Don't understand the question's wording (1 Viewer)

[leehuan]

So the question just said to show that Q = (cos, -sin // sin cos) is orthogonal, and that x (in R²) and Qx are equidistant. Easy.

What does it mean here to say that Q acts as a rotation on R²?

 

[InteGrand]

So the question just said to show that Q = (cos, -sin // sin cos) is orthogonal, and that x (in R²) and Qx are equidistant. Easy.

What does it mean here to say that Q acts as a rotation on R²?

Given any point x in R², the point Qx is the point x rotated by some constant angle (what angle?) counter-clockwise about the origin. Such a matrix Q is called a rotation matrix for this reason.

This fact can be deduced by either geometrical means or through complex numbers. Here's how it's done via complex numbers.

We can think of a point x = (x,y) in the plane as z = x + iy.

Then let z' = x' + iy' be the point obtained by rotating z counterclockwise by angle θ about the origin. From common HSC 4U complex numbers knowledge, we know z' = (cos(θ) + i*sin(θ))*(x + iy), so

z' = (cos(θ).x – sin(θ).y) + i*(sin(θ).x + cos(θ).y) = x' + i*y'. Now we can equate coefficients.

So x' := (x', y') = (cos(θ).x – sin(θ).y, sin(θ).x + cos(θ).y) = Q (x,y) (by definition of matrix multiplication and the definition of Q)

= Qx.

In other words, the point x' that is a rotation of x by angle θ counterclockwise about the origin is given by Qx.

 

[leehuan]

(edit after realising edit above)

 

[InteGrand]

Er... hmm how do I prove that there is a rotation then? I haven't been taught dot product yet (ignoring my 1 day spent on it during holidays).

I edited it in using complex numbers. It is clear though that it has to be some rotation for a given fixed x (at the very least a trivial one of angle 0) from the fact that the length is preserved, which means it's constrained to the circle it started on.

 

[KingOfActing]

I'm not sure if this is rigorous but if A and B are equidistant from C, then A can be rotated around C by some angle to get B (imagine a circle with center C and radius AC).

 

[leehuan]

Oh right. That sounds logical. But yeah I wonder what can be said about the proof behind it though.

Just thinking about it logically it has to be a rotation by theta specifically so...

 

[dan964]

Er... hmm how do I prove that there is a rotation then? I haven't been taught dot product yet (ignoring my 1 day spent on it during holidays).

I might quickly define it (the joys of second year maths in first year when they dump it on you)....
The dot product (I think called such for only and possibly ) or inner product (more generally is)
generally for


Alternatively

We can define
as the length of the vector x.