Let `veca=hati-hatj, vecb=hatj-hatk, vecc=hatk-hati. If hatd` is a unit vector such that `veca.hatd=0=[vecb, vecc, vecd]` then hatd` equals (A) `+-(hati+hatj-2hatk)/sqrt(6)` (B) `+-(hati+hatj-hatk)/sqrt(3)` (C) `+-(hati+hatj+hatk)/sqrt(3)` (D) `+-hatk`

Options

A. `+- (hati + hatj - 2hatk)/sqrt6`
B. `+- (hati + hatj - hatk)/sqrt3`
C. `+- (hati + hatj + hatk)/sqrt3`
D. `+- hatk`

Answer

a

Explanation

Let ` vecd= xhati + yhatj + zhatk`
where `x^(2) + y^(2) +z^(2)=1`
( `vced ` being a unit vector)
`veca .vced=0`
` Rightarrow x-y =0 or x=y`
`[vecb vecc vecd]=0`
`Rightarrow |{:(0,1,-1),(-1,0,1),(x,y,z):}|=0`
or x+y +z=0
or 2x + z=0
or z= -2x
From (i), (ii), and (iii) we have
`x^(2) +x^(2)+4x^(2)=1`
`x = +- 1/sqrt6`
`vecd=+-(1/sqrt6hati1/sqrt6hatj-2/sqrt6hatk)`
`= +- ((hati+hatj -2hatk)/sqrt6)`