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 =?

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Answer 1 · 2016-10-01T08:50:31

#p+q=488#

#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 · 2016-10-01T14:44:13

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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