The torque $\overrightarrow{\tau }$  on a body about a given point is found to be equal to $\overrightarrow{A}\times \overrightarrow{L},$  where $\overrightarrow{A}$  is a constant vector and $\overrightarrow{L}$  is the angular momentum of the body about that point. From this, it follows that :

$\overrightarrow{\tau }=\overrightarrow{A}\times \overrightarrow{L}or\frac{d\overrightarrow{L}}{dt}=\overrightarrow{A}\times \overrightarrow{L}$

Thus $\frac{d\overrightarrow{L}}{dt}$  is perpendicular to both $\overrightarrow{A}$  and $\overrightarrow{L}$

$\overrightarrow{L}.\overrightarrow{L}=L^2$

$\overrightarrow{L}.\frac{d\overrightarrow{L}}{dt}+\frac{d\overrightarrow{L}}{dt}.\overrightarrow{L}=2L\frac{dL}{dt}$

or            $\overrightarrow{L}\frac{d\overrightarrow{L}}{dt}=L\frac{dL}{dt}$

But         $\overrightarrow{L}.\frac{d\overrightarrow{L}}{dt}=0$

So,          $\frac{dL}{dt}=0$

So, $\left|\overrightarrow{L}\right|$  does not change with time.

Since $\frac{d\overrightarrow{L}}{dt}\perp \overrightarrow{L}$  and $\left|\overrightarrow{L}\right|$  is not changing with time, thus, direction of $\overrightarrow{L}$  is changing but its magnitude is constant, Also, $\overrightarrow{\tau }$  is perpendicular to $\overrightarrow{L}$  at all points.

If              $\overrightarrow{L}=\left(a\cos \theta \right)\hat{i}+\left(a\sin \theta \right)\hat{j}$

then      $\tau =\left(a\cos \theta \right)\hat{i}-\left(a\sin \theta \right)\hat{j}$

So that $\overrightarrow{L}.\overrightarrow{\tau }=0$

$\overrightarrow{A}\perp \overrightarrow{\tau },let\overrightarrow{A}=A\hat{k}$

So that $\overrightarrow{L}.\overrightarrow{A}=0$

$\overrightarrow{A}\perp \overrightarrow{\tau }$

Component of $\overrightarrow{L}$  along $\overrightarrow{A}$  is zero

or component of $\overrightarrow{L}$  along $\overrightarrow{A}$  is always constant.

It may be concluded that $\overrightarrow{\tau }$ , $\overrightarrow{A}$  and $\overrightarrow{L}$  are always mutually perpendicular