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 =?

1 Answer
Chitram Banerjee
Jun 24, 2016

488

Explanation:

let #vecb# is along #vecx#. So the the vector #vec{OP}# is given as:

#vec{OP} = 3sqrt3/2(e^t+e^-t)vecy +(4(e^t-e^-t)-3/2(e^t+e^-t))vecx#

hence the distance of P from origin is given as the magnitude of #vec(OP)#. Which is:

#l=##abs(vec(OP)) = sqrt((3sqrt3/2(e^t+e^-t))^2 + (4(e^t-e^-t)-3/2(e^t+e^-t))^2) = sqrt(27/4 (e^(2t)+e^(-2t)+2)+16(e^(2t)+e^(-2t)-2)+9/4(e^(2t)+e^(-2t)+2)-12(e^(2t)-e^(-2t)))#
=#sqrt(13e^(2t)+37e^(-2t)-14) = sqrt((sqrt13 e^(t)-sqrt37e^(-t))^2 + 2(sqrt(37xx13)-7))#

for the minimum length, #(sqrt13 e^(t)-sqrt37e^(-t)) =0#, hence,

#p=37xx13 = 481# and #q =7#

hence, #p+q = 488#

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