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    <p>Say you want to rotate A toward B.</p> <p>Take the cross product AxB = C and normalize it.</p> <p><strike>Now break A into two components, one parallel to C and one normal:</strike></p> <pre><code>... </code></pre> <p><strike>Now construct a vector normal to A and C (with the right sense):</strike></p> <pre><code>... </code></pre> <p><strike>Now you can construct the rotated vector:</strike></p> <pre><code>... </code></pre> <p><b>EDIT</b><br> I feel like an idiot. The correct (and more simple) derivation is</p> <pre><code>F = C x A G = cos(theta) A + sin(theta) F </code></pre> <p><b>EDIT:</b><br> This works by simple geometry. C is normal to the plane containing A and B. F is in the plane, and normal to A. So any vector in the plane is a linear combination of A and F; that is, any vector Z in the plane can be constructed as Z = aA + bF, where a and b are numbers, and any such sum will be in the plane. F also has the same magnitude as A, so if we construct</p> <pre><code>G = cos(theta) A + sin(theta) F </code></pre> <p>what we get is a vector with the same magnitude, but separated from A by an angle theta. (This is not immediately obvious, but if you play around with it a little you'll see that it works.)</p> <p>Using your example:</p> <pre><code>A = (2, 3, 3) (magnitude = 4.69) B = (2, -3, -2) C = AxB = (3, 10, -12) (magnitude = 15.906) Now normalize: C = (0.189, 0.629, -0.754) (magnitude = 1.0) F = CxA = (4.149, -2.075, -0.692) (magnitude = 4.69) theta = 20 degrees G = cos(theta) A + sin(theta) F = (3.299, 2.109, 2.583) (magnitude = 4.69) </code></pre> <p>G is in the same plane as A and B (normal to C), and the angle between A and G is 20 degrees. (The angle between A and B is 124.7 degrees, the angle between G and B is 104.7 degrees.)</p>
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    1. COA . (AxB) = 0, so D is 0. Why is it calculated at all?
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    2. COAlso, that results in E = A, F = C x A, which makes F parallel to B. So in effect you're just weighing the vectors A and B using cos(theta) and k*sin(theta). Does not work.
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    3. CO@Ollie Etuaho: thank you, I was wrong, I have corrected my answer. (Your first comment was correct, your second incorrect.)
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