Understanding of Camera Up Vector Assume eye position e, gaze direction g, view-up vector t, right-handed base. As the graphs shows:

  

$w=-\frac{g}{||g||}$
$u=\frac{t\times w}{||t\times w||}$
$v=w\times u$

I am really confused about the 2nd equation $u=\frac{t\times w}{||t\times w||}$.
What if the camera rotate around its gaze direction? The v rotates while the t stays the same.
Then the vector t will not in the plane that w and v lies in which means the second equation is wrong.

I know I have something misunderstood.
Could someone please give me some clues?
What if the camera rotate around its gaze direction? The v rotates while the t stays the same.


The camera's orientation is defined by two directions: g and t. If the camera "rotates around its gaze direction", g will not be changing; that is what it means to "rotate around its gaze direction". If g isn't changing, and the camera's orientation is defined by two directions, then you can only get a different orientation if t is changing.

Therefore, your second statement is a contradiction of the first. Changing t is precisely how you cause the camera to "rotate around its gaze direction".

To put it a different way, there are changes to t which will not affect the values of u, v, or w. However, none of those changes will cause the camera to "rotate around its gaze direction".