Get range of 3D object given lowest and highest point and angle

Question

How do you get 3D range of object (highlighted in red below) given its lowest (PL) and highest (PH) (x,y,z) coordinates and orientation of object? Image below is a top view of a box.

Answer 1

I've found a way already

Given:
1)$$H_1 = L\sin(\theta) + W\cos(\theta)$$
2)$$H_2 = L\cos(\theta) + W\sin(\theta)$$

Then...

3)$$L = \frac{H_2 - W\sin\theta} {\cos(\theta)}$$

Substitute... $$H_1 = \frac{H_2-W\sin\theta} {\cos\theta}sin\theta + W\cos\theta$$ $$H_1 = \frac{H_2\sin\theta} {\cos\theta} - \frac{W\sin^2\theta}{\cos\theta} + Wcos\theta$$ $$H_1 = \frac{H_2\sin\theta} {\cos\theta} - \frac{W(1-\cos^2\theta)}{\cos\theta} + W\cos\theta$$ $$H_1 = \frac{H_2\sin\theta} {\cos\theta} - \frac{W}{\cos\theta} + W\cos\theta + W\cos\theta$$ $$H_1 = \frac{H_2\sin\theta} {\cos\theta} - W\left(\frac{1}{\cos\theta} + \cos\theta + \cos\theta\right)$$ $$W\left(\frac{1}{\cos\theta} + \cos\theta + \cos\theta\right) = \frac{H_2\sin\theta} {\cos\theta} - H_1$$ $$W = \frac{(H_2\tan\theta - H_1){\frac{1-2\cos^2\theta}{\cos\theta}}$$ 4)$$W = \frac{(H_2\tan\theta - H_1)\cos\theta}{1-2\cos^2\theta}$$

Since you have W, substitute W of formula 3 to get L.

Once you have L, you can get the points easily.

P.S. This does not apply if angle is 45 degrees.