`veca ,vecb and vecc` are unimdular and coplanar. A unit vector `vecd` is perpendicualt to them , `(veca xx vecb) xx (vecc xx vecd) = 1/6 hati - 1/3 hatj + 1/3 hatk` , and the angle between `veca and vecb is 30^(@)` then `vecc` isA. `(hati-2 hatj + 2 hatk)//3`B. `(-hati +2hatj - 2 hatk)//3`C. `(-hati +2hatj - hatk)//3`D. `(-2hati -2hatj + hatk)//3`

`veca ,vecb and vecc` are unimdular and coplanar. A unit vector `vecd`

1.	`veca ,vecb and vecc` are unimdular and coplanar. A unit vector `vecd` is perpendicualt to them , `(veca xx vecb) xx (vecc xx vecd) = 1/6 hati - 1/3 hatj + 1/3 hatk` , and the angle between `veca and vecb is 30^(@)` then `vecc` isA. `(hati-2 hatj + 2 hatk)//3`B. `(-hati +2hatj - 2 hatk)//3`C. `(-hati +2hatj - hatk)//3`D. `(-2hati -2hatj + hatk)//3`
Answer» Correct Answer - a,b Given `1/6hati -1/3 hatj + 1/3hatk = (veca xx vecb) xx (veccxxvecd)` `[ veca vecb vecd]vecc - [veca vecb vecc] vecd` `[veca vecb vecd]vecc` `[ veca, vecb and vecc` are coplanar] `[veca vecb vecd] = (veca xx vecb).vecd` ` \|veca xx vecb\| \|vecd\| cos theta` `(therefore vecd bot veca,vecd,bot vecb, therefore vecd\|\|vecaxxvecb)` `ab sin 30^(@) .1. (+-1) ( theta = 0 or pi) ` `1.1. 1/2 .1(+-1)= +- 1/2` form (i) m we have `vecc= +- (1/3 hati -2/3hatj +2/3hatk) =+- (hati -2hatj + 2hatk)/3`

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