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Find the value of a so that the volume o...

Find the value of `a` so that the volume of the parallelepiped formed by vectors ` hat i+a hat j+k , hat j+a hat ka n da hat i+ hat k` becomes minimum.

A

`-(1)/(sqrt(3))`

B

`(1)/(sqrt(3))`

C

`sqrt(3)`

D

`-sqrt(3)`

Text Solution

Verified by Experts

The correct Answer is:
B

Given vecotors are `hati + lambda hatj + hatk, hatj+ lambda hat k and lambda hat i + hat k` which forms a parallelopiped
`:.` Volume of the parallelopiped is `V||{:(1,lambda,1),(0,1,lambda),(lambda,0,1):}||=1+lambda^(3)-lambda`
`rArr V= lambda^(3)-lambda+1`
On differentating. w.r.t. lambda, we get
`(dV)/(dlambda)=3lambda^(2)-1`
For maxima of minima, `(dV)/(dlambda)=0`
`rArr lamda+-(1)/(sqrt(3))`
`and (d^(2)V)/(dlambda^(2))=6lambda={{:(2sqrt(3)ge0,"for", lambda=(1)/(sqrt(3))),(2sqrt(3)le0, "for", lambda=-(1)/(sqrt(3))):}`
`:. (d^(2)V)/(d lambda^(2))` is poisitive for `lambda=(1)/(sqrt(3))`, so volume 'V' is minimum for `lambda=(1)/(sqrt(3))`
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