How do I get the angles between 2 vector3 points

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#1

how to find “alpha” and “beta” given 2 vector3 points

image
image849×841 27.4 KB

#2 
alpha = math.atan2(z, x)
beta = math.atan2(y, z)
angle_between = math.acos(math.clamp(Vector1:Dot(Vector2) / (Vector1.Magnitude * Vector2.Magnitude)), -1, 1)
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#3

what is z? what is x? what is y?

#4

They are the x, y, z components of the vector

#5

components of which vector? the middle or the top one

#6

The image you posted only shows 1 vector. It looks like the blue and red dotted ones represent the projections of that green vector onto those respective planes. In my first post x, y, and z are the coordinates of the green vector.

#7

im finding the angles between the vectors of bottom left (the origin) to the top right

#8

the angles of the origin and the tip of green’s arrow is what im finding, not to be mistaken with blue, green is the combination of red and blue

#9

With respect to what other vector? You can’t compute an angle if one of the vectors has no magnitude.

#10

image
image1376×1076 156 KB

here is an image to explain what i mean

i want to find 2 angles

the X and Y angles, X being right and left, Y being up and down

so combined, they can be used to locate the green brick

#11

Alright, sorry for the confusion. You basically want to convert a set of cartesian coordinates into spherical coordinates.

Screenshot (5)
Screenshot (5)1174×866 177 KB 

local point = --grey part position (direction and magnitude of the brown line)

-- the orange line in the image represents the projection of the brown vector onto the XZ plane.
local xzProjection = point * Vector3.new(1, 0, 1)

-- let theta be the angle between the blue unit vector and the orange projection vector.
-- also known as the 'azimuth 'or longitude
local theta = math.atan2(xzProjection.X, xzProjection.Z)

-- let phi be the angle between the projection and the original brown vector
-- also the 'polar angle' or latitude
local phi = math.atan2(point.Y, xzProjection.Magnitude)

-- finally, our 'r' value is just the distance from the origin
local r = point.Magnitude

You can also re-construct the position using CFrame operations, too.
Because the coordinate system is kind of weird on roblox, I had to rotate the azimuth by 180 degrees and make r negative but this should give you the original position:

local cf = CFrame.fromOrientation(phi, theta + math.pi, 0) * CFrame.new(0, 0, -r)
local position = cf.Position
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