Question
A particle moves on a given straight line with a constant speed v. At a certain time it is at a point P on its straight line path. O is a fixed point. Show that $\overrightarrow{\text{OP}}\times\overrightarrow{\text{v}}$ is independent of the position P.

Answer

The particle moves on the straight line PP’ at speed v. From the figure,$\overrightarrow{\text{OP}}\times\overrightarrow{\text{v}}=(\text{OP})\sin\theta \ \hat{\text{n}}=\text{v(OP)}\sin\theta \ \hat{\text{n}}=\text{v(OQ)} \ \hat{\text{n}}$
It can be seen from the figure, $\text{OQ = OP}\sin\theta=\text{OP}'\sin\theta'$ So, whatever may be the position of the particle, the magnitude and direction of $\overrightarrow{\text{OP}}\times\vec{\text{v}}$ remain constant.$\therefore\overrightarrow{\text{OP}}\times\vec{\text{v}}$ remain constant.

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