I have a unit vector (direction) which represent a rotation around Y axis,

and would like to turn this into a quaternion to represent the same rotation.

I use GLM library, but interested any library agnostic solution.

Should I pack the direction vector to a matrix then convert to quaternion? Is there a better way?

UPDATE: As I said the matrix solution works already, but there must be a shorter way, as i don't need all dimension only one (around Y):

glm::quat q = glm::conjugate(glm::toQuat(
    glm::lookAt(glm::vec3(from.getX(), from.getY(), from.getZ()),
                glm::vec3(from.getX(), from.getY(), from.getZ()) + direction,
                glm::vec3(0, 1, 0)
  • \$\begingroup\$ Is the unit vector in the xz plane? (ie. y = 0, only its x & z components might be non-zero) \$\endgroup\$ – DMGregory Sep 30 '17 at 21:16
  • \$\begingroup\$ @DMGregory yes, correct \$\endgroup\$ – Avi Oct 1 '17 at 19:41

From http://www.euclideanspace.com/maths/geometry/rotations/conversions/angleToQuaternion/

qx = ax * sin(angle/2)
qy = ay * sin(angle/2)
qz = az * sin(angle/2)
qw = cos(angle/2)

But since your vector represents the rotation, and is not the axis of rotation, we need to compute the angle. Your axis of rotation is just 0,1,0

angle = atan2( vector.x, vector.z ) // Note: I expected atan2(z,x) but OP reported success with atan2(x,z) instead! Switch around if you see 90° off.
qx = 0
qy = 1 * sin( angle/2 )
qz = 0
qw = cos( angle/2 )

NOTE: this even works for non-unit vectors, as atan2 will compute the correct angle for any length vector, as long as it is not zero.

  • \$\begingroup\$ We can do one better and skip computing the angle directly, by noting that x & z already encode the sine and cosines we need to use the half-angle trig identities... \$\endgroup\$ – DMGregory Sep 30 '17 at 23:29
  • \$\begingroup\$ If you know cos(a), how does that help you to know cos(a/2) then, without doing cos( acos(a) / 2 ) which seems costly, and also would require unit vector. \$\endgroup\$ – Bram Oct 1 '17 at 3:31
  • \$\begingroup\$ By using the half-angle trig identities I mention in the second half of my comment. eg. cos(a/2) = sqrt((1 + cos(a))/2) for angles in the range (-π...+π) — still need two square roots instead of two trig and one inverse trig, so it's not a gigantic savings by any means, it's just kind if a cute pattern. ;) \$\endgroup\$ – DMGregory Oct 1 '17 at 10:35
  • \$\begingroup\$ @Bram thanks, it works and much shorter than my matrix solution ... one minor thing: if i rotate a model with the resulting quaternion, it seems to always add extra 90 to the correct rotation... my model is originally facing +Z axis ... but i can fix that substracting a constant -90 degree \$\endgroup\$ – Avi Oct 1 '17 at 19:48
  • 1
    \$\begingroup\$ @Bram that did it, angle should be angle = atan2( vector.x, vector.z) ... thank you very much, this is a beautiful whole answer! \$\endgroup\$ – Avi Oct 2 '17 at 18:06

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