Let `vecu = 2hati - hatj + hatk, vecv = -3hatj + 2hatk` be vectors in `R^(3)` and `vecw` be a unit vector in the xy-plane. Then the maximum possible value of `|(vecu xx vecv).vecw|` is-

Options

A. `sqrt(5)`
B. `sqrt(12)`
C. `sqrt(13)`
D. `sqrt(17)`

Answer

D

Explanation

`vecuxxvecv = (2hati - hatj + hatk)xx(-3hatj + 2hatk)`
`-6hatk - 4hatj - 2hati + 3hati = hati - 4hatj - 6hatk`
Let `vecw = a hati + bhatj , a^(2) + b^(2) = 1 , a cos theta , b = sintheta`
`vecu xx vecv.vecw = a - 4b = costheta - 4 sin theta`
max. value `= sqrt(1^(2) +(-4)^(2)) = sqrt(17)`