Let `veca , vecb and vecc` be vectors forming right- hand triad . Let `vecP= (vecbxxvecc)/([vecavecb vecc]), vecq (vecc xx veca)/([veca vecb vecc])and vecr= (vecaxxvecb)/([veca vecb vecc]) If x cupR^(+)` then

Options

A. `x [veca vecb vecc] + ([vecp vecqvecr])/x ` has least value 2
B. `x^(2) [veca vecb vecc]^(2) + ([vecp vecqvecr])/x^(2) ` has least value `(3//2^(2//3))`
C. `[vecp vecq vecr] gt 0 `
D. none of these

Answer

a,c

Explanation

we have ` [vecp vecq vecr] = 1/ ([ veca vecb vecc])` therefore,
` [vecp vecq vecr] gt 0`
a. ` x gt 0, x [veca vecb vecc] + ([vecp vecq vecr])/x ge 2`
( using ` A.M. ge G.M.`)
b . Similarly, use `A.M. ge G.M`