*Collinear singularities:* another singularity case occurs when $a2\xd7b2$ and $a3\xd7b3$ are collinear. These two vectors are defined as orthogonal to two other vectors; hence, we can assume that these two vectors are collinear when the vectors they are defined with are coplanar. Moreover, vectors $b2$ and $b3$ are in the plane of the platform. Therefore, the coplanar condition can be reduced to seeking the conditions for which $a2$ and $a3$ are in the plane of the platform. To this end, vectors $b20\xd7b30$ and $a2\xd7a3$ are calculated and the rotation matrix which aligns these two vectors is calculated. Then, the rotation matrix of the degree-of-freedom allowed to remain in the singular configuration is computed. This matrix is the matrix of the rotation around vector $a2\xd7a3$ with a free angle. However, in this particular state, $a2\xd7a3$ is equal to $0$ because $a2$ and $a3$ are aligned. It is sufficient to calculate the rotation matrix defined by the cosine between $a2$ and its projection in the reference plane of the platform $a2p=(13\u2212(((b20\xd7b30)\u22c5(b20\xd7b30)T)/(||b20\xd7b30||2)))a2$ around the axis $a2p\xd7a2$. So, the platform is free to rotate around $(b2\xd7b3)$ and $a2$ and the singularity state is maintained. The method used to locate the alignment singularity loci can be used here too but two matrices rather than one must be introduced to represent the freedom allowed for the robot to remain in the singularity state. As before, $Q1$ positions the platform in the collinear singularity state and $Q2$ is the free rotation around $(b2\xd7b3)$ and finally $Q3$ is the free rotation around $a2$. We get $Q=Q1Q2Q3$ which is equated to $R(\varphi ,\theta ,\sigma )$. Similar to the previous case, five equations among nine are simple to solve and yield
Display Formula

(A9)${\varphi =\varphi \theta =\xb1\pi 2\sigma =\sigma $