How do I get the exit angle of a body?

Question

I have a 3D environment with vectors $(x, y, z)$.

For example:

• Room size $10 \times 10 \times 10$
• Bulb in the position $(3,5,10)$
• Measuring points: $(5,5,0), (1,1,0), (5, 0, 5)$, etc.

A light bulb emits a certain amount of light $(I)$ in one direction, for example in the horizontal angle $20$, and vertical angle $40$, emits $100$ candles $(I)$.

As you can see in the picture I have a light bulb that emits light, and a point where I will measure it $(P)$.

From the bulb comes out in the direction of $P$, a line $(I)$.

That is, from the bulb, at a horizontal angle and a vertical angle comes line $(I)$.

The question is if I only have the positions of the bulb and the measuring point, how do I obtain that horizontal and vertical angle of the bulb?

Thanks

Answer 1

If you know the location of the point C and the height $h$ above which the bulb rests then with trigonometry you know that $$\tan y = \frac{\| CP \|}{h}$$

The angular location of P on the xz plane is

$$ \tan g = \frac{P_x-C_x}{P_z-C_z} $$ where $x$ is perpendicular to the road and $z$ is along the sidewalk.