Question

Let ` vecu = u_(1) hati + u_(2) hatj` be a unit vector in xy plane and ` vecw = 1/sqrt6 (hati +hatj +2hatk)` . Given that there exists a vector `vecc` " in " `R_(3)` " such that `|vecu xx vecv|=1` and `vecw `. `(vecu xx vecv) =1` , then
A. `|u_(1)|=|u_(2)|`
B. `|u_(2)|=2|u_(2)|`
C. ` 2|u_(1)|=|u_(2)|`
D. ` |u_(1)|-3|u_(2)|`

Answer 1

Correct Answer - A
we have ,
` veca.vecb= veca.vecc = vec0`
` Rightarrow veca bot vecb, veca bot vecc`
` Rightarrow veca || (vecb xx vecc)`
` Rightarrow veca = +- (vecb xx vecc)/(|vecb xx vecc|) =+- (|vecb xx vecc|)/(|vecb||vecc| sin pi/3) = +- 2 ( vecb xx vecc)`