$\overrightarrow{\text{OP}}\times\overrightarrow{\text{v}}=(\text{OP})\sin\theta \ \hat{\text{n}}=\text{v(OP)}\sin\theta \ \hat{\text{n}}=\text{v(OQ)} \ \hat{\text{n}}$
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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