Let u ^ = u 1 i ^ + u 2 j ^ + u 3 k ^ be a unit vector in R 3 and w ^ = 1 6   i ^ +   j ^ + 2 k ^ …

Let u^=u1i^+u2j^+u3k^ be a unit vector in R3 and w^=16 i^+ j^+2k^. Given that there 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 ?
  1. There is exactly one choice for such v→
  2. There are infinitely many choice for such v→
  3. If u^ lies in the xy - plane then u1=u2
  4. If u^ lies in the xz - plane then 2u1=u3

Solution

As | w ^ |=| u ^ |=1=| u ^ × v → |
Let ϕ is angle between w ^   & ( u ^ × v → )
w^ u^×v→cos⁡ϕ=1  ⇒ ϕ=0
⇒ u ^ × v → = w ^
⇒ There may be infinite vectors v → