Computing the altitude angle

OK, it might be I finally understood your comment below about the result being independent of the y angle, and about how it relates to the two vectors. It seems you are not really interested in two vectors and the angle between these two, but instead you're interested in the difference vector and the angle that one forms against the horizontal plane. In a horizontal coordinate system (often used in astronomy), that angle would be called “altitude” or “elevation”, as opposed to the “azimuth” you compute with the formula in your (edited) question. “altitude” closely relates to the “tilt” of your camera, whereas “azimuth” relates to “panning”.

We still have a 2D problem. One coordinate of the 2D vector is the y coordinate of the difference vector. The other coordinate is the length of the vector after projecting it on the horizontal plane, i.e. sqrt(x*x + z*z). The final solution would be

x = A.x - B.x
y = A.y - B.y
z = A.z - B.z
alt = toDegrees(atan2(y, sqrt(x*x + z*z)))
az = toDegrees(atan2(-x, -z))

The order (A - B as opposed to B - A) was chosen such that “A above B” yields a positive y and therefore a positive altitude, in accordance with your comment below. The minus signs in the azimuth computation above should replace the + 180 in the code from your question, except that the range now is [-180, 180] instead of your [0, 360]. Just to give you an alternative, choose whichever you prefer. In effect you compute the azimuth of B - A either way. The fact that you use a different order for these two angles might be somewhat confusing, so think about whether this really is what you want, or whether you want to reverse the sign of the altitude or change the azimuth by 180°.

Orthogonal projection

For reference, I'll include my original answer below, for those who are actually looking for the angle of rotation around some fixed x axis, the way the original question suggested.

If this x angle you mention in your question is indeed the angle of rotation around the x axis, as the camera example suggests, then you might want to think about it this way: set the x coordinate to zero, and you will end up with 2D vectors in the y-z plane. You can think of this as an orthogonal projection onto said plain. Now you are back to a 2D problem and can tackle it there.

Personally I'd simply call atan2 twice, once for each vector, and subtract the resulting angles:

toDegrees(atan2(A.z, A.y) - atan2(B.z, B.y))

The x=0 is implicit in the above formula simply because I only operate on y and z.

I haven't fully understood the logic behind your single atan2 call yet, but the fact that I have to think about it this long indicates that I wouldn't want to maintain it, at least not without a good explanatory comment.

I hope I understood your question correctly, and this is the thing you're looking for.

Use cross product of the two vectors to get the normal of the plane formed by the two vectors. Then check the dotproduct between that and the original plane normal to see if they are facing the same direction.

angle = acos(dotProduct(Va.normalize(), Vb.normalize()));
cross = crossProduct(Va, Vb);
if (dotProduct(Vn, cross) < 0) { // Or > 0
  angle = -angle;
}

This is not an answer to the stated question, but more like an extended comment; specifically, that there is an even easier way to find out , the minimum distance between the line segment and point .

The question implies that OP is interested in whether a spherical shell, let's say of radius , centered at , intersects the line segment between and .

1. If , and , then both points and are within the spherical shell. Because the shell is convex, the line segment between and is also completely inside the spherical shell (and therefore there is no intersection per se).

   In practice, the checks are better written as and , i.e. and

   Otherwise,

2. If , then point is within the spherical shell. Because isn't, the line segment between and must pass through the spherical shell.

   Otherwise,

3. If , then point is within the spherical shell. Because isn't, the line segment between and must pass through the spherical shell.

   Otherwise,

4. Calculate the distance squared between point and the line that passes through points and (using e.g. Point-Line distance from Wolfram MathWorld) :

   

   i.e.

   

   Now, if , the entire line passing through and is outside the spherical shell, and there cannot be any intersection.

   Otherwise,

5. Calculate the relative position of the point closest to on the line passing through and :

   

   If , then the closest point to on the line is between and , and it is either inside () or on () the spherical shell.

   This test can also be written as

   Otherwise, there is no intersection. (The line that passes through and does intersect the -radius spherical shell centered at , but the intersections do not occur in the segment between and .)

If I counted right, the maximum cost of the above tests, total, is 25 multiplications, one division, and 46 additions or subtractions (but much fewer multiplications and additions or subtractions if you use temporary variables so you don't do the same operations repeatedly). On a computer, that is roughly comparable to the work done to evaluate a single trigonometric function; so if you use that as the metric for "easy" (I do), this is definitely "easier".

If you're going to check with Wolfram Alpha, you can ask your question somewhat more directly:

angle between (sqr(3)/3, sqr(3)/3, sqr(3)/3) and (sqr(2)/2,sqr(2)/2,0)

You'll see that the answer is $\arccos(\sqrt(2/3))\approx0.61$, which is exactly what you got.

Are you looking for this ?

θ_radian = arccos ( (​P⋅Q) / ​(∣P∣∣Q∣)​ ​​) with vectors P and Q

θ_radian = θ_degree * π / 180

EDIT 0-360 range

angle = angle * 360 / (2*Math.PI);
if (angle < 0) angle = angle + 360;