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I was given the following information: There is a frame $B$ which is rotated relative to frame $A$ and they share the same origin. The following vectors are in frame $A$: $(2, 1, 2)$ and $(-1, -2, 2)$. These two vectors also happen to lay on the $x$ axis and $z$ axis of Frame $B$ respectively.

I am tasked with finding the axis of rotation, and I also need to find the angle at which the frame $B$ was rotated in relation to frame $A$.

Through my knowledge I was able to find that the angle at which the frame was rotated was approximately 48.2 degrees, but I am unsure if this is correct and am unable to find the axis of rotation. I feel like I am missing something critical yet simple here.

Kolmin
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John
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  • Finding the axis is equivalent to finding the fixed points of the rotation. Solve Rx=x. – Jean-François Gagnon Oct 06 '15 at 22:54
  • Very closely related: http://math.stackexchange.com/questions/1125203/finding-rotation-axis-and-angle-to-align-two-3d-vector-bases and http://math.stackexchange.com/questions/624348/finding-rotation-axis-and-angle-to-align-two-oriented-vectors – David K Oct 06 '15 at 22:59
  • Note that if you multiply each vector by the scalar $\frac13$, the resulting vectors are two basis vectors from an orthonormal basis. – David K Oct 06 '15 at 23:03

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Because they line up with the new axes the vectors $(2,1,2)$ and $(-1,-2,2)$ will (once normalized) be the first and third rows of your rotation matrix respectively. Now you need to find the $y$-axis.

Why is this? The two vectors you are given (along with the third you must find) form an orthogonal basis for your new frame. Normalizing them provides an orthonormal basis. Now take any vector in the old frame. How would you represent it in the new frame? Because the basis is orthonormal you only need to find how much of this vector falls in each of the basis directions. This is just the inner product of the vector with each basis element. In a rotational matrix this is what each row vector is doing.

Once you have the matrix you just need to find the vector space that doesn't move. This is just solving the equation $Mx=x$ or rather $(M-I)x=0$.

Luke
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