A particle moves on a given straight line with a constant speed ν. At a certain time it is at a point P on its straight line path. O is a fixed point. Show that \[\vec{OP} \times \vec{\nu}\] is independent of the position P.
Solution
The particle moves on the straight line XX' at a uniform speed ν.
In
\[∆ POQ\]
OQ = OP sin θ
\[\vec{OP} \times \nu = \left( OP \right) \nu \sin \theta \hat n \]
\[ = \nu \left( OP \right) \sin \theta \hat n\]
\[ = \nu\left( OQ \right)\hat n\]
This product is always equal to the perpendicular distance from point O. Also, the direction of this product remains constant.
So, irrespective of the the position of the particle, the magnitude and direction of \[\vec{OP} \times \vec{\nu}\] remain constant. \[\therefore \vec{OP} \times \vec{\nu}\] is independent of the position P.
