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## Extracting basis vectors from a quaternion

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### #1maxest  Members

Posted 02 May 2014 - 10:43 AM

A 3x3 matrix describes orientation and/or rotation with three (basis) vectors placed in the matrix vectors (or columns). Such a matrix can be converted to a quaternion and vice-versa. I'm wondering how to extract the three basis vectors from the quaternio, without converting the quaternion to a matrix. Is that even possible?

### #2Brother Bob  Moderators

Posted 02 May 2014 - 10:55 AM

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Since the matrix is the set of basis vectors you're looking for, converting the quaternion to a matrix or to basis vectors would have to be equivalent. If you get the matrix, you automatically have the basis vectors; if you have the basis vectors, you automatically have the matrix. A possible exception, of course, would be if you only need one or two of the basis vectors, but that's equivalent to a partial matrix conversion.

### #3maxest  Members

Posted 02 May 2014 - 11:01 AM

Yes but is there any other way? I mean, to sort of "read" the basis vectors straight from the quaternion? Or to put it this way. From a quaternion we can easily read the axis of rotation and the angle. I think that axis-angle and quaternion are equivalent beings. If so, then that means that from axis-angle we could also somewhat "read" the basis vectors. But I don't see how.

### #4Brother Bob  Moderators

Posted 02 May 2014 - 11:20 AM

But my point is, you want the basis vectors, and the matrix consists exclusively of the basis vectors. No matter how you read the basis vectors from the quaternion, a matrix can always be created exactly the same way and you get the matrix for free when you have the basis vectors. There's no way to read the basis vectors that doesn't give you the matrix, because the collection of basis vectors is the matrix.

### #5Álvaro  Members

Posted 02 May 2014 - 12:10 PM

Of course Brother Bob is right, but perhaps this explanation might help you. A rotation is a mapping from R^3 to R^3. When a matrix is used to represent a rotation, its columns are the images of (1,0,0), (0,1,0) and (0,0,1). If you have the rotation represented as a quaternion, you can compute those columns the same way you would rotate any other vector (something like q * i * conj(q), q * j * conj(q) and q * k * conj(q)).

### #6maxest  Members

Posted 02 May 2014 - 03:02 PM

Okay... I think I get your point but still something bugs me.

Let's put it this way. We have axis-angle representation and want to get the basis-vectors (or matrix if you like). One of the basis vectors is the axis of rotation. How do I compute the other two vectors? The only information I have left is that these two vectors lie on the plane perpendicular to the axis of rotation. And of course I have the angle, which definitely is needed to get the second vector but I'm not sure how.

### #7Brother Bob  Moderators

Posted 02 May 2014 - 03:21 PM

The axis of rotation is not a basis vector. This is only true if you are rotating about one of the principal axes, but not in general.

### #8NickW  Members

Posted 02 May 2014 - 03:57 PM

Just take each of the identity basis vectors (1, 0, 0), (0, 1, 0) and (0, 0, 1) and rotate them with the quaternion.  This will give you a new set of basis vectors that should correspond to the columns in a rotation matrix.

Edited by NickW, 02 May 2014 - 03:58 PM.

### #9maxest  Members

Posted 09 November 2015 - 10:15 AM

Late but still - thanks .

Thumbs up given.

### #10Alundra  Members

Posted 09 November 2015 - 10:22 AM

Here the code to compute the forward, right and up vector from a quaternion :
CVector3 CQuaternion::ComputeForwardVector() const
{
const float x2 = 2.0f * x;
const float y2 = 2.0f * y;
const float z2 = 2.0f * z;
const float x2w = x2 * w;
const float y2w = y2 * w;
const float x2x = x2 * x;
const float z2x = z2 * x;
const float y2y = y2 * y;
const float z2y = z2 * y;
return CVector3( z2x + y2w, z2y - x2w, 1.0f - ( x2x + y2y ) );
}

CVector3 CQuaternion::ComputeRightVector() const
{
const float y2 = 2.0f * y;
const float z2 = 2.0f * z;
const float y2w = y2 * w;
const float z2w = z2 * w;
const float y2x = y2 * x;
const float z2x = z2 * x;
const float y2y = y2 * y;
const float z2z = z2 * z;
return CVector3( 1.0f - ( y2y + z2z ), y2x + z2w, z2x - y2w );
}

CVector3 CQuaternion::ComputeUpVector() const
{
const float x2 = 2.0f * x;
const float y2 = 2.0f * y;
const float z2 = 2.0f * z;
const float x2w = x2 * w;
const float z2w = z2 * w;
const float x2x = x2 * x;
const float y2x = y2 * x;
const float z2y = z2 * y;
const float z2z = z2 * z;
return CVector3( y2x - z2w, 1.0f - ( x2x + z2z ), z2y + x2w );
}

### #11clb  Members

Posted 09 November 2015 - 12:02 PM

Code snippets like the above should definitely come with a comment disambiguating the used conventions. Inherently a quaternion (nor a matrix) does not have forward, right or up vectors, but it is the semantic conventions that give such meanings. What is 'Up' in one engine can be 'Forward' in the conventions of another engine. Quickly glancing, in the conventions used by the above snippet, the function ComputeRightVector is the image of (1,0,0) or +X vector when transformed (=conjugated) by the quaternion, and ComputeUpVector is the image of (0,1,0)/+Y vector, and ComputeForwardVector is the image of (0,0,1)/+Z vector transformed by the quaternion. The coordinate system is assumed to be left-handed. Were those the assumed conventions?

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### #12Alundra  Members

Posted 09 November 2015 - 01:18 PM

This is true, the coordinate system used was needed to be say.

Edited by Alundra, 09 November 2015 - 01:18 PM.

### #13JohnnyCode  Members

Posted 14 November 2015 - 09:23 AM

Just take each of the identity basis vectors (1, 0, 0), (0, 1, 0) and (0, 0, 1) and rotate them with the quaternion.

Good luck on doing that without the quaternion's matrix.

Alundra- that is just an expanded routine of the actual quaternion to matrix conversion.

### #14Álvaro  Members

Posted 14 November 2015 - 07:04 PM

Good luck on doing that without the quaternion's matrix.

You don't need luck if you know what you are doing, or if you have the formulas I posted above in this very thread.

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