Derivation rotated axis using an angle and axis of rotation

Question

This question is hard to title, I am reading Kuipers' book, in a section called Angle and axis of rotation he developed a method of how to constructor the rotation matrix given two coordinate frames relative to each other, one is initial frame and second one is the rotated frame, then he places the axis of rotation and the angle of rotation, using a geometrically approach he finds out each rotated axis of the rotated frame

Starting from this figure, to compute, for example, the y rotated axis is just a matter of summing the vectors $\overline{Ob} + \overline{be} + \overline{ey}$

In the next figure he show how these three vectors are derived, for example to compute $y$ rotated axis vector

My problem is that i can't figure out why $\overline{be}$ is $(\overline{OY} - \overline{Ob})cos\phi$ instead of $(\overline{OY} - \overline{Ob})sin\beta cos\phi$

Answer 1

After many days I finally did it, indeed, you multiply $(\overline{OY} - \overline{Ob})$ times $sin\beta*cos\phi$ but only if $(\overline{OY} - \overline{Ob})$ vector is unit length!, so the correct writing would be

$$ \frac{(\overline{OY} - \overline{Ob})}{\vert(\overline{OY} - \overline{Ob})\vert}(sin\beta cos\phi) $$

But if you do some algebra magic, this can be rewritten as

$$ (\overline{OY} - \overline{Ob})cos\phi $$

Like it is expressed in the book.

The key that is $v$ is unit vector, so $v_1^2 + v_2^2 + v_3^2 = 1$, also keep in mind that for a given angle, ie: $\theta$, $cos(\theta)^2 + sin(\theta)^2=1$ with theses at hand

First we compute $\overline{Ob}$ vector, that is vector $v$ scaled to length $b$, that is $cos\beta = v \cdot Y$, remember again, that $v$ and $Y$ are unit vectors, so

$$ \overline{Ob} = v*cos\beta = (v_1cos\beta, v_2cos\beta, v_3cos\beta) $$

Now, we can compute $\overline{OY} - \overline{Ob}$

$$\eqalign{ &= (0,1,0) - (v_1cos\beta, v_2cos\beta, v_3cos\beta) \\ &= (-v_1cos\beta, 1 - v_2cos\beta, -v_3cos\beta) }$$

Next, we compute $\vert(\overline{OY} - \overline{Ob})\vert$, here we exploited the fact that $cos\beta = v_2 = v \cdot Y$

$$\eqalign{ &= \sqrt((-v_1cos\beta)^2 + (1 - v_2cos\beta)^2 + (-v_3cos\beta)^2) \\ &= \sqrt((v_1v_2)^2 + (1 - v_2v_2)^2 + (v_3v_2)^2) \\ &= \sqrt((v_1v_2)^2 + (1 - v_2^2)^2 + (v_3v_2)^2) \\ &= \sqrt(v_1^2v_2^2 + 1 - 2v_2^2 + v_2^4 + v_3^2v_2^2) \\ &= \sqrt(v_2^2(v_1^2 + v_2^2 + v_3^3 - 2) + 1) \\ &= \sqrt(v_2^2(1 - 2) + 1) \\ &= \sqrt(1 - v_2^2) \\ &= \sqrt(1 - cos\beta^2) \\ &= \sqrt(sin\beta^2) \\ &= sin\beta \\ }$$

So knowing that its length is $sin\beta$, we can simplify our vector $\overline{be}$ like it is in the book

$$\eqalign{ \frac{(\overline{OY} - \overline{Ob})}{\vert(\overline{OY} - \overline{Ob})\vert}(sin\beta cos\phi) &= \\ \frac{(\overline{OY} - \overline{Ob})}{sin\beta}(sin\beta cos\phi) &= \\ (\overline{OY} - \overline{Ob})cos\phi &= \\ }$$

The same reasoning is done when computing $Y \times v$ vector, that is

$$\eqalign{ \begin{vmatrix} i & j & k \\ 0 & 1 & 0 \\ v_1 & v_2 & v_3 \end{vmatrix} = iv_3 - kv_1 }$$

Now to compute $\vert{Y \times v}\vert$ length, we use the fact that $v_1^2 + v_2^2 + v_3^2 = 1$

$$\eqalign{ &= \sqrt(v_3^2 + (-v_1)^2) \\ &= \sqrt(1-v_1^2 - v_2^2 + 1 - v_2^2 - v_3^2) \\ &= \sqrt(-v_1^2 - v_2^2 - v_3^2 + 2 - v_2^2) \\ &= \sqrt(-1 + 2 - v_2^2) \\ &= \sqrt(1 - v_2^2) \\ &= \sqrt(1 - cos\beta^2) \\ &= \sqrt(sin\beta^2) \\ &= sin\beta \\ }$$

Hope it helps for anyone reading Kuiper's book.