rotation matrix and vector - understand step calculation

Question

I have an extremely equation, but I just don't understand which step they made to get to the last line.

${\bf W}$ and ${\bf V}$ are all 3d vectors. A is a rotation matrix. How did they get that scalar multiplication, regular multiplication and that subtraction?

$$\hat {\bf W}_1=A\hat {\bf V}_1$$ $$\left(\frac{\hat{{\bf W}}_2-(\hat{\bf W}_1\cdot\hat{\bf W}_2)\hat{\bf W}_1}{|\hat{\bf W}_2-(\hat{\bf W}_1\cdot\hat{\bf W}_2)\hat{\bf W}_1|}\right) =A\left(\frac{\hat{{\bf V}}_2-(\hat{\bf V}_1\cdot\hat{\bf V}_2)\hat{\bf V}_1}{|\hat{\bf V}_2-(\hat{\bf V}_1\cdot\hat{\bf V}_2)\hat{\bf V}_1|}\right) $$

I think it has something to do with this, but I am not sure: $$\hat {\bf s}_2=\frac{\hat {\bf W}_1\times\hat {\bf W}_2}{|\hat{\bf W}_1\times\hat{\bf W}_2|}$$

Any idea's?

source p408: http://www.malcolmdshuster.com/Pub_2004c_J_dirangs_AAS.pdf

Answer 1

$\newcommand{vec}[1]{\hat{\bf #1}}$ In the linked paper, Equation $(18)$,

$$\left(\frac{\vec W_2-(\vec W_1\cdot\vec W_2)\vec W_1} {|\vec W_2-(\vec W_1\cdot\vec W_2)\vec W_1|}\right) = A \left(\frac{\vec V_2-(\vec V_1\cdot\vec V_2)\vec V_1} {|\vec V_2-(\vec V_1\cdot\vec V_2)\vec V_1|}\right), $$

is apparently not derived from Equation $(17)$, $\hat {\bf W}_1=A\hat {\bf V}_1$. Rather, it is derived directly from Equations $(\mbox{12abcdef})$ and $(13)$.

Equations $(12\mbox{abc})$ state that $$\vec r_1 = \vec V_1, \quad \vec r_2 = \frac{\vec V_1 \times \vec V_2}{|\vec V_1 \times \vec V_2|}, \quad\mbox{and}\quad \vec r_3 = \vec r_1 \times \vec r_2.$$

One can use this description to work out how to express $\vec r_3$ in terms of $\vec V_1$ and $\vec V_2$, or use this identity for the vector triple product:

$$\bf A \times (\bf B \times \bf C) = \bf B(\bf A \cdot \bf C) - \bf C(\bf A \cdot \bf B)$$

(using $\bf A$ for one of the vectors, since that is how this formula is usually presented; no relationship to the matrix $A$) to write

$$\vec V_1 \times (\vec V_1 \times \vec V_2) = \vec V_1(\vec V_1 \cdot \vec V_2) - \vec V_2(\vec V_1 \cdot \vec V_1) = (\vec V_1 \cdot \vec V_2)\vec V_1 - \vec V_2. $$

Therefore $$\vec r_3 = \vec V_1 \times \left(\frac{\vec V_1 \times \vec V_2}{|\vec V_1 \times \vec V_2|}\right) = \frac{(\vec V_1\cdot\vec V_2)\vec V_1 - \vec V_2}{|\vec V_1 \times \vec V_2|}. $$

The construction of $\vec r_3$ (the cross product of two unit vectors) guarantees that $\vec r_3$ is a unit vector, implying that $|\vec V_1 \times \vec V_2| = |\vec V_2 - (\vec V_1 \cdot \vec V_2)\vec V_1|,$ an equation you can confirm by squaring both sides and using the facts $|\vec V_1 \times \vec V_2| = \sin\theta$ and $\vec V_1 \cdot \vec V_2 = \cos\theta$ (since $\vec V_1$ and $\vec V_2$ are unit vectors) where $\theta$ is the angle between the two vectors.

It comes out then that

$$-\vec r_3 = \frac{\vec V_2 - (\vec V_1 \cdot \vec V_2)\vec V_1} {|\vec V_2 - (\vec V_1 \cdot \vec V_2)\vec V_1|}. \tag{$\spadesuit$}$$

The geometric explanation is that $\vec r_2$ is a unit vector perpendicular to the plane of $\vec V_1$ and $\vec V_2$, and $\vec r_3$ is perpendicular to both $\vec r_1$ and $\vec r_2,$ that is, it is back in the plane of $\vec V_1$ and $\vec V_2$, but with no component in the $\vec V_1$ direction. As explained in the first few lines of text after equation $(18)$, the idea was to remove the component of $\vec V_2$ in the direction of $\vec V_1$ (in other words, project $\vec V_2$ onto a line perpendicular to $\vec V_1$) and then rescale the resulting vector so that it is of unit length. As we see from Equation $(\spadesuit)$ above, however, $r_3$ is exactly opposite the vector described in this construction. I suspect that one of the cross-products in Equations $(12\mbox{abc})$ should have been reversed in order to implement the authors' intent.

Similarly, from Equations $(12\mbox{def})$ we find that

$$-\vec s_3 = \frac{\vec W_2 - (\vec W_1 \cdot \vec W_2)\vec W_1} {|\vec W_2 - (\vec W_1 \cdot \vec W_2)\vec W_1|}.$$

Then since $A(-\vec r_3) = -A\vec r_3,$ Equation $(18)$ is then simply a restatement of Equation $(13)$ for the case $k = 3,$ that is,

$$\vec s_3 = A \vec r_3.$$

Answer 2

You have $$ \begin{align} \mathbf W_2-(\mathbf W_1\cdot\mathbf W_2)\mathbf W_1 &= A \mathbf V_2-(A \mathbf V_1\cdot A\mathbf V_2)A \mathbf V_1 \\ &= A \mathbf V_2-(\mathbf V_1\cdot \mathbf V_2)A \mathbf V_1 \\ &= A (\mathbf V_2-(\mathbf V_1\cdot \mathbf V_2) \mathbf V_1) \end{align} $$ where the first equals sign is by definition, the second is because $A$ is a rotation matrix (and so preserves dot product), and the third is because multiplication by $A$ is a linear map.

If we write $\mathbf X = \mathbf V_2-(\mathbf V_1\cdot \mathbf V_2) \mathbf V_1$, all we now need to show is $$ \frac{A\mathbf X}{|A\mathbf X|} = A\frac{\mathbf X}{|\mathbf X|} $$ Here we have $|A\mathbf X|=|\mathbf X|$ because (again) $A$ is a rotation matrix, and finally you can either multiply by the scalar $\frac1{|\mathbf X|}$ before or after you multiply by $A$.