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

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`