The three vectors `hat i+hat j,hat j+hat k, hat k+hat i` taken two at a time form three planes, The three unit vectors drawn perpendicular to these planes form a parallelopiped of volume:

Options

A. `1/3`
B. `4`
C. `(3sqrt(3))/4`
D. `4/(3sqrt(3))`

Answer

D

Explanation

Let `veca=hati+hatj, vecb=hatj+hatk` and `vec=hatk+hatk`. Further let
`vec(alpha)=(vecaxxvecb)/(|vecaxxvecb|),vec(beta)=(vecbxxvecc)/(|vecbxxvecc|)` and `vec(gamma)=(veccxxveca)/(|veccxxveca|)`
`:.` Required volume
`=[(vec(alpha), vec(beta), vec(gamma))]`
`=1/(|vecaxxvecb||vecbxxvecc||veccxxveca|)=[(vecaxxvecb, vecbxxvecc, vecxxveca)]`
`=([(veca, vecb, vecc)]^(2))/(|vecaxxvecb||vecbxxvecc||veccxxveca|) [ :' [(vecaxxvecb, vecbxxvecc, veccxxveca)], =[(veca, vecb, vecc)]^(2)]`
Now,
`[(veca, vecb, vecc)]=|(1,1,0),(0,1,1),(1,0,1)|=1-1(0-1)=2`
And
`vecaxxvecb=(hati+hatj)xx(hatj+hatk)=hatk-hatj+hati`
`vecbxxvecc=(hatj+hatk)xx(hatk+hati)=hati-hatk+hatj`
and `veccxxveca=(hatk+hati)xx(hati+hatj)=hatj-hati+hatk`
`:.|vecaxxvecb|=|vecbxxvecc|=|veccxxveca|=sqrt(3)`
Hence required volume `=(2^(2))/(3sqrt(3))=4/(3sqrt(3))`