OMG, Rodrigues' Rotation Formula

Discussion in 'Physics & Math' started by Awoken, Jun 27, 2014.

  1. Awoken Registered Member

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    71
    Rodrigues' Rotation Formula

    Please someone out there help me understand this formula.
    A warning though, I have never worked with Matrices before and this is all new math to me, all I know is I need to figure out this formula so as to be able to incorporate it into a program of mine.

    I'm hoping someone has heard of this formula before and knows how to use it with real world examples.
    I've searched long and hard to find an actual workable example, but it's all just mumbo-jumbo to me as they never show an actual example with tangible numbers, just symbols, terms and expressions that mean a whole lot to people who know what they're doing.

    Here's what I've got, I know the point location of my axis normal, it is (0, 0.7875, -0.615625). The 'plane' which I want to rotate a point about on, passes through the origin at (0, 0, 0).

    If I understand my research, the vector from the origin to the axis normal point is perpendicular to the plane. On the plane I have a point at a right angle to the axis normal and equal distance from the origin as the axis normal point is. it is (0, 0.615625, 0.7875) I'm gonna call this point "point A".

    Question, would it be right of me to assume that point A is located at (0, 0, 1) relative to the plane I'm referring to? When I picture this in my head it seems to make sense.

    Anyways, I want to rotate the position of point A along the plane 30 degrees, I don't really much care which direction unless it makes a big difference in solving. If I rotate point A (0, 0, 1) 30 degrees I get a new location for my point relative to the plane at (0.5, 0, 0.866025). Anyways how would I figure out what the location for point A is if my axis of rotation is the one mentioned above and the point in question is the one mentioned above. I've come across Rodrigues' Rotation Formula which many people seem to say is a piece of pie to work with but means nothing to me.

    I've got NO background in matrices or any other terminology other than what I've briefly mentioned here and even then my knowledge is sketchy at best.

    Anyone who knows how to solve the problem your help would be much appreciated. I need to translate this formula into workable javascript.

    PLEASE do not link me to some page explaining the formula, I've come across many and they all explain the formula using terminology that the educated in this matter understand but I don't. Unless you know of a page showing a real world example dumbed down to a level that a novice could grapple with.

    Thanks in Advance.
     
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  3. rpenner Fully Wired Valued Senior Member

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    I do something similar to Rodrigues' Rotation Formula in some of my discussions about relativity.

    For any complex number, z,
    \(\begin {eqnarray} e^z & = & 1 + z + \frac{z^2}{2} + \frac{z^3}{6} + \frac{z^4}{24} + \dots \\ & = & \sum_{k=0}^{\infty} \frac{z^k}{k!} \\ & = & \sum_{\ell=0}^{\infty} \frac{z^{4 \ell}}{(4 \ell) !} + \sum_{\ell=0}^{\infty} \frac{z^{4 \ell + 1}}{(4 \ell + 1) !} + \sum_{\ell=0}^{\infty} \frac{z^{4 \ell + 2}}{(4 \ell + 2) !} + \sum_{\ell=0}^{\infty} \frac{z^{4 \ell + 3}}{(4 \ell + 3) !} \\ & = & \frac{\cosh z + \cos z}{2} + \frac{\sinh z + \sin z}{2} + \frac{\cosh z - \cos z}{2} + \frac{\sinh z - \sin z}{2} \\ & = & \frac{\cos \, iz \; + \; \cos \, z}{2} + \frac{ -i \, \sin \, iz \; + \; \sin \, z}{2} + \frac{\cos \, iz \; - \; \cos \, z}{2} + \frac{- i \, \sin iz \; - \; \sin \, z}{2}\end{\eqnarray} \) ​

    So if z is pure imaginary, \(z = i \theta\), this reduces to:
    \(\begin {eqnarray} e^{i \theta} & = & \frac{\cos (-\theta) \; + \; \cos (i \theta)}{2} + \frac{ -i \, \sin (-\theta) \; + \; \sin (i \theta)}{2} + \frac{\cos (-\theta) \; - \; \cos (i \theta)}{2} + \frac{- i \, \sin (-\theta) \; - \; \sin (i \theta)}{2} \\ & = & \cos (-\theta) \; - \; i \, \sin( - \theta ) + \frac{ \cos (i \theta) \; + \; \sin (i \theta) \; - \; \cos (i \theta)\; - \; \sin (i \theta)}{2} \\ & = & \cos \, \theta \; + \; i \, \sin \, \theta \end{\eqnarray} \) ​

