Let u^=u1i^+u2j^+u3k^ be a unit vector in ℝ3 and ω^=16(i^+j^+2k¯). Given that three exists a vector v→ in ℝ3 such that |u^×v→|=1 and w.^(u^×v→)=1 which of the following statement(s) is (are) correct?
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a
There is exactly one choice for such v→
b
There are infinitely many choices for such v→
c
if u^ lies in the xy plane then u1=u2
d
if u^ lies in the xz plane then 2u1=u3
answer is B.
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Detailed Solution
u=1,w=1,u×v=1⇒vsinu,v=1w.^u^×v→=1⇒wu×vcosw,u×v=1⇒cos ω,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→ existsif u^=u1i^+u2j^ and u^.w^=0 ⇒u1+u2=0⇒u1=u2if u^=u1i^+u3k^ and u^.w^=0 ⇒u1+2u3=0⇒u1=2u3