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I am reading Kuipers' book, I am struggling with following problem

Consider an XYZ coordinate frame, with the vector $v_0 = (1,1,1)$ through an angle $\phi = 2\pi/3$. It is geometrically clear that such a rotation results in a new frame in which the new x-axis coincides with former Y-axis, the new y-axis coincides with the former Z-axis, and the new z-axis coincides with the initial X-axis.

$(1,0,0) \rightarrow (0,0,1)$

$(0,1,0) \rightarrow (1,0,0)$

$(0,1,0) \rightarrow (0,1,0)$

My question is, could you figure out without any calculations, just by drawing lets say, deduce that $(1,0,0)$ will transform to $(0,0,1)$?, because the author says, "It is geometrically clear that such a rotation results in a new frame ..." In my case, I have to calculate the rotation matrix, write some code and verified numerically that indeed $(1,0,0)$ will transform to $(0,0,1)$ but I can't imagine in my head how each rotation transform the vectors.

user2321651
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1 Answers1

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The author is referring to the rotation through an angle $2\pi/3$ about the vector $v_0$ as what he talks about in section 3.5.1 on page 56. Therefore, if we draw this image enter image description here And look at it in the opposite direction of $v_0$, which is enter image description here you can see that a $2\pi/3$ rotation about $v_0$ is a rotation of the equilateral triangle (look at the vertices). This will result in the transforms you quoted.

justadzr
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    Hi Yourong Zang, sorry my ignorance, i know that this question is silly to answer, but how do you can assure that these tree vectors form an equilateral triangle in the plane perpendicular to the axis vector? – user2321651 Dec 13 '19 at 03:57
  • @user2321651 If I understand your question correctly, you are asking for how to know that points $A, B, C$ form an equilateral triangle on the plane they lie on. Is that right? – justadzr Dec 13 '19 at 05:41
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    @user2321651 In my opinion, to geometrically see this without further rigorous proofs, you can connect the point which $v_0$ is pointing at, $(1,1,1)$ and $A, B, C$ respectively. Then you will be able to see that there are three congruent triangles and thus $A, B, C$ form an equilateral triangle. All these can be done geometrically. (I believe that you are referring to vectors $\left<(0,0,0), A\right>$, $\left<(0,0,0), B\right>$, $\left<(0,0,0), C\right>$ in your comment) – justadzr Dec 13 '19 at 05:47
  • Ok, I got it, thanks for your help mister. – user2321651 Dec 13 '19 at 14:33
  • @user2321651 You are welcome – justadzr Dec 13 '19 at 14:37