Finding quaternion representing the rotation from one vector to another

The problem as stated is not well-defined: there is not a unique rotation for a given pair of vectors. Consider the case, for example, where u = <1, 0, 0> and v = <0, 1, 0>. One rotation from u to v would be a pi / 2 rotation around the z-axis. Another rotation from u to v would be a pi rotation around the vector <1, 1, 0>.

Quaternion q;
vector a = crossproduct(v1, v2);
q.xyz = a;
q.w = sqrt((v1.Length ^ 2) * (v2.Length ^ 2)) + dotproduct(v1, v2);

Don't forget to normalize q.

Richard is right about there not being a unique rotation, but the above should give the "shortest arc," which is probably what you need.

I'm not much good on Quaternion. However I struggled for hours on this, and could not make Polaris878 solution work. I've tried pre-normalizing v1 and v2. Normalizing q. Normalizing q.xyz. Yet still I don't get it. The result still didn't give me the right result.

In the end though I found a solution that did. If it helps anyone else, here's my working (python) code:

def diffVectors(v1, v2):
""" Get rotation Quaternion between 2 vectors """
v1.normalize(), v2.normalize()
v = v1+v2
v.normalize()
angle = v.dot(v2)
axis = v.cross(v2)
return Quaternion( angle, *axis )

A special case must be made if v1 and v2 are paralell like v1 == v2 or v1 == -v2 (with some tolerance), where I believe the solutions should be Quaternion(1, 0,0,0) (no rotation) or Quaternion(0, *v1) (180 degree rotation)

Half-Way Vector Solution

I came up with the solution that I believe Imbrondir was trying to present (albeit with a minor mistake, which was probably why sinisterchipmunk had trouble verifying it).

Given that we can construct a quaternion representing a rotation around an axis like so:

q.w == cos(angle / 2)
q.x == sin(angle / 2) * axis.x
q.y == sin(angle / 2) * axis.y
q.z == sin(angle / 2) * axis.z

And that the dot and cross product of two normalized vectors are:

dot     == cos(theta)
cross.x == sin(theta) * perpendicular.x
cross.y == sin(theta) * perpendicular.y
cross.z == sin(theta) * perpendicular.z

Seeing as a rotation from u to v can be achieved by rotating by theta (the angle between the vectors) around the perpendicular vector, it looks as though we can directly construct a quaternion representing such a rotation from the results of the dot and cross products; however, as it stands, theta = angle / 2, which means that doing so would result in twice the desired rotation.

One solution is to compute a vector half-way between u and v, and use the dot and cross product of u and the half-way vector to construct a quaternion representing a rotation of twice the angle between u and the half-way vector, which takes us all the way to v!

There is a special case, where u == -v and a unique half-way vector becomes impossible to calculate. This is expected, given the infinitely many "shortest arc" rotations which can take us from u to v, and we must simply rotate by 180 degrees around any vector orthogonal to u (or v) as our special-case solution. This is done by taking the normalized cross product of u with any other vector not parallel to u.

Pseudo code follows (obviously, in reality the special case would have to account for floating point inaccuracies -- probably by checking the dot products against some threshold rather than an absolute value).

Also note that there is no special case when u == v (the identity quaternion is produced -- check and see for yourself).

// N.B. the arguments are _not_ axis and angle, but rather the
// raw scalar-vector components.
Quaternion(float w, Vector3 xyz);


Quaternion get_rotation_between(Vector3 u, Vector3 v)
{
// It is important that the inputs are of equal length when
// calculating the half-way vector.
u = normalized(u);
v = normalized(v);


// Unfortunately, we have to check for when u == -v, as u + v
// in this case will be (0, 0, 0), which cannot be normalized.
if (u == -v)
{
// 180 degree rotation around any orthogonal vector
return Quaternion(0, normalized(orthogonal(u)));
}


Vector3 half = normalized(u + v);
return Quaternion(dot(u, half), cross(u, half));
}

The orthogonal function returns any vector orthogonal to the given vector. This implementation uses the cross product with the most orthogonal basis vector.

