If $\overrightarrow{\mathrm{F}} \mathrm{is} \mathrm{the} \mathrm{force} \mathrm{acting} \mathrm{on} \mathrm{a} \mathrm{particle} \mathrm{having} \mathrm{position} \mathrm{vector} \overrightarrow{\mathrm{r}} \mathrm{and} \overrightarrow{\mathrm{\tau }}$ be the torque of this force about the origin, then

Torque is an axial vector, i.e., its direction is always perpendicular to the plane containing vectors $\overrightarrow{\mathrm{r}} \mathrm{and} \overrightarrow{\mathrm{F}}$.

$\overrightarrow{\mathrm{\tau }} = \overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}$

Torque is perpendicular to both $\overrightarrow{\mathrm{r}} \mathrm{and} \overrightarrow{\mathrm{F}}$

$\overrightarrow{\mathrm{\tau }}.\overrightarrow{\mathrm{r}} = 0$

$\overrightarrow{\mathrm{F}}.\overrightarrow{\mathrm{\tau }} = 0$