    As it turns out, a real, anti-symmetric 3×3 matrix, \(M\), has certain nice properties that remind us of the multiplication of pure imaginary numbers:
    \( \begin {eqnarray} M^0 & = & \begin{pmatrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{pmatrix} ^ 0 & = & \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} & = & I \\ M^1 & = & \begin{pmatrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{pmatrix} ^ 1 & = & \begin{pmatrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{pmatrix} & = & \sqrt{x^2 + y^2 + z^2} S \\ M^2 & = & \begin{pmatrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{pmatrix} ^ 2 & = & \begin{pmatrix} -y^2 - z^2 & xy & xz \\ xy & -x^2 - z^2 & yz \\ xz & yz & - x^2 - y^2 \end{pmatrix} & = & -(x^2 + y^2 + z^2 ) C & = & \begin{pmatrix} x \\ y \\ z \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} ^ { \textrm{T} } - \left( \begin{pmatrix} x \\ y \\ z \end{pmatrix} ^ { \textrm{T} } \begin{pmatrix} x \\ y \\ z \end{pmatrix} \right) I \\ M^3 & = & \begin{pmatrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{pmatrix} ^ 3 & = & -(x^2 + y^2 + z^2 ) \begin{pmatrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{pmatrix} & = & -\left(x^2 + y^2 + z^2 \right)^{\tiny \frac{3}{2}} S \\ M^4 & = & \begin{pmatrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{pmatrix} ^ 4 & = & -(x^2 + y^2 + z^2 ) \begin{pmatrix} -y^2 - z^2 & xy & xz \\ xy & -x^2 - z^2 & yz \\ xz & yz & - x^2 - y^2 \end{pmatrix} & = & \left(x^2 + y^2 + z^2 \right)^{\tiny \frac{4}{2}} C \\ M^5 & = & \begin{pmatrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{pmatrix} ^ 5 & = & (x^2 + y^2 + z^2 )^2 \begin{pmatrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{pmatrix} & = & \left(x^2 + y^2 + z^2 \right)^{\tiny \frac{5}{2}} S \end{\eqnarray} \) ​

    This simplifies if \(x^2 + y^2 + z^2 = 1\) because then \(S = M\) and \(C = -M^2\).

    So if we define matrix exponentiation of square matrices similarly to the above definition for real numbers, we calculate:

    \(\begin {eqnarray} e^M & = & I + \left(x^2 + y^2 + z^2 \right)^{\tiny \frac{1}{2}} S + \frac{1}{2} \left(x^2 + y^2 + z^2 \right)^{\tiny \frac{2}{2}} C - \frac{1}{6} \left(x^2 + y^2 + z^2 \right)^{\tiny \frac{3}{2}} S - \frac{1}{24} \left(x^2 + y^2 + z^2 \right)^{\tiny \frac{4}{2}} C + \dots \\ & = & I - C + \sum_{\ell=0}^{\infty} \frac{\left(x^2 + y^2 + z^2 \right)^{\frac{2 \ell}{2}}}{(2 \ell) !} C + \sum_{\ell=0}^{\infty} \frac{\left(x^2 + y^2 + z^2 \right)^{\frac{2 \ell + 1}{2}}}{(2 \ell + 1) !} S \\ & = & I + \left( \cos \sqrt{x^2+y^2+z^2} \; - \; 1 \right) C + \sin \sqrt{x^2+y^2+z^2} S \\ & = & I + \frac{\sin \sqrt{x^2 + y^2 + z^2}}{\sqrt{x^2 + y^2 + z^2}} M + \frac{1 \; - \; \cos \sqrt{x^2+y^2+z^2} }{x^2 + y^2 + z^2} M^2 \end{\eqnarray} \) ​

    So how do apply this if we have point A (0, 197/320, 252/320) and we want to rotate it 30° about the origin, using the vector as the axis of rotation: ( 0, 252/320, -197/320 ) and want the coordinates of the new point?

