Simplify algebraic vector expressions with dot product and cross product

42 Views Asked by At

I am trying to derive the bending force of a discrete curve, which requires the derivative of angles between two vectors represented tangent half angle. I follow this note for my derivation: https://www.cs.utexas.edu/users/evouga/uploads/4/5/6/8/45689883/turning.pdf (section 3), and I am stuck at simplifying the expression shown as the last step in the note. Basically it says:

$v_2 \times z - \frac{((v_1 \times v_2) \cdot z)(v_1 + v_2)}{1 + v_1 \cdot v_2} = v_1 \times z$

where $v_1$ and $v_2$ are arbitrary unit vectors, and $z$ is the unit vector perpendicular to the $v_1 v_2$ plane. I have no idea why it can be simplified to the expression on the RHS. Does anyone can help?

1

There are 1 best solutions below

0
On

Sometimes it's best not to be so general. Changing coordinates (if you insist on thinking of it that way), take \begin{align*} v_1 &= \vec i \\ v_2 &= \cos\theta \vec i + \sin\theta\vec j \\ z &= \vec k. \end{align*} Now it's just simple algebra to check that (I've multiplied through by $-1$ to make it slightly simpler) $$\cos\theta\vec j - \sin\theta\vec i + \frac{\sin\theta\big((1+\cos\theta)\vec i + \sin\theta\vec j\big)}{1+\cos\theta} = \vec j.$$ Just multiply by $1+\cos\theta$ and simplify.