The lengths of two opposite edges of tetrahedron are  a and b and their S.D is d and angle between them  is $\theta .$ If volume of tetrahedron is $\frac{1}{k}abd\sin \theta$ then k is

$$\begin{gathered} From the diagram \\ \left|\overline{AB}\right|=a\Rightarrow \left|\overline{\beta } -\overline{\alpha } \right|=a and \left|\overline{OC}\right|=b\Rightarrow \left|\overline{r}\right|=b \\ Now equations of two lines are \\ \overline{r}=0+t\overline{r } and \overline{r}=\alpha +s(\overline{\beta } -\overline{\alpha }) \\ Now shotest dis\tan ce =\frac{(\overline{\alpha } -0).(\overline{\beta } -\overline{\alpha } )\times \overline{r}}{\left|(\overline{\beta } -\overline{\alpha } )\times \overline{r}\right|} \\ \Rightarrow d=\frac{\left[\overline{\alpha } \overline{\beta } -\overline{\alpha } \overline{r} \right]}{\left|\overline{\beta } -\overline{\alpha } \right|\left|\overline{r}\right|\sin \theta } \\ \Rightarrow d=\frac{\left[\overline{\alpha } \overline{\beta } \overline{r} \right]}{ab\sin \theta } \\ \Rightarrow abd \sin \theta =\left[\overline{\alpha } \overline{\beta } \overline{r} \right] \\ \Rightarrow abd \sin \theta =6V \\ V=\frac{1}{6}abd \sin \theta \\ \therefore By comparision k=6 \end{gathered}$$