Consider a screw $\{ {\cal V} \}$ of non-zero resultant $\bf V$. Let $\{ {\cal S}_1 \}$ be a slider whose axis is not parallel to
$\bf R$ and which satisfies
$\{ {\cal S}_1 \} \cdot \{ {\cal V} \} \neq 0$. Then there exist a unique slider $\{ {\cal S}_2 \}$ and a unique scalar
$\lambda$ such that
\[
\{ {\cal V} \} = \lambda \{ {\cal S}_1 \}+ \{ {\cal S}_2 \}
\]
Given $\{ {\cal S}_1 \}$ and a scalar $\lambda$, define the screw $\{ {\cal W} \} = \{ {\cal V} \} -\lambda \{ {\cal S}_1 \}$.
Can a value of $\lambda$ be found such that $\{ {\cal W} \}$ defines a slider? The resultant of $\{ {\cal W} \}$ is non-zero since
it is the sum of two non-collinear vectors. Hence a necessary and sufficient condition for $\{ {\cal W} \}$ to be a slider
can be stated by imposing $\{ {\cal W} \}\cdot \{ {\cal W} \} =0$:
\[
\{ {\cal V} \}\cdot \{ {\cal V} \} -2 \lambda \{ {\cal V} \}\cdot \{ {\cal S}_1 \}+ \lambda^2 \{ {\cal S}_1 \} \cdot \{ {\cal S}_1 \} =0
\]
However, we have $\{ {\cal S}_1 \} \cdot \{ {\cal S}_1 \} =0$ since $\{ {\cal S}_1 \}$ is a slider. This gives a unique value of
$\lambda$:
\[
\lambda = \frac{1}{2} \frac{ \{ {\cal V} \}\cdot \{ {\cal V} \} } { \{ {\cal V} \}\cdot \{ {\cal S}_1 \} }
\]
This value of $\lambda$ corresponds to unique slider $\{ {\cal W} \}= \{ {\cal S}_2 \}$.
