# 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 =?

Oct 1, 2016

$p + q = 488$

#### Explanation:

$\vec{O P} = 2 \cosh \left(t\right) \vec{a} + 2 \sinh \left(t\right) \vec{b}$

${\left\lVert \vec{O P} \right\rVert}^{2} = 4 \left({\left\lVert \vec{a} \right\rVert}^{2} {\cosh}^{2} \left(t\right) + 2 \left\langle\vec{a} , \vec{b}\right\rangle \cosh \left(t\right) \sinh \left(t\right) + {\left\lVert \vec{b} \right\rVert}^{2} {\sinh}^{2} \left(t\right)\right)$ or

${\left\lVert \vec{O P} \right\rVert}^{2} = 4 \left({3}^{2} {\cosh}^{2} \left(t\right) + 2 \cdot 3 \cdot 4 \cdot \cos \left(\frac{2 \pi}{3}\right) \cosh \left(t\right) \sinh \left(t\right) + {4}^{2} {\sinh}^{2} \left(t\right)\right)$ and after some simplifications

${\left\lVert \vec{O P} \right\rVert}^{2} = 50 \cosh \left(2 t\right) - 24 \sinh \left(2 t\right) - 14$

Now, $\min {\left\lVert \vec{O P} \right\rVert}^{2}$ is at the stationary points determined by

$\frac{d}{\mathrm{dt}} {\left\lVert \vec{O P} \right\rVert}^{2} = - 48 \cosh \left(2 t\right) + 100 \sinh \left(2 t\right) = 0$

with real solution at

$t = \frac{1}{4} {\log}_{e} \left(\frac{37}{13}\right)$ so

$\min \left\lVert \vec{O P} \right\rVert = \sqrt{2} \sqrt{\sqrt{481} - 7}$

Here $p = 481$ and $q = 7$ and finally

$p + q = 488$

Oct 1, 2016

Choosing x-axis in the direction of the vector $a$,

$a = 3 \left(\cos 0 , \sin 0\right) = 3 \left(1 , 0\right)$ and

b=4(cos (2/3pi), sin (2/3pi)= 4(-1/2, sqrt 3/2).

Now, the vector in the addition for the vector $O P$ are scalar

multiples of $a \mathmr{and} b$, and so, these are collinear with $a \mathmr{and} b$,

respectively.

And so, the angle in between is $\frac{2}{3} \pi \mathmr{and} 2 \pi - \frac{2}{3} \pi$.

Now, $| O P {|}^{2} = {\left(3 \left({e}^{t} + {e}^{-} t\right)\right)}^{2} + {\left(4 \left({e}^{t} - {e}^{-} t\right)\right)}^{2}$

$- \left(2\right) \left(3\right) \left(4\right) \left({e}^{t} + {e}^{-} t\right) \left({e}^{t} - {e}^{-} t\right) \cos \left(\frac{2}{3} \pi\right)$

to Cesareo.

Of course, I can add that vector $O P$ for this minimum length

$O P = \sqrt{2 \sqrt{481} - 14}$ is

$\sqrt{2 \sqrt{481} - 14} \left(\cos {76.25}^{o} , \sin {75.25}^{o}\right)$

$= 5.465 \left(\cos {76.25}^{o} , \sin {75.25}^{o}\right)$, nearly.

The angle ie what this makes with vector $a$.

A graphical depiction could enhance the merits of this nonpareil

problem.
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