Question

I have program in which I am drawing some shapes such as rectangle/circle/triangle etc. in 3D space, equidistant from the centre.

Example of two points which are equidistant from the centre (0,0,0) are

x1: 93, 313, 123
x2: -263, -135, -186

and there may be more points as well.

At these two points, I am drawing the shapes by something like

glRectangle(x1, width, height);

The shapes (rectangles/triangles etc.) come out very well at the starting vertex (x1 or x2) equi-distant from the centre. However, the shapes have wrong orientation in space as to what I am looking for.

I want centre of the plane of in which the shape lies to to be prependicular to an imaginary line coming from the centre. This probably can be solved by rotation of the plane but I am not sure about the approach to adopt. For the current example, let’s say I have the vertex of the shape but how do I figure out the plane in which the the rectangle is and rotate that plane to make it prependicular (to the line coming from the centre of the space (0,0,0))

Answer 1

If the center needs to be perpendicular to the vector C (Line between coord-center and rect-center), it should work like this (pseudo-code):

A = C x UP_VECTOR (CROSS-PRODUCT to calculate helper vector pointing in "sideway" direction of "rectangle")
B = C x A (CROSS-PRODUCT to get the local "up/down" vector of the "rectangle")
A = normalize(A)
B = normalize(B)
P1 = width/2 * A + height/2 * B
P2 = -width/2 * A + height/2 * B
P3 = -width/2 * A + -height/2 * B
P4 = width/2 * A + -height/2 * B

Hopefully this is understandable and right (didn’t try it and had some trouble finding the right vocabulary)

Some additional informations:
A and B are helper vectors and are the local up and left directions on the rectangle, after normalizing they can be used to calculate the 4 points of the rectangle accordingly.
Also be careful if no rectangle is displayed it’s probably related to the drawing order so just reverse the order then if front and backfaces are wrong…