    Note that the rotation axis is normal to a plane, but is not a unit vector. So we save ourselves a lot of pain by computing the x,y,z of the normalized vector where \(x^2 + y^2 + z^2 = 1\). Then :
    \( x_{\hat{n}} = 0, \quad y_{\hat{n}} = \frac{252}{\sqrt{102313}}, \quad z_{\hat{n}} = -\frac{197}{\sqrt{102313}}, \\ M = S = \begin{pmatrix} 0 & \frac{197}{\sqrt{102313}} & \frac{252}{\sqrt{102313}} \\ -\frac{197}{\sqrt{102313}} & 0 & 0 \\ -\frac{252}{\sqrt{102313}} & 0 & 0 \end{pmatrix} \\ -M^2 = C = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \frac{38809}{102313} & \frac{49644}{102313} \\ 0 & \frac{49644}{102313} & \frac{63504}{102313} \end{pmatrix} \\ A' = e^{ 30^{\circ} M} A = \left( I + \sin \, 30^{\circ} \; M + \left( 1 \; - \; \cos \, 30^{\circ} \right) M^2 \right) A = \left( I + \frac{1}{2} \; M + \frac{ 2 \; - \; \sqrt{3}}{2} M^2 \right) A \\ A' = \begin{pmatrix} \frac{\sqrt{3}}{2} & \frac{197\sqrt{102313}}{204626} & \frac{252\sqrt{102313}}{204626} \\ \frac{-197\sqrt{102313}}{204626} & \frac{127008 + 38809 \sqrt{3}}{204626} & \frac{-99288 + 49644 \sqrt{3}}{204626} \\ \frac{-252\sqrt{102313}}{204626} & \frac{-99288 + 49644 \sqrt{3}}{204626} & \frac{77618 + 63504 \sqrt{3}}{204626} \end{pmatrix} \begin{pmatrix} 0 \\ \frac{197}{320} \\ \frac{252}{320} \end{pmatrix} = \begin{pmatrix} \frac{\sqrt{102313}}{640} \\ \frac{197 \sqrt{3}}{640} \\ \frac{252 \sqrt{3}}{640} \end{pmatrix} \approx \begin{pmatrix} 0.499788 \\ 0.533147 \\ 0.681995 \end{pmatrix}\)​

    Bonus points: prove that this is a rotation matrix that preserves the length of any line segment.

    (Yes, I am using sleazy notation to mix degrees with complex exponentials where the natural unit of angle is radians...)

    Rodrigues' Rotation Formula is a way of getting this same result without the overhead of computing matrices, so it's more efficient when you just want to rotate a few points or know the dot and cross products with the rotation axis vector already.

    For rotations about the origin, we can treat the coordinates A as if they were components of a vector:

    \(\vec{A}' = \cos \theta \, \vec{A} + \frac{\sin \theta}{\sqrt{\vec{n} \cdot \vec{n}}} \vec{n} \times \vec{A} + \left( 1 - \cos \theta \right) \frac{ \vec{n} \cdot \vec{A} }{ \vec{n} \cdot \vec{n} } \vec{n} \)

    (if your rotation axis vector is normalized, this works also: \(\vec{A}' = \cos \theta \, \vec{A} + \sin \theta \, \hat{n} \times \vec{A} + \left( 1 - \cos \theta \right) \left( \hat{n} \cdot \vec{A} \right) \hat{n} \) )


    which gives:
    \(\vec{A}' = \frac{\sqrt{3}}{2} \begin{pmatrix} 0 \\ \frac{197}{320} \\ \frac{252}{320} \end{pmatrix} + \frac{160}{\sqrt{102313} } \begin{pmatrix} 0 \\ \frac{252}{320} \\ -\frac{197}{320} \end{pmatrix} \times \begin{pmatrix} 0 \\ \frac{197}{320} \\ \frac{252}{320} \end{pmatrix} = \begin{pmatrix} 0 \\ \frac{197\sqrt{3}}{640} \\ \frac{252\sqrt{3}}{640} \end{pmatrix} + \frac{160}{\sqrt{102313} } \begin{pmatrix} \frac{252^2 + 197^2}{320^2} \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} \frac{\sqrt{102313}}{640} \\ \frac{197 \sqrt{3}}{640} \\ \frac{252 \sqrt{3}}{640} \end{pmatrix} \)

    http://en.wikipedia.org/wiki/Rodrigues'_rotation_formula
    http://mathworld.wolfram.com/RodriguesRotationFormula.html
     
