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quaternion basis products

Started by July 25, 2003 03:50 PM
0 comments, last by leinad 21 years, 6 months ago
given i2 = j2 = k2 = -1 = i*j*k therefore i = j*k = -k*j j = k*i = -i*k k = i*j = -j*i I see how we arive at i = j*k, b/c i2 = i*j*k, so i*i/i = i*j*k/i, and finally i = j*k. But I don''t see how you get the sign when you revert the order of multiplication. FYI, I''m getting this out of Andre Lamothe''s great new book. I still can''t find (in this 1700 pg book) where he show you how to use quaternions for camera manipulation.
i = j*k
j*i = j*j*k
j*i = -k
-j*i = k
k = -j*i

(etc)

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