# Two non collinear position vectors veca & vecb are inclined at an angle (2pi)/3,where |veca|=3 & |vecb|=4 . A point P moves so that vec(OP)=(e^t+e^-t)veca +(e^t-e^-t)vecb. The least distance of P from origin O is sqrt2sqrt(sqrtp-q) then p+q =?

Jun 24, 2016

488

#### Explanation:

let $\vec{b}$ is along $\vec{x}$. So the the vector $\vec{O P}$ is given as:

$\vec{O P} = 3 \frac{\sqrt{3}}{2} \left({e}^{t} + {e}^{-} t\right) \vec{y} + \left(4 \left({e}^{t} - {e}^{-} t\right) - \frac{3}{2} \left({e}^{t} + {e}^{-} t\right)\right) \vec{x}$

hence the distance of P from origin is given as the magnitude of $\vec{O P}$. Which is:

$l =$$\left\mid \vec{O P} \right\mid = \sqrt{{\left(3 \frac{\sqrt{3}}{2} \left({e}^{t} + {e}^{-} t\right)\right)}^{2} + {\left(4 \left({e}^{t} - {e}^{-} t\right) - \frac{3}{2} \left({e}^{t} + {e}^{-} t\right)\right)}^{2}} = \sqrt{\frac{27}{4} \left({e}^{2 t} + {e}^{- 2 t} + 2\right) + 16 \left({e}^{2 t} + {e}^{- 2 t} - 2\right) + \frac{9}{4} \left({e}^{2 t} + {e}^{- 2 t} + 2\right) - 12 \left({e}^{2 t} - {e}^{- 2 t}\right)}$
=$\sqrt{13 {e}^{2 t} + 37 {e}^{- 2 t} - 14} = \sqrt{{\left(\sqrt{13} {e}^{t} - \sqrt{37} {e}^{- t}\right)}^{2} + 2 \left(\sqrt{37 \times 13} - 7\right)}$

for the minimum length, $\left(\sqrt{13} {e}^{t} - \sqrt{37} {e}^{- t}\right) = 0$, hence,

$p = 37 \times 13 = 481$ and $q = 7$

hence, $p + q = 488$