Rotating vectors and matrices

[Val di Vera]

I want to find the rotations needed to rotate one unit vector into another unit vector and then use these rotations to rotate a 3x3 matrix.

For example: I want to determine the rotations needed to rotate [1 0 0] into [-0.342, -0.938, 0.0566] and apply the same rotation to the matrix $$M$$ =

(1 0 0)
(0 2 0)
(0 0 3)

The way I've thought of doing this is to:

1. Rotate [1 0 0] about the z-axis by the angle arctan( $$\frac{0.938}{0.342}$$ ) to get [-0.3425 -0.9395 0]. Apply the same rotation to $$M$$.
2. Take the cross product between [-0.3425 -0.9395 0] and [-0.342 -0.938 0.0566] to get a new axis of rotation $$\hat{r}$$.
3. The new angle of rotation should be $$\hat{\theta}$$ = arctan($$\frac{0.0566}{\sqrt{0.3425^{2} + 0.9395^{2}}}$$).
4. Apply Rodriguez's rotation formula by $$\hat{\theta}$$ about $$\hat{r}$$ to $$M$$

I hope it's clear what I'm trying to do. If anyone can confirm that I'm doing this correctly, or come up with a better way of doing this, I'd very much appreciate it.

Thanks!

 

[pkleinod]

Hello and welcome to PF!
To rotate one vector into another one, you
need only a single rotation in the plane containing the two vectors.
Here is a general method of finding such a single rotation.

The angle of rotation is a
bivector whose direction specifies the plane of rotation and whose
magnitude specifies how much to rotate.

Let $$a$$ and $$b$$ be two unit vectors in 3D space.
Since any unit vector multiplied by itself is just equal to the
square of its magnitude, $$a^2=b^2=1$$ , so it follows that
$$a = ab^2 = (ab) b = (a \cdot b + a \wedge b) b.$$
i.e. the geometric product $$ab$$ rotates the vector $$b$$
into the vector $$a$$ . The product may be written in terms of the
angle of rotation, the bivector $$\mathbf A$$ .
Write $${\mathbf A} = \mid {\mathbf A} \mid \widehat{\mathbf A}$$ , where
$$\mid {\mathbf A}\mid$$ is the magnitude of the
rotation angle and $$\widehat{\mathbf A}$$ is the unit
bivector specifying the plane of rotation. Using the fact that
$$\widehat{\mathbf A}^2 = -1$$ , the product can be expressed as
$$ab = e^{\mathbf A}= \cos{\mathbf A} + \sin{\mathbf A}.$$
$$ab = \cos{\theta} + \widehat{\mathbf A } \sin{\theta}.$$
It only remains to identify the scalar and bivector parts of this with
$$a\cdot b$$ and $$a\wedge b$$ (which you know) in order
to get the sine and cosine of the rotation angle and the plane of
rotation.

Any vector in the rotation plane, $$\widehat{\mathbf A}$$,
may therefore be rotated through the angle $$\theta$$ by
pre-multiplying it with the geometric product $$ab$$ . Any
vector perpendicular to this plane remains unaltered by this
multiplication; hence, to rotate some arbitrary vector $$x$$ in the same
way that you rotated the vector $$b$$ , you must first find
its components parallel and perpendicular to the plane of rotation:
$$x = x_\parallel + x_\perp$$
The rotated vector is then
$$x' = x_\perp + ab x_\parallel$$ .
You can get the two components from
$$x_\parallel = (x\cdot \widehat{\mathbf A})\widehat{\mathbf A}^{-1}$$
$$x_\perp = (x\wedge \widehat{\mathbf A})\widehat{\mathbf A}^{-1}$$ .

There is an alternative way to rotate an arbitrary vector. Let
$$R = \cos{\theta/2} + \widehat{\mathbf A} \sin{\theta/2}$$
$$R^\dagger = \cos{\theta/2} - \widehat{\mathbf A } \sin{\theta/2}$$
The rotated vector is then
$$x' = R^\dagger x R$$ .

That's it, but it may be useful to spell this out somewhat. Let
$${e_1,e_2,e_3}$$ be a set of orthonormal vectors spanning the
space. These have the properties $$e_i^2=1$$
and $$e_i e_j = -e_j e_i$$ . The unit vectors defining the rotation
are then
$$a = a_1e_1 + a_2e_2 + a_3e_3$$
$$b = b_1e_1 + b_2e_2 + b_3e_3$$
The dot product and wedge products are
$$a\cdot b = a_1b_1 + a_2b_2 + a_3b_3 = \cos{\theta}$$
$$a\wedge b = {\mathbf A} = A_3 e_1e_2 + A_1 e_2e_3 + A_2 e_3e_1$$
where $$A_3=a_1b_2 - a_2b_1$$ , with similar expressions for
$$A_1$$ and $$A_2$$ . The magnitude of the bivector
$$\mathbf A$$ is
$$\mid {\mathbf A} \mid = \sqrt{A_1^2+A_2^2 + A_3^2}=\sin{\theta}$$
and the unit plane of rotation is
$$\widehat{\mathbf A}= {\mathbf A} / \sin{\theta}.$$
This should be enough to get you started.