from rotations around axes to normal and direction

44 Views Asked by At

I have a rectangle in 3D-space that is only defined by his width, height and 3 angles. These 3 angles are angles around X,Y and Z axis. But they are not X,Y and Z axis of the rectangle , but of the world, so rotations applied to the rectangle won't have repercussions on the coordinate system that is used for the rotations (I hope it is understandable)...

My goal is to find the up vector, the directionnal vector and the normal vector of my rectangle using only the informations I have. Is it mathematically possible? If yes, what is the way to achieve that?

(In my head, this is understandable, but maybe because of my lack of english vocabulary and/or my twisted spirit this is crap. In that case, I will try to rephrase my question).

Thanks!

Edit : Here's how I see up, direction and normal vector (I took the bad habit to name them unproperly) enter image description here

1

There are 1 best solutions below

5
On

When you have a square, then, its height and width must be same (unless basis vectors of space are transformed). Second, how are the three angles defined? Are they defined with respect to an edge?

If the angles are defined with respect to the adjacent edges, and the coordinate of the common vertices. you can easily calculate their unit vectors. The up direction will be obtained by the cross product of the unit vectors of the two edges. I hope that 'up' and 'normal' vector refers to the same parameter.

'Directional' vector is not quite understandable. Are you going to define any function defined over that squared surface? And you want to obtain direction of optimal change? Please clarify.

Edit: Direction cosines of the edge $\vec{DA}$ is represented in the figure, and it's length is known. Similarly, for edge $\vec{DC}$ as well. $\vec{DA}$ and $\vec{DC}$ are your up and directional vectors (Can be according to your definition). Normal vector can be obtained by $\vec{DA} \times \vec{CD}$ (Choose these two vectors accordingly, because $\vec{DA} \times \vec{DC} = -\vec{DC} \times \vec{DA}$). Pictorial representation of Direction Cosines