Let ` vecu = u_(1)hati + u_(2)hatj +u_(3)hatk` be a unit vector in ` R^(3) and vecw = 1/sqrt6 ( hati + hatj + 2hatk)` , Given that there exists a vector `vecv " in " R^(3)` such that ` | vecu xx vecv| =1 and vecw . ( vecu xx vecv) =1` which of the following statements is/are correct ?

Options

A. a. there is exactly one choice for such `vecv`
B. b. there are infinitely many choices for such `vecv`
C. c. if `hatu` lies in the xy - plane then `|u_(1)|=|u_(2)|`
D. d. if `hatu` lies in the xz-plane then `2|u_(1)|=|u_(3)|`

Answer

B, C

Explanation

b.,c.
`|hatuxxvecv|=1`
`implies |vecv|sin theta=1, " where"theta` is angle between `hatu and vecv`
also `hatw. (hatuxxvecv)=1`
`implies |hatw||hatu||hatv|sin theta cosalpha=1,`where `alpha` is angle beteen `hatw and (hatuxxvecv)`
`implies 1,1 (1) cos alpha=1`
`implies alpha=0`
`implies hatuxxvecv=lamdahatw`where `lamdagt0`
`implies |{:(hati,hatj,hatk),(u_(1),u_(2),u_(3)),(v_(1),v_(2),v_(3)):}|=(lamda)/(sqrt(6))(hati+hatj+2hatk)`
`implies (u_(2)v_(3)-u_(3)v_(2))hati+(u_(3)v_(1)-u_(1)v_(3)) hatj+(u_(1)v_(2)v_(1)) hatk=(lamda)/(sqrt(6))(hati+hatj+2hatk)`
`vecv` is a vector such that `(hatuxxvecv)` is parallel to `hatw`
`u_(3)=0impliesu_(2)v_(3)=(lamda)/(sqrt(6))and -u_(1)v_(3)=(lamda)/(sqrt(6))implies |u_(2)|=|u_(1)|`
` u_(2)=0implies -u_(3)v_(2)=(lamda)/(sqrt(6))and u_(1)v_(1)=(2lamda)/(sqrt(6))implies |u_(1)|=2|u_(3)|`