## Appendix A Derivation of the Precision Measurement Optimization Model  
### Derivation of Equations (1) and (2)  
To determine the relationship between the coverage width \(W\) and the orientation angle \(\beta\), selecting the optimal direction of the survey line that maximizes the detection coverage at the center of the marine area is crucial. This process requires the use of a Cartesian coordinate system. By deriving the plane equations for the seabed slope \(A\) and the multibeam plane \(B\) when the ship is at the center (distance zero), the line \(l_{1}\) representing the coverage width \(W\) can be established. From there, using the sine theorem, the coverage width \(W\) is calculated. **Definition A1**. : _General Equation of a Plane in Space [29]. Any plane in space can be defined by a point \(M_{0}(x_{0},y_{0},z_{0})\) on the plane and its orientation vectors \(a=(X_{1},Y_{1},Z_{1})\) and \(b=(X_{2},Y_{2},Z_{2})\). Thus, any plane can be represented by the following equation:_  
\[ \begin{vmatrix}x-x_{0}&y-y_{0}&z-z_{0}\\ X_{1}&Y_{1}&Z_{1}\\ X_{2}&Y_{2}&Z_{2}\end{vmatrix} =0\]  
_The above results are expanded in:_  
\[Ax+By+Cz+D=0\]  
_where_  
\[A= \begin{vmatrix}Y_{1}&Z_{1}\\ Y_{2}&Z_{2}\end{vmatrix} ,\quad B= \begin{vmatrix}Z_{1}&X_{1}\\ Z_{2}&X_{2}\end{vmatrix} ,\quad C= \begin{vmatrix}X_{1}&Y_{1}\\ X_{2}&Y_{2}\end{vmatrix} ,\quad D=- \begin{vmatrix}x_{0}&y_{0}&z_{0}\\ X_{1}&Y_{1}&Z_{1}\\ X_{2}&Y_{2}&Z_{2}\end{vmatrix} \]  
_Since vectors \(a\) and \(b\) are not collinear, \(A\), \(B\), and \(C\) are not all zero, indicating that any plane in space can be represented by a linear equation in \(x\), \(y\), and \(z\)._  
**Definition A2**. : _If a point \(M_{0}(x_{0},y_{0},z_{0})\) and a non-zero vector \(\mathbf{n}=(A,B,C)\) are given in space, then the plane passing through \(M_{0}\) and perpendicular to \(\mathbf{n}\) is uniquely determined. The vector \(\mathbf{n}\), being perpendicular to the plane, is called the normal vector of the plane. Thus, the point-normal form of the equation of the plane [30] can be expressed as:_  
\[A(x-x_{0})+B(y-y_{0})+C(z-z_{0})=0\]  
**Theorem A1**. : _The direction vector of the line of intersection of two planes \(A\) and \(B\) can be determined by the cross product of their normal vectors [31]._  
Proof of Theorem A1. : Consider two planes \(A\) and \(B\) with equations:  
\[A:A_{1}x+B_{1}y+C_{1}z+D_{1} =0,\] \[B:A_{2}x+B_{2}y+C_{2}z+D_{2} =0.\]  
In the Cartesian coordinate system, the normal vectors of the planes are \(\mathbf{n}_{1}=(A_{1},B_{1},C_{1})\) and \(\mathbf{n}_{2}=(A_{2},B_{2},C_{2})\). The direction vector \(\mathbf{v}\) of their line of intersection can be computed as:  
\[\mathbf{v}=\mathbf{n}_{1}\times\mathbf{n}_{2}=\left( \begin{array}{cc|cc}B_ {1}&C_{1}\\ B_{2}&C_{2}\end{array} \right| \begin{array}{cc|cc}C_{1}&A_{1}\\ C_{2}&A_{2}\end{array} \right| \begin{array}{cc|cc}A_{1}&B_{1}\\ A_{2}&B_{2}\end{array} \right)\]  
First, the equations for the seabed slope plane \(A\), the multibeam plane \(B\), and the survey line direction plane \(C\) are determined. As shown in Figure 5, after establishing the Cartesian coordinate system, the projection diagrams of these three planes are obtained:  
As shown in Figure A1a, \(n_{1}\) is the normal vector of the seabed slope \(A\), where \(n_{1}=(\sin\alpha,0,\cos\alpha)\). Since the seabed slope \(A\) passes through the origin, the point-normal form of the equation for plane \(A\) is given by:  
\[x\cdot\sin\alpha+z\cdot\cos\alpha=0\]  
As illustrated in Figure A1b, \(n_{2}\) is the normal vector of plane \(B\), which passes through the origin and lies on the \(XOY\) plane. The direction of the survey line is perpendicular to plane \(B\); hence, the angle \(\beta\) between \(n_{2}\) and the \(x\)-axis is given by \(n_{2}=(\cos\beta,\sin\beta,0)\).