If `veca=hati+hatj+hatk, vecb=4hati+3hatj+4hatk` and `vecc=hati+alphahatj+betahatk`
are linearly dependent vectors and `|vecc|=sqrt(3)` then:

Options

A. (a) `alpha=1,beta=-1`
B. (b) `alpha=1,beta=+-1`
C. (c) `alpha=+-1,beta=+-1`
D. (d) `alpha=+-1,beta=1`

Answer

D

Explanation

The given vectors are linearly dependent, hence there exist scalars x,y and z not all zero, such that
`xa+yb+zc=0`
i.e., `x(hati+hatj+hatk)+y(4hati+3hatj+4hatk)+z(hati+alphahatj+betahatk)=0`
i.e., `(x+4y+z)hati+(x+3y+alphaz)hatj+(x+4y+betaz)hatk=0`
`impliesx+4y+z=0, x+3y+alphaz=0,x+4y+betaz=0`
For non-trivial solution `|(1,4,1),(1,3,alpha),(1,4,beta)|=0implies beta=1`
`|c|^(2)=3implies1+alpha^(2)+beta^(2)=3`
`impliesalpha^(2)=2-beta^(2)=2-1=1`
`thereforealpha=+-1`
Trick `|c|=sqrt(1+alpha^(2)+beta^(2))=sqrt(3)`
`implies alpha^(2)+beta^(2)=2`
`because` a,b and c are linearly dependent, hence `|(1,1,1),(4,3,4),(4,alpha,beta)|=0`
`impliesbeta=1`
`thereforealpha^(2)=1impliesalpha=+-1`.