Question

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 =?

Answer 1

`p+q=488`

Explanation:

`vec (OP) = 2cosh(t)vec a+2sinh(t)vec b`

`norm(vec (OP))^2=4(norm veca^2cosh^2(t)+2<< vec a,vec b >> cosh(t)sinh(t)+norm vecb^2sinh^2(t))` or

`norm(vec (OP))^2= 4(3^2cosh^2(t)+2cdot 3cdot 4 cdot cos((2pi)/3)cosh(t)sinh(t)+4^2sinh^2(t))` and after some simplifications

`norm(vec (OP))^2=50 cosh(2 t) - 24 sinh(2 t)-14`

Now, `min norm(vec (OP))^2` is at the stationary points determined by

`d/(dt)norm(vec (OP))^2 = -48 cosh(2 t)+ 100 sinh(2 t) = 0`

with real solution at

`t = 1/4 log_e(37/13)` so

`min norm(vec (OP)) = sqrt(2)sqrt(sqrt(481)-7)`

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

`p+q=488`

Answer 2

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

`a=3(cos 0, sin 0)= 3(1, 0)` and

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

Now, the vector in the addition for the vector `OP` are scalar

multiples of `a and b`, and so, these are collinear with `a and b`,

respectively.

And so, the angle in between is `2/3pior 2pi-2/3pi`.

Now, `|OP|^2=(3(e^t+e^-t))^2+(4(e^t-e^-t))^2`

`-(2)(3)(4)(e^t+e^-t)(e^t-e^-t)cos(2/3pi)`

Beyond this, I follow the previous answer, with due compliments

to Cesareo.

Of course, I can add that vector `OP` for this minimum length

`OP= sqrt(2sqrt 481-14)` is

`sqrt(2sqrt 481-14)(cos 76.25^o, sin 75.25^o)`

`=5.465(cos 76.25^o, sin 75.25^o)`, 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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