The torque τ→  on a body about a given point is found to be equal to A→×L→,  where A→  is a constant vector and L→  is the angular momentum of the body about that point. From this, it follows that :

τ→=A→×L→  or  dL→dt=A→×L→ Thus dL→dt  is perpendicular to both A→  and L→ L→.L→=L2 L→.dL→dt+dL→dt.L→=2LdLdt or            L→dL→dt=LdLdt But         L→.dL→dt=0 So,          dLdt=0 So, |L→|  does not change with time.Since dL→dt⊥L→  and |L→|  is not changing with time, thus, direction of L→  is changing but its magnitude is constant, Also, τ→  is perpendicular to L→  at all points.If              L→=(acosθ)i^+(asinθ)j^ then      τ=(acosθ)i^−(asinθ)j^ So that L→.τ→=0 A→⊥τ→, let A→=Ak^ So that L→.A→=0 A→⊥τ→ Component of L→  along A→  is zeroor component of L→  along A→  is always constant.It may be concluded that τ→ , A→  and L→  are always mutually perpendicular