How to get a rotation quaternion from two vectors?

Question

I have two vectors $\vec a$ and $\vec b$ in 3d space. Both of these vectors has length (magnitude) 1 and begin from the origin, so $\vec a$ can be turned into $\vec b$ by a unit quaternion $q$:

$$\vec b = q^{-1} \vec a q$$

So the question is: how can I get this quaternion $q$ that turns vector $\vec a$ into vector $\vec b$? Speaking about why I need this, actually the problem's to rotate a bunch of points, while $\vec a$ and $\vec b$ just set the rotation. And I had to do this exactly using quaternion math, not matrixes or angles.

Any help on how I can solve this would be appreciated, but the better way is to get a rotation quaternion directly without finding a matrix and converting it into a quaternion. Thank in advance!

NOTE: The angle between these two vectors can't be greater than 90°.

Answer 1

I'll use $\hat{a}$ and $\hat{b}$ for the two unit vectors ($\lVert\hat{a}\rVert = 1$ and $\lVert\hat{b}\rVert = 1$. The axis of rotation is then their cross product, and the cosine of the angle is their dot product: $$\begin{aligned} \vec{n} &= \hat{a} \times \hat{b} \\ \hat{n} &= \frac{\vec{n}}{\lVert\vec{n}\rVert} \\ \cos\theta &= \hat{a} \cdot \hat{b} = c, \quad 0 \le \theta \le 180° \\ \cos\frac{\theta}{2} &= \sqrt{\frac{1 + c}{2}} \\ \sin\frac{\theta}{2} &= \sqrt{\frac{1 - c}{2}} \\ \end{aligned}$$ An unit quaternion $\mathbf{q} = (r; i, j, k)$ describes rotation around unit axis vector $\hat{n} = (n_x, n_y, n_z)$ by angle $\theta$, when $$\left\lbrace ~ \begin{aligned} \mathbf{q}_r &= \cos\frac{\theta}{2} = \sqrt{\frac{1 + c}{2}} \\ \mathbf{q}_i &= n_x \sin\frac{\theta}{2} = n_x \sqrt{\frac{1 - c}{2}} \\ \mathbf{q}_j &= n_y \sin\frac{\theta}{2} = n_y \sqrt{\frac{1 - c}{2}} \\ \mathbf{q}_k &= n_z \sin\frac{\theta}{2} = n_z \sqrt{\frac{1 - c}{2}} \\ \end{aligned} \right.$$ As you see, all you need to remember is to normalize the cross product (rotation axis $\vec{n}$) by dividing it by its 2-norm, $\lVert\vec{n}\rVert = \sqrt{\vec{n}\cdot\vec{n}} = \sqrt{n_x^2 + n_y^2 + n_z^2}$), and to halve the rotation angle; I included the two half-angle formulas in the first set for $0 \le \theta \le 180°$.

Also note that the result indeed is an unit quaternion, $\lVert\mathbf{q}\rVert = \sqrt{q_r^2 + q_i^2 + q_j^2 + q_k^2} = 1$. (If weren't isn't, you can always normalize it safely, without introducing any directional bias, by dividing it by its length (that square root). The $q_r = 1$, $q_i = q_j = q_k = 0$ represents no rotation, in case the division yields a non-finite value.)

Answer 2

Let the angle between $a$ and $b$ be $\theta$. Now, there's a formula for the quaternion multiplication that relates it to the cross product and cosine and sine of $\theta$. Note that throughout I'm using the usual representation of $a,b$ and $b\times a$ and $a\times b$ as quaternions with real part zero. (They have the same coefficients as their vector counterparts in front of $i,j,k$.)

The identity I speak of is $$ba=-\cos(\theta)+\sin(\theta)(b\times a)$$

But if we rewrite this,

$$-\cos(\theta)+\sin(\theta)(b\times a)=\\ -\cos(\theta)-\sin(\theta)(a\times b)=\\ -(\cos(\theta)+\sin(\theta)(a\times b)) $$

We recognize this quaternion is precisely (up to a negative sign, which doesn't matter of course!) the rotation around $a\times b$, but unfortunately the angle is two times what we need! So we want this to all wind up with the angle being half of what it is.

What we need is a unit vector bisecting the angle between $a$ and $b$ so that we can obtain the same thing with $\theta/2$, and luckily the answer is geometrically obvious:

$$c=\frac{a+b}{\|a+b\|}$$

Since $q=ca=-(\cos(\theta/2)+\sin(\theta/2)(a\times b))$ it will do exactly what you want: rotate $a$ onto $b$ by rotating the plane $a$ and $b$ span. (So there is no mistake, I refer to $x\mapsto qxq^{-1}$)

To reiterate, the formula is:

$$q=\frac{(a+b)a}{\|a+b\|}$$

There are corner cases, of course, when $a$ and $b$ are linearly independent but I gather you are restricting yourself to a valid case.