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  5. Awoken Registered Member

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    71
    Wow

    Right on, tonight I just wanted to see if anyone responded to this behemoth of a question, Thank you good sir.

    I'll have questions, many questions, tomorrow when I start to try and understand all this stuff.
     
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  7. Awoken Registered Member

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    71
    O.k, so I got up this morning and started to stare at your answer some more. I think I've come to understand a few things.
    Because my rotations always happen about the origin then I think what you're telling me is I don't need Rodrigues' Rotation, but instead can use the following instead?

    Hopefully that's true, because already I can make more sense of it.

    K, I think I'm reading this formula right. It's broken up into portions that are equal right?

    \( \vec{A}' = \frac{\sqrt{3}}{2} \begin{pmatrix} 0 \\ \frac{197}{320} \\ \frac{252}{320} \end{pmatrix} + \frac{160}{\sqrt{102313} } \begin{pmatrix} 0 \\ \frac{252}{320} \\ -\frac{197}{320} \end{pmatrix} \times \begin{pmatrix} 0 \\ \frac{197}{320} \\ \frac{252}{320} \end{pmatrix} \) is equal to \( \begin{pmatrix} 0 \\ \frac{197\sqrt{3}}{640} \\ \frac{252\sqrt{3}}{640} \end{pmatrix} + \frac{160}{\sqrt{102313} } \begin{pmatrix} \frac{252^2 + 197^2}{320^2} \\ 0 \\ 0 \end{pmatrix} \) which is equal to the answer which is \( \begin{pmatrix} \frac{\sqrt{102313}}{640} \\ \frac{197 \sqrt{3}}{640} \\ \frac{252 \sqrt{3}}{640} \end{pmatrix} \)

    If I'm right, then...

    Could you be of assistants with this part ?
    How did these two multiply together to get the end result? Or am I'm doing it all wrong?

    \( \begin{pmatrix} 0 \\ \frac{252}{320} \\ -\frac{197}{320} \end{pmatrix} \times \begin{pmatrix} 0 \\ \frac{197}{320} \\ \frac{252}{320} \end{pmatrix} = \begin{pmatrix} \frac{252^2 + 197^2}{320^2} \\ 0 \\ 0 \end{pmatrix} \)

    I think what i want is the cross product? I'm researching and rewritting my responce over and over.

    But if infact this is the only formula I need to use, then I think you've really helped simplify things a lot.
     
  8. rpenner Fully Wired Valued Senior Member

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    4,833
    Let A and B be two vectors. The dot product of A and B is normally given in matrix notation as \(\vec{A} \cdot \vec{B} = A^{\textrm{T}} B\). So what is \(\vec{A}' \cdot \vec{B}'\)?

    \(A' = e^M A, \quad B' = e^M B, \quad \left( A' \right) ^{\textrm{T}} B' = \left( e^M A \right) ^{\textrm{T}} e^M B = A ^{\textrm{T}} \left( e^M \right) ^{\textrm{T}} e^M B \).

    So what is \( \left( e^M \right) ^{\textrm{T}}\)? Well if you remember:

