Let `veca,vecb and vecc` be the non zero vectors such that `(vecaxxvecb)xxvecc=1/3 |vecb||vecc|veca.` if theta is the acute angle between the vectors `vecb and veca` then theta equals (A) `1/3` (B) `sqrt(2)/3` (C) `2/3` (D) `2sqrt(2)/3`

Options

A. `(2sqrt2)/3`
B. `sqrt2/3`
C. `2/3`
D. `1/3`

Answer

A

Explanation

Given `(vecaxxvecb)xxvecc=1/3 |vecb||vecc|veca`
Clearly `veca` and `vecb` are noncollinear
`rArr (veca.vecc)vecb-(vecb.vecc)veca=1/3 |vecb||vecc|veca`
`therefore veca.vecc=0` and `-vecb.vecc=1/3 |vecb||vecc|rArr cos theta =(-1)/3`
`therefore sin theta =sqrt(1-1/9)=(2sqrt2)/3` [ `theta` is acute angle between `vecb` and `vecc` ]