Jtdylktuy

I have looked at another page which explains how to rotate a point around the origin for three axis but I don't understand how you do it for just two axis.link

I would like to rotate a point at (x,y,z) around the origin with a rotation around the X-axis and Y-axis. However I would like to go against the right hand rule and define the z-as as being depth, and x and y being the same as on a 2d coordinate plane. Also as Z goes in, the z value increases.

So, the new coordinate would be (x',y',z')

What will the rotation matrix look like, when there is no rotation on the z axis.

An easy way to find what a Transformation does to a set, is to find out what the transformation does to each axis.

Now, let $C={e_1,e_2}$ be the Canonial Basis for $mathbb{R}^2$. ($e_i$ is the vector with 1 in the $ith$ position and 0 in the rest)

$C=begin{pmatrix}
1 & 0\
0 & 1
end{pmatrix}$ would be the representation of this.

Now suppose you want to rotate the whole system about $theta$. The new coordinates of $e_1$ will be: $(costheta,sintheta)$ and given $e_2$ is the orthogonal vector to $e_1$ with both having positive direction, the new coordinates of $e_2$ will be $(-sintheta,costheta)$. (If you have any doubt about how this happens, take a look on the coordinates of a vector over the unitary circumference)

In a more formal way: Let $T:mathbb{R}^2longrightarrowmathbb{R}^2$ the rotation about $theta$, hence:

$[T]=begin{pmatrix}
costheta & -sintheta\
sintheta & costheta
end{pmatrix}$,

where $[T]$ is the matrix associated to the transformation T.

Now, a result in Geommetry and Linear Algebra gives us that:

$T(x)=[T][x]$, where $x$ is a vector on the space over where the Transformation is applied.

Therefore, if $x=(x_1,x_2) in mathbb{R}^2$:

$T(x)=[T][x]=begin{pmatrix}
costheta & -sintheta\
sintheta & costheta
end{pmatrix}begin{pmatrix} x_1\
x_2
end{pmatrix}$.

Your question is about a generalization of what was said before:

Let $B={e_1,e_2,e_3}$ be the Canonial Basis for $mathbb{R}^3$.

$B=begin{pmatrix}
1 & 0 & 0\
0 & 1 & 0\
0 & 0 & 1
end{pmatrix}$,

and $S:mathbb{R}^3longrightarrowmathbb{R}^3$ the rotation about $theta$ without $Z$ axis rotating (i.e, rotation only in XY plane), hence:

$[S]=begin{pmatrix}
costheta & -sintheta & 0\
sintheta & costheta & 0 \
0 & 0 & 1
end{pmatrix}$

and if $y=(y_1,y_2,y_3) in mathbb{R}^3$, then

$S(y)=[S][y]=begin{pmatrix}
costheta & -sintheta & 0\
sintheta & costheta & 0 \
0 & 0 & 1
end{pmatrix}begin{pmatrix} y_1\
y_2 \
y_3
end{pmatrix}$.

¿can you deduce why is $[S]$ equal to this now that I've given you the details of how this works in $mathbb{R}^2$?

Hint: think of this as only a rotation in $mathbb{R}^2$.