    So by virtue of M being anti-symmetric, \(M^{\textrm{T}} = - M\) and \(\left( M^2 \right)^{\textrm{T}} = \left( M^{\textrm{T}}\right)^2 = \left( - M \right)^2 = M^2\) and so:
    \( \begin{eqnarray} \left( e^M \right) ^{\textrm{T}} & = & \left( I + \frac{\sin \sqrt{x^2 + y^2 + z^2}}{\sqrt{x^2 + y^2 + z^2}} M + \frac{1 \; - \; \cos \sqrt{x^2+y^2+z^2} }{x^2 + y^2 + z^2} M^2 \right) ^{\textrm{T}} \\ & = & I^{\textrm{T}} + \frac{\sin \sqrt{x^2 + y^2 + z^2}}{\sqrt{x^2 + y^2 + z^2}} M^{\textrm{T}} + \frac{1 \; - \; \cos \sqrt{x^2+y^2+z^2} }{x^2 + y^2 + z^2} \left( M^2 \right) ^{\textrm{T}} \\ & = & I - \frac{\sin \sqrt{x^2 + y^2 + z^2}}{\sqrt{x^2 + y^2 + z^2}} M + \frac{1 \; - \; \cos \sqrt{x^2+y^2+z^2} }{x^2 + y^2 + z^2} M^2 \\ & = & e^{-M} \end{eqnarray} \)​

    And because \( - \frac{1}{x^2 + y^2 + z^2} M^2 = \frac{1}{\left( x^2 + y^2 + z^2 \right)^2} M^4 = C\) and \(1 - \sin^2 \theta = \cos^2 \theta\):
    \(\begin{eqnarray} \left( e^{M} \right)^{\textrm{T}} e^M & = & e^{-M} e^M & = & \left( I - \frac{\sin \sqrt{x^2 + y^2 + z^2}}{\sqrt{x^2 + y^2 + z^2}} M + \frac{1 \; - \; \cos \sqrt{x^2+y^2+z^2} }{x^2 + y^2 + z^2} M^2 \right) \left( I + \frac{\sin \sqrt{x^2 + y^2 + z^2}}{\sqrt{x^2 + y^2 + z^2}} M + \frac{1 \; - \; \cos \sqrt{x^2+y^2+z^2} }{x^2 + y^2 + z^2} M^2 \right) \\ & = & I^2 & + & \frac{\sin \sqrt{x^2 + y^2 + z^2}}{\sqrt{x^2 + y^2 + z^2}} M + \frac{1 \; - \; \cos \sqrt{x^2+y^2+z^2} }{x^2 + y^2 + z^2} M^2 \\ & & & - & \frac{\sin \sqrt{x^2 + y^2 + z^2}}{\sqrt{x^2 + y^2 + z^2}} M - \frac{\sin^2 \sqrt{x^2 + y^2 + z^2}}{x^2 + y^2 + z^2} M^2 - \frac{ \sin \sqrt{x^2 + y^2 + z^2} \left(1 \; - \; \cos \sqrt{x^2+y^2+z^2} \right) }{\left(x^2 + y^2 + z^2\right) ^ {\frac{3}{2}}} M^3 \\ & & & +& \frac{1 \; - \; \cos \sqrt{x^2+y^2+z^2} }{x^2 + y^2 + z^2} M^2 + \frac{ \sin \sqrt{x^2 + y^2 + z^2} \left(1 \; - \; \cos \sqrt{x^2+y^2+z^2} \right) }{\left(x^2 + y^2 + z^2\right) ^ {\frac{3}{2}}} M^3 + \frac{1 \; - \;2 \cos \sqrt{x^2+y^2+z^2} \; + \; \cos^2 \sqrt{x^2+y^2+z^2} }{\left( x^2 + y^2 + z^2 \right)^2} M^4 \\ & = & I & + & \frac{1 \; - \; 2 \cos \sqrt{x^2+y^2+z^2} \; + \; 1 \; - \; \sin^2 \sqrt{x^2+y^2+z^2} }{x^2 + y^2 + z^2} M^2 + \frac{1 \; - \;2 \cos \sqrt{x^2+y^2+z^2} \; + \; \cos^2 \sqrt{x^2+y^2+z^2} }{\left( x^2 + y^2 + z^2 \right)^2} M^4 \\ & = & I & - & \left( 1 \; - \;2 \cos \sqrt{x^2+y^2+z^2} \; + \; \cos^2 \sqrt{x^2+y^2+z^2} \right) C + \left( 1 \; - \;2 \cos \sqrt{x^2+y^2+z^2} \; + \; \cos^2 \sqrt{x^2+y^2+z^2} \right) C \\ & = & I \end{eqnarray} \)​

    Thus \(\vec{A}' \cdot \vec{B}' = A ^{\textrm{T}} \left( e^M \right) ^{\textrm{T}} e^M B = A ^{\textrm{T}} I B = A ^{\textrm{T}} B = \vec{A} \cdot \vec{B}\) so the transformation by \(e^M\) preserves all dot-products. Therefore it preserves the dot-product of a vector with itself. Therefore it preserves all lengths. Therefore it preserves all angles between three points. Therefore it is an isometry of Euclidean geometry exactly as expected of a rotation.
     
