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I am having some issues constructing a rotation matrix from a camera direction vector.

The Up vector is computed from the camera direction as such:

up = cross(cross(direc,(0,1,0)),direc)

How do i given the up vector and a camera direction vector compute the rotation vector, that tells the camera how it should be rotated such that it points toward my object?

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  • \$\begingroup\$ What are you looking for exactly, how to create a view matrix facing your object or the transformation matrix for one object to face another? \$\endgroup\$ – Tim Apr 23 '16 at 11:58
  • \$\begingroup\$ Of you have the rotation of the camera, why not use that instead of what you're currently doing? \$\endgroup\$ – Bálint Apr 23 '16 at 12:00
  • \$\begingroup\$ This is a bit related to my other post: gamedev.stackexchange.com/questions/120343/… The situation is the same, but i am not sure how i should construct it using vectors. \$\endgroup\$ – Lamda Apr 23 '16 at 12:03
  • \$\begingroup\$ You're going to need a camera Position, unless I'm mistaken. \$\endgroup\$ – Patrick Hughes Apr 23 '16 at 21:04
  • \$\begingroup\$ The position of the camera, is any position on the sphere. \$\endgroup\$ – Lamda Apr 23 '16 at 21:46
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NOTE: Edited because it was likely too verbose (source).

A rotation matrix actually always defines an orthonormal basis. What this means is each column defines one of your original axes in its rotated state. For example, consider a simple rotation matrix around the z-axis (more on rotation matrices here). Let's say we plug in Pi / 2, in other words, we rotate counterclockwise at a right angle.

| cos(Pi/2) -sin(Pi/2) 0 |   | 0 -1  0 |    x-axis: ( 0, 1, 0)
| sin(Pi/2)  cos(Pi/2) 0 | = | 1  0  0 |    y-axis: (-1, 0, 0)
|    0          0      1 |   | 0  0  1 |    z-axis: ( 0, 0, 1)

Which is exactly what you'd expect. The z-axis stays the same. The x-axis rotates 90 degrees, to where the y-axis used to be, and the y-axis rotates 90 degrees further from that, to the negative x-axis.

In camera coördinates, I'm going to assume the following (which is free of choice):

  • The x-axis (column 1) points to the right of your camera: the right vector.
  • The y-axis (column 2) points upwards from your camera: the up vector.
  • The z-axis (column 3) points straight out of your camera: the forward vector.

You already have your forward vector (I'm assuming that's the 'direc' vector you're talking about). However, the way you're calculating your up vector doesn't seem right. If you know the right hand rule, apply it twice for each of your cross products, and you'll end up with some vector pointing towards the negative y-axis, which is probably not what you want.

There's actually a few ways to calculate your remaining two vectors, but it's not quite clear what you're going for, so I'll give you the most straightforward (but least flexible) approach. We'll calculate the up vector always pointing as close towards the sky (which is to say, towards (0, 1, 0)) as possible. First make sure your forward vector is normalized. Then do this:

UP = (-Fx * Fy, Fx² + Fz², -Fy * Fz)
RIGHT = cross(UP, FORWARD)

with (Fx, Fy, Fz) being the forward vector. Normalize your up and right vector, plug all three of them in their respective columns in your rotation matrix, and you're done.

There's two drawbacks to this approach:

  • It bugs out when your camera points perfectly up or down, so make sure that never happens by limiting your viewing angle.
  • You can't easily 'roll' your camera around the z-axis (imagine tilting your head left or right).

Otherwise this should work fine.

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  • \$\begingroup\$ It's probably because your answer is very long. I've seen that happen to new users. We like your enthusiasm, but this question could probably be answered with a much shorter answer, and maybe a link to some information about linear algebra. \$\endgroup\$ – Dietrich Epp Apr 25 '16 at 23:54
  • \$\begingroup\$ Yeah, I had an inkling, but still, it would be nice to have a suggested word count or something. Just spouting 'this is spam' as an error seems incredibly unhelpful. What do you suggest, edit and rewrite or do I leave it as it is? \$\endgroup\$ – Zeno Apr 26 '16 at 10:07
  • \$\begingroup\$ I went ahead and typed up a shorter version. Any comments are appreciated, to make sure I handle this better in the future. \$\endgroup\$ – Zeno Apr 26 '16 at 10:46

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