Vector3 orthogonal(Vector3 v)
{
float x = abs(v.x);
float y = abs(v.y);
float z = abs(v.z);


Vector3 other = x < y ? (x < z ? X_AXIS : Z_AXIS) : (y < z ? Y_AXIS : Z_AXIS);
return cross(v, other);
}

Half-Way Quaternion Solution

This is actually the solution presented in the accepted answer, and it seems to be marginally faster than the half-way vector solution (~20% faster by my measurements, though don't take my word for it). I'm adding it here in case others like myself are interested in an explanation.

Essentially, instead of calculating a quaternion using a half-way vector, you can calculate the quaternion which results in twice the required rotation (as detailed in the other solution), and find the quaternion half-way between that and zero degrees.

As I explained before, the quaternion for double the required rotation is:

q.w   == dot(u, v)
q.xyz == cross(u, v)

And the quaternion for zero rotation is:

q.w   == 1
q.xyz == (0, 0, 0)

Calculating the half-way quaternion is simply a matter of summing the quaternions and normalizing the result, just like with vectors. However, as is also the case with vectors, the quaternions must have the same magnitude, otherwise the result will be skewed towards the quaternion with the larger magnitude.

A quaternion constructed from the dot and cross product of two vectors will have the same magnitude as those products: length(u) * length(v). Rather than dividing all four components by this factor, we can instead scale up the identity quaternion. And if you were wondering why the accepted answer seemingly complicates matters by using sqrt(length(u) ^ 2 * length(v) ^ 2), it's because the squared length of a vector is quicker to calculate than the length, so we can save one sqrt calculation. The result is:

q.w   = dot(u, v) + sqrt(length_2(u) * length_2(v))
q.xyz = cross(u, v)

And then normalize the result. Pseudo code follows:

Quaternion get_rotation_between(Vector3 u, Vector3 v)
{
float k_cos_theta = dot(u, v);
float k = sqrt(length_2(u) * length_2(v));


if (k_cos_theta / k == -1)
{
// 180 degree rotation around any orthogonal vector
return Quaternion(0, normalized(orthogonal(u)));
}


return normalized(Quaternion(k_cos_theta + k, cross(u, v)));
}

From algorithm point of view , the fastest solution looks in pseudocode

 Quaternion shortest_arc(const vector3& v1, const vector3& v2 )
{
// input vectors NOT unit
Quaternion q( cross(v1, v2), dot(v1, v2) );
// reducing to half angle
q.w += q.magnitude(); // 4 multiplication instead of 6 and more numerical stable


// handling close to 180 degree case
//... code skipped


return q.normalized(); // normalize if you need UNIT quaternion
}

Be sure that you need unit quaternions (usualy, it is required for interpolation).

NOTE: Nonunit quaternions can be used with some operations faster than unit.

Why not represent the vector using pure quaternions? It's better if you normalize them first perhaps.
q₁ = (0 u_x u_y u_z)'
q₂ = (0 v_x v_y v_z)'
q₁ q_rot = q₂
Pre-multiply with q₁^-1
q_rot = q₁^-1 q₂
where q₁^-1 = q₁^conj / q_norm
This is can be thought of as "left division". Right division, which is not what you want is:
q_rot,right = q₂^-1 q₁

Some of the answers don't seem to consider possibility that cross product could be 0. Below snippet uses angle-axis representation:

//v1, v2 are assumed to be normalized
Vector3 axis = v1.cross(v2);
if (axis == Vector3::Zero())
axis = up();
else
axis = axis.normalized();


return toQuaternion(axis, ang);

The toQuaternion can be implemented as follows:

static Quaternion toQuaternion(const Vector3& axis, float angle)
{
auto s = std::sin(angle / 2);
auto u = axis.normalized();
return Quaternion(std::cos(angle / 2), u.x() * s, u.y() * s, u.z() * s);
}

If you are using Eigen library, you can also just do:

Quaternion::FromTwoVectors(from, to)

Working just with normalized quaternions, we can express Joseph Thompson's answer in the follwing terms.

Let q_v = (0, u_x, v_y, v_z) and q_w = (0, v_x, v_y, v_z) and consider

q = q_v * q_w = (-u dot v, u x v).

So representing q as q(q_0, q_1, q_2, q_3) we have

q_r = (1 - q_0, q_1, q_2, q_3).normalize()

According to the derivation of the quaternion rotation between two angles, one can rotate a vector u to vector v with

function fromVectors(u, v) {


d = dot(u, v)
w = cross(u, v)


return Quaternion(d + sqrt(d * d + dot(w, w)), w).normalize()
}

If it is known that the vectors u to vector v are unit vectors, the function reduces to

function fromUnitVectors(u, v) {
return Quaternion(1 + dot(u, v), cross(u, v)).normalize()
}

Depending on your use-case, handling the cases when the dot product is 1 (parallel vectors) and -1 (vectors pointing in opposite directions) may be needed.

The Generalized Solution

function align(Q, u, v)
U = quat(0, ux, uy, uz)
V = quat(0, vx, vy, vz)
return normalize(length(U*V)*Q - V*Q*U)

To find the quaternion of smallest rotation which rotate u to v, use

align(quat(1, 0, 0, 0), u, v)

Why This Generalization?

R is the quaternion closest to Q which will rotate u to v. More importantly, R is the quaternion closest to Q whose local u direction points in same direction as v.

This can be used to give you all possible rotations which rotate from u to v, depending on the choice of Q. If you want the minimal rotation from u to v, as the other solutions give, use Q = quat(1, 0, 0, 0).

Most commonly, I find that the real operation you want to do is a general alignment of one axis with another.

// If you find yourself often doing something like
quatFromTo(toWorldSpace(Q, localFrom), worldTo)*Q
// you should instead consider doing
align(Q, localFrom, worldTo)

Example

Say you want the quaternion Y which only represents Q's yaw, the pure rotation about the y axis. We can compute Y with the following.

Y = align(quat(Qw, Qx, Qy, Qz), vec(0, 1, 0), vec(0, 1, 0))


// simplifies to
Y = normalize(quat(Qw, 0, Qy, 0))

Alignment as a 4x4 Projection Matrix

If you want to perform the same alignment operation repeatedly, because this operation is the same as the projection of a quaternion onto a 2D plane embedded in 4D space, we can represent this operation as the multiplication with 4x4 projection matrix, A*Q.

I = mat4(
1, 0, 0, 0,
0, 1, 0, 0,
0, 0, 1, 0,
0, 0, 0, 1)


A = I - leftQ(V)*rightQ(U)/length(U*V)


// which expands to
A = mat4(
1 + ux*vx + uy*vy + uz*vz, uy*vz - uz*vy, uz*vx - ux*vz, ux*vy - uy*vx,
uy*vz - uz*vy, 1 + ux*vx - uy*vy - uz*vz, uy*vx + ux*vy, uz*vx + ux*vz,
uz*vx - ux*vz, uy*vx + ux*vy, 1 - ux*vx + uy*vy - uz*vz, uz*vy + uy*vz,
ux*vy - uy*vx, uz*vx + ux*vz, uz*vy + uy*vz, 1 - ux*vx - uy*vy + uz*vz)


// A can be applied to Q with the usual matrix-vector multiplication
R = normalize(A*Q)

//LeftQ is a 4x4 matrix which represents the multiplication on the left
//RightQ is a 4x4 matrix which represents the multiplication on the Right
LeftQ(w, x, y, z) = mat4(
w, -x, -y, -z,
x,  w, -z,  y,
y,  z,  w, -x,
z, -y,  x,  w)


RightQ(w, x, y, z) = mat4(
w, -x, -y, -z,
x,  w,  z, -y,
y, -z,  w,  x,
z,  y, -x,  w)