  9. Awoken Registered Member

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    I think what i want is the cross product? I'm researching and rewritting my responce over and over.
     
  10. Awoken Registered Member

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    71
    O.k, so cross products are as such, let vector column U = ( a,b,c ) and V = ( x,y,z ) then U x V = ( bz-cy, cx-az, ay-bx ) So then bz = 252/320 * 252/320 and cy = -197/320*197/320

    BUT because it's bz-cy the two negatives cancel out and you're left with 252^2 + 197^2 / 320. and 0 times anything else =0, so that's how you end up with the answer.

    O.k I haven't plugged this into code yet so I can't see the result. But it seems really easy to compute, especially with what you've provided me. Thanks a ton.

    Now on to my next question for you good sir.

    So the axis of rotation is perpendicular to the the plane and the point in question rotates about the plane. One could say there is also another plane which all three, the origin, point A and the axis point fall on. and on this plane, call it plane z, the axis point and point A are 90 degrees from one another and equal distances from one another. How would I then rotate the axis and point A 30 degrees along plane z?
     
  11. rpenner Fully Wired Valued Senior Member

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    4,833
    No, this is Rodrigues' Rotation formula for rotations of vectors, or equivalently for points about the coordinate origin.

    If you want to rotate point A about point O with axis vector \(\vec{n}\) and angle \(\theta\) then you want:
    \(A' = O + \cos \theta \, (A - O) + \frac{\sin \theta}{\sqrt{\vec{n} \cdot \vec{n}}} \vec{n} \times (A-O) + \left( 1 - \cos \theta \right) \frac{ \vec{n} \cdot (A-O) }{ \vec{n} \cdot \vec{n} } \vec{n} \)​
    But because the coordinates of the origin are (0,0,0) we have numerical congruence between the coordinates of point A and the components of the vector \(\vec{A} = A - O = A - (0,0,0)\) so we get the simpler formula above.

    That's just the Cross Product -- a peculiarity about 3-D vectors is that there is a way to multiply two vectors and get a third vector which is orthogonal to both of them:

    \( \begin{pmatrix} n_x \\ n_y \\ n_z \end{pmatrix} \times \begin{pmatrix} A_x \\ A_y \\ A_z \end{pmatrix} \equiv \begin{pmatrix} n_y A_z - n_z A_y \\ n_z A_x - n_x A_z \\ n_x A_y - n_y A_x \end{pmatrix}\)

    So Rodrigues' Rotation formula for rotations of vectors when written out for all the components is:

    \(\begin{pmatrix} A'_x \\ A'_y \\ A'_z \end{pmatrix} = \cos \theta \begin{pmatrix} A_x \\ A_y \\ A_z \end{pmatrix} + \frac{ \sin \theta }{ \sqrt{ n_x^2 + n_y^2 + n_z^2 } } \begin{pmatrix} n_y A_z - n_z A_y \\ n_z A_x - n_x A_z \\ n_x A_y - n_y A_x \end{pmatrix} + (1 - \cos \theta ) \frac{ n_x A_x + n_y A_y + n_z A_z }{ n_x^2 + n_y^2 + n_z^2 } \begin{pmatrix} n_x \\ n_y \\ n_z \end{pmatrix} \)​

    15-minute video: http://www.khanacademy.org/math/lin.../v/linear-algebra--cross-product-introduction
    http://tutorial.math.lamar.edu/Classes/CalcII/CrossProduct.aspx
     
  12. Awoken Registered Member

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    71
    YEEES. My camera now rotates about my sphere perfectly. thank you again for all your help.
     

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