The rotation matrix about y axis should look like
$$\left[ \begin{array}{ccc} \cos\frac{\pi}{2} & 0 &\sin\frac{\pi}{2}\\ 0 & 1 & 0\\ -\sin\frac{\pi}{2} & 0 &\cos\frac{\pi}{2}\\ \end{array} \right] = \left[ \begin{array}{ccc} 0 & 0 &1\\ 0 & 1 & 0\\ -1& 0 &0\\ \end{array} \right] $$
Now, I pictured this rotation, and it should look like rotation about y axis, counterclockwise
I drew $(1,1,1)$ vector in original coordinate system, and in the new coordinate system, the vector should be $(-1,1,1)$
BUT! the matrix calculation gives otherwise.
$$ \left[ \begin{array}{ccc} 0 & 0 &1\\ 0 & 1 & 0\\ -1& 0 &0\\ \end{array} \right] \left[ \begin{array}{c} 1\\ 1\\ 1 \end{array} \right] =\left[ \begin{array}{c} 1\\ 1\\ -1 \end{array} \right] $$
Am I missing something?
Okay, it seems like I should rotate vector instead of coordinate system.
Then, How can you explain rotation about z axis?
The rotation matrix is $$ \left[ \begin{array}{ccc} \cos\frac{\pi}{2} &\sin\frac{\pi}{2} & 0\\ -\sin\frac{\pi}{2} &\cos\frac{\pi}{2} & 0\\ 0 & 0 & 1 \end{array} \right] = \left[ \begin{array}{ccc} 0 & 1 &0\\ -1 & 0 & 0\\ 0& 0 &1\\ \end{array} \right] $$
But... calculation is correct. $$ \left[ \begin{array}{ccc} 0 & 1 &0\\ -1 & 0 & 0\\ 0& 0 &1\\ \end{array} \right] \left[ \begin{array}{c} 1\\ 1\\ 1 \end{array} \right] =\left[ \begin{array}{c} 1\\ -1\\ 1 \end{array} \right] $$
IF you rotate the vector, then in the picture, it should be $(-1,1,1)$
Oh. god.... I think I am looking at the wrong source.
My book tells
well, clearly, googling gives me
I think I solved my case. Thank you all.
$\endgroup$ 83 Answers
$\begingroup$It seems that you are confusing active and passive rotations.
Your matrix $R$, applied to a vector $v$, give you a rotated vector $v'=Rv$ that is expressed in the same reference system than $v$ and this is called an active transformation.
If you don't change the vector but rotate the axis of the referenc system, than this is called a passive transformation and it is represented by the inverse matrix $R^{-1}$, in the sense that the coordinates of the old vector in the new reference systems are $v''=R^{-1}v$.
$\endgroup$ 3 $\begingroup$@Emilio Novati Explained very well about the rotation. But I spent some time draw this picture, so I'll post it.
You misunderstood the rotation matrix. You should rotate the vector counterclockwisely, instead of rotate the coordinate system.
I believe that you are viewing the problem backwards. The matrix $$ A=\begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ -1 & 0 & 0 \end{pmatrix} $$ rotates a vector into a new, rotated frame, according to what you are describing with your images. However, what you are asking is, "What matrix brings me from the rotated frame back to the original frame", which would be the inverse of $A$, i.e. $$ A^{-1} = \begin{pmatrix} 0 & 0 & -1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix}. $$
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