Let two non-collinear vectors `veca` and `vecb` inclined at an angle `(2pi)/(3)` be such that `|veca|=3` and `vec|b|=2`. If a point P moves so that at any time t its position vector `vec(OP)` (where O is the origin) is given as `vec(OP) = (t+1/t)veca+(t-1/t)vecb` then least distance of P from the origin is

Options

A. `sqrt(2sqrt(133)-10)`
B. `sqrt(2sqrt(133)+10)`
C. `sqrt(5+sqrt(133))`
D. none of these

Answer

B

Explanation

We have `|vec(OP)|^(2)= (t+1/t)^(2)|veca|^(2)+(t+1/t)^(2) |vecb|^(2)+2(t^(2)-1/t^(2))|veca||vecb|cos(2pi)/(3)`
`therefore |vec(OP)|^(2) = 9(t+1/t)^(2)3.2.(-1/2)`
`=9(t+1/t)^(2)+4(t-1/t)^(2)+2(t^(2)-1/t^(2))3.2(-1/2)`
`=9(t^(2)+1/t^(2)+2)+4(t^(2)+1/t^(2)-2)-6(t^(2)-1/t^(2))`
`=7r^(2)+19/t^(2)+10`
`rArr |vec(OP)|^(2) ge2. (sqrt(7t^(2)).19/t^(2))+10 (therefore A.M geG.M)`
`therefore` minimum value of `|vec(OP)|= sqrt(10+2sqrt(133))`