Quaternion: Difference between revisions

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→‎Approximating 3D Translations using Double quaternions

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 {{hidden | Algebraic Verification |
-Let <math>q=1-\frac{d}{2}\mathbf{n}</math> and <math>q^* = 1+\frac{d}{2}\mathbf{n}</math>.
+Suppose our point is <math>\mathbf{x} \in \mathbb{R}^3</math> and our translation has direction <math>\mathbf{d} \in \mathbb{R}^3</math> with distance <math>d</math>.
-Then
+Let <math>q=1+\frac{d}{2}\mathbf{n}</math> and <math>q^* = 1-\frac{d}{2}\mathbf{n}</math>.
+<math>
+\begin{aligned}
+q(1+\mathbf{x})(q^*)^* &= (1+\frac{d}{2}\mathbf{n})(1+\mathbf{x})(1+\frac{d}{2}\mathbf{n})\\
+&= [(1 - \frac{d}{2} \mathbf{n} \cdot \mathbf{x}) + (\frac{d}{2}\mathbf{n} + \mathbf{x} + \frac{d}{2}\mathbf{n} \times \mathbf{x})](1+\frac{d}{2}\mathbf{n})\\
+&= \left(1 - \frac{d}{2} \mathbf{n} \cdot \mathbf{x} - ((\frac{d}{2}\mathbf{n} + \mathbf{x} + \frac{d}{2}\mathbf{n} \times \mathbf{x})) \cdot \frac{d}{2} \mathbf{n}\right) + \left(\frac{d}{2}\mathbf{n} + \mathbf{x} + \frac{d}{2}\mathbf{n} \times \mathbf{x} +\frac{d}{2}\mathbf{n} + (\frac{d}{2}\mathbf{n} + \mathbf{x} + \frac{d}{2}\mathbf{n} \times \mathbf{x})\times(\frac{d}{2}\mathbf{n})\right)\\
+&= c + \left(d\mathbf{n} + \mathbf{x} + \frac{d}{2}\mathbf{n} \times \mathbf{x} + (\frac{d}{2}\mathbf{n}\times(\frac{d}{2}\mathbf{n}) + \mathbf{x}\times(\frac{d}{2}\mathbf{n}) + (\frac{d}{2}\mathbf{n} \times \mathbf{x})\times(\frac{d}{2}\mathbf{n}))\right)\\
+&= c + \left(d\mathbf{n} + \mathbf{x} + \frac{d}{2}\mathbf{n} \times \mathbf{x} + (0 + \mathbf{x}\times(\frac{d}{2}\mathbf{n}) + 0)\right)\\
+&= c + \left(d\mathbf{n} + \mathbf{x} + \frac{d}{2}\mathbf{n} \times \mathbf{x} - \frac{d}{2}\mathbf{n}\times \mathbf{x}\right)\\
+&= c + \left(\mathbf{x} + d\mathbf{n}\right)\\
+\end{aligned}
+</math>
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