Find the value of a so that the volume of the parallelopiped formed by vectors `hati+ahatj+hatk,hatj+ahatkandahati+hatk` becomes minimum.

Options

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

Answer

C

Explanation

Let `V` be the volume of the parallelopiped formed by the given vectors.
Let `vec(alpha)=hati+ahatj+hatk, vec(beta)=hatj+ahatk` and `hat(gamma)=ahati+hatk`. Then,
`[(vec(alpha),vec(beta),vec(gamma))]=|(1,a,1),(0,1,a),(a,0,1)|=1+a^(3)-a`
`:.V=|[(vec(alpha),vec(beta),vec(gamma))]|`
Clearly `V` will be minimum when `lamda=[(vec(alpha),vec(beta),vec(gamma))]=1+a^(3)-a` is minimum.
Now, `lamda=1+a^(3)-a`
`=(d lamda)/(da)=3a^(2)-1` and `(d^(2)lamda)/(da^(2))=6a`
For maximum or minimum value of `lamda` we must have
`(d lamda)/(da)=0impliesa=+-1/(sqrt(3))`
Clearly `((d^(2)lamda)/(da^(2)))_(a=1//sqrt(3))=6/(sqrt(3))gt0`
So `lamda` is minimum for `a=1/(sqrt(3))`
Hence `V` is minimum for `a=1//sqrt(3)`.