Question

Let `vecu=u_(1)hati +u_(3)hatk` be a unit vector in xz-plane and ` vecq = 1/sqrt6 ( hati + hatj + 2hatk) `. If there exists a vector `vecv` in such that ` |vecu xx vecu|=1 and vecw . (vecu xx vecc)` .Then
A. `|u_(1)|=|u_(3)|`
B. `|u_(1)|= 2 |u_(3)|`
C. ` |u_(1)|=2|u_(3)|`
D. `2|u_(1)|=|u_(3)|`

Answer 1

Correct Answer - B
we observe that `vecw` is a unit vector.
Now, ` |vecu xx vecv| =1 and vecw . (vecu xx vecv) = 1`
` Rightarrow vecw = vecu xx vecv`
` Rightarrow 1/sqrt6 (hati +hatj + 2hatk) = |{:(hati,hatj,hatk),(u_(1),0,u_(3)),(v_(1),(v_(2),(v_(3)):}|`
` Rightarrow 1/sqrt6 ( hati + hatj +2hatk) = -u_(3)v_(2)hati-(u_(1)v_(3)-u_(3)v_(1))hatj +u_(1)v_(2)hatk`
` Rightarrow u_(3) v_(2)= -1/sqrt6 ,-(u_(1)v_(3)- u_(3)v_(1)) = 1/sqrt6 and u_(1)v_(2)= 2/sqrt6`
` Rightarrow u_(1)v_(2) = -2 u_(3)v_(2)`
` Rightarrow u_(1) = -2u_(3) Rightarrow |u_(1)| = 2|u_(3)|`