 

[Simon Bridge]

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 $\sin\theta$

 

[phyzz]

Yes, cross product

Thanks for the tip!

 

[Simon Bridge]

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?

 

[phyzz]

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!

 

[HallsofIvy]

Yes, the "associative" law holds for the cross product: [itex](a\vec{u}\times (b\vec{v})= (ab)(\vec{u}\times\vec{v})[/tex]