The volume of the parallelepiped whose coterminous edges are represented by the vectors `2vecb xx vecc, 3vecc xx veca` and `4veca xx vecb` where `veca=(1+sintheta)hati+costhetahatj+sin2thetahatk` , `vecb=sin(theta+(2pi)/(3))hati+cos(theta+(2pi)/(3))hatj+sin(2theta+(4pi)/(3))hatk`, `vecc=sin(theta-(2pi)/(3))hati+cos(theta-(2pi)/(3))hatj + sin(2theta-(4pi)/(3))hatk` is 18 cubic units, then the values of `theta`, in the interval `(0,pi/2)`, is/are

Options

A. `pi/9`
B. `(2pi)/(9)`
C. `pi/3`
D. `(4pi)/9`

Answer

A, B, D

Explanation

Volume `=|[2vecb xx vecc 3 vecb]|=18`
`rArr 24[vecavecbvecc]^(2)=18`
`rArr [|vecavecbvecc|]=sqrt(3)/2`
Now, `[vecavecbvecc]= |{:(1+sintheta,costheta,sin2theta),(sintheta+(2pi)/(3),costheta+(2pi)/(3),sin2theta+(4pi)/(3)),(sintheta-(2pi)/(3), costheta-(2pi)/(3), sin2theta-(4pi)/(3)):}|`
Applying `R_(1) to R_(1) + R_(2)+R_(3)` and expanding
`|[vecavecbvecc]|=sqrt(3)|cos3theta|=sqrt(3)/2`
`rArr cos3theta=+-1/2 rArr 3theta=pi/3,(2pi)/(3), (4pi)/(3)`
`rArr theta=pi/9, (2pi)/9, (4pi)/9`