Most Efficient way to solve for Euler Angles

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To efficiently solve for the Euler angles (yaw, pitch, and roll) needed to align a laser with a fixed target in 3D space, it is essential to calculate the vector from the laser's initial position to the target. The vector can be expressed as v = (x1 - x0)i + (y1 - y0)j + (z1 - z0)k, and its length is determined using the formula r = √((x1 - x0)² + (y1 - y0)² + (z1 - z0)²). The unit vector in the direction of the target is then found by dividing the vector by its length. The Euler angles can be derived from the cosine of the angles, where cos(θx) = (x1 - x0)/r, cos(θy) = (y1 - y0)/r, and cos(θz) = (z1 - z0)/r. Providing the Cartesian coordinates of the initial point and the target, along with the laser's current Euler angles, will facilitate the calculations needed to achieve the desired alignment.
Robconway
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Hi guys, could you please help me out?

Essentially, a laser is pointing at a certain point in 3 dimensional space. There is a fixed target that the laser is specifically supposed to point to. My job is to find the most efficient way to solve for the euler angles(yaw pitch and roll) in order to sync the laser with the target point. What is the best way to do this? The values known are the direct distance from the laser to to the target, the distance from the laser to the initial point, and the distance from the initial point to the fixed target. Thank you guys, it would help a lot!
 
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You need more or different info. One set of info that would be sufficient is the cartesian coordinates of the initial point and the target with respect to the laser plus the yaw, pitch, and roll of the laser in that coordinate system. Another sufficient set is yaw, pitch, and roll of the laser plus the yaw and pitch of the target position with respect to some fixed coordinate system. It's not clear exactly what your particular situation is, but the info you've given is not sufficient to solve the problem.
 
A vector pointing from (x_0, y_0, z_0) to (x_1, y_1, z_1) can be written as \vec{v}= (x_1- x_0)\vec{i}+ (y_1- y_0)\vec{j}+ (z_1- z_0)\vec{j}. Its length is
|\vec{v}|= r= \sqrt{(x_1-x_0)^2+ (y_1- y_0)^2+ (z_1- z_0)2}

Finally, the unit vector in that direction is
\frac{\vec{v}}{|\vec{v}|}= \frac{x_1- x_0}{r}\vec{i}+ \frac{y_1- y_0}{r}\vec{j}+ \frac{z_1- z_0}{r}\vec{k}

So the Euler angles are given by
cos(\theta_x)= \frac{x_1- x_0}{r}
cos(\theta_y)= \frac{y_1- y_0}{r}
cos(\theta_z)= \frac{z_1- z_0}{r}

That is, the Euler angles for a given direction are the arccosines of the components of a unit vector in that direction.
 
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