Convert a Unit Vector to a Quaternion

Question

So I'm very new to quaternions, but I understand the basics of how to manipulate stuff with them. What I'm currently trying to do is compare a known quaternion to two absolute points in space. I'm hoping what I can do is simply convert the points into a second quaternion, giving me an easy way to compare the two.

What I've done so far is to turn the two points into a unit vector. From there I was hoping I could directly plug in the i j k into the imaginary portion of the quaternion with a scalar of zero. From there I could multiply one quaternion by the other's conjugate, resulting in a third quaternion. This third quaternion could be converted into an axis angle, giving me the degree by which the original two quaternions differ by.

Is this thought process correct? So it should just be [ 0 i j k ]. I may need to normalize the quaternion afterwards, but I'm not sure about that.

I have a bad feeling that it's not a direct mapping from a vector to a quaternion. I tried looking at converting the unit vector to an axis angle, but I'm not sure this would work, since I don't know what angle to give as an input.

Answer 1

Notation

Quaternions are defined in a four-space with bases {1, i, j, k}. Hamilton famously carved the fundamental relationship into the stone of the Brougham Bridge in Dublin:

i² = j² = k² = i j k = -1.

There are many equivalent quaternion parameterizations, but here I'll use a {scalar, vector} form.

1.) A = {a0, a} and B = {b0, b}, where A and B are quaternions, a0 and b0 are scalars, and a and b are three-vectors.

2.) X = { 0, x } is a vector quaternion.

3.) The (non-commutative) quaternion product derives directly from the properties of i, j and k above, A⊗B = {a0 b0 - a.b, a0 b + b0 a + a x b}

4.) The quaternion conjugate is A^* = {a0, - a}

5.) The conjugate of a quaternion product is the product of the conjugates in reverse order.
(A⊗B)^* = B^*⊗A^*

6.) The conjugate of a vector quaternion is its negative. X^* = {0, -x } = -X

7.) The quaternion norm is |A| = √(A⊗A^*) = √( a0² + a.a )

8.) A unit quaternion is one that has a norm of 1.

9.) A unit three-vector x = {x₁, x₂, x₃} with x . x = 1 is expressible as a unit vector quaternion X = { 0, x }, |X| = 1.

10.) The spherical rotation of a quaternion vector X by an angle θ about a unit vector axis n is Q⊗X⊗Q^*, where Q is the quaternion {cos(θ/2), sin(θ/2) n }. Note that |Q| = 1.

Notice the form of the quaternion vector product. Given vector quaternions X₁ = { 0, x₁ ) and X₂ = { 0, x₂ }, the quaternion product is X₂⊗X₁^* = { x₁.x₂, x₁ × x₂ }. The quaternion reunites the dot product as the scalar part and cross product as the vector part, divorced over a hundred years ago. Neither of these products is invertible, but the quaternion is in the way described below.

Inversion

Find the spherical transform quaternion Q₁₂ to rotate vector X₁ to align with vector X₂.

From above

X₂ = Q₁₂⊗X₁⊗Q₁₂^*

Multiplying both sides by X₁^*,

X₂⊗X₁^* = Q₁₂⊗X₁⊗(Q₁₂^*⊗X₁^*)

Remember that the rotation axis n derives from the cross product x₁×x₂, so n . x₁ = 0. and Q^*⊗X^* = (X⊗Q)^* = X^*⊗Q, leaving

X₂⊗X₁^* = Q₁₂⊗X₁⊗X₁^*⊗Q₁₂ = Q₁₂⊗Q₁₂

So the quaternion transform can be solved directly as

Q₁₂ = √(X₂⊗X₁^*)

You're on your own for the quaternion square root. There are lots of ways to do it, and the best will depend on your application, considering speed and stability.

--hth,
Fred Klingener