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. `alpha=1, beta =-1`
B. `alpha= 1,beta =+-1`
C. `alpha=- 1,beta = +-1`
D. ` alpha= +- 1,beta =1`

Answer

D

Explanation

Since `vec(a) , vec(b) ,vec(c ) ` are linearly dependent vectors .
`rArr [vec(a) ,vec(b) vec(c )]=0`
` rArr |{:(1,,1,,1),(4,,3,,4),(1,,alpha,,beta ):}|=0`
applying `C_(2) to C_(2)- C_(1), C_(3) to C_(3) - C_(1)`
`|{:(1,,0,,0),(4,,-1,,0),(1,,alpha-1,,beta-1):}|=0 rArr - (beta - 1) =0 rArr beta =1`
Also `|vec(c )| = sqrt(3)`
`rArr 1+alpha + beta^(2) =3" ""[given " c = hat(i) + alpha hat(j) + beta hat(k)"]"`
`rArr 1+ alpha^(2) +1 =3 rArr alpha^(2) =1rArr alpha = +-1`