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Q.

 Let u^=u1i^+u2j^+u3k^ be a unit vector in 3 and ω^=16(i^+j^+2k¯). Given that three  exists a vector v in 3 such that |u^×v|=1 and  w.^(u^×v)=1 which of the following  statement(s) is (are) correct? 

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a

 if u^ lies in the xy plane then u1=u2

b

 There are infinitely many choices for such v

c

 if u^ lies in the xz plane then 2u1=u3

d

 There is exactly one choice for such v

answer is B, C.

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Detailed Solution

u=1,w=1,u×v=1vsinu,v=1w.^u^×v=1wu×vcosw,u×v=1cosω,u×v=1hence w^ is perpendicular to both u^ and v  

As it is given there exists a vector v w^ must be  to u^

hence infinitely many such v exists

if u^=u1i^+u2j^  and  u^.w^=0 u1+u2=0u1=u2

if u^=u1i^+u3k^ and  u^.w^=0 u1+2u3=0u1=2u3

 

 

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