1.

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

Answer» Correct Answer - a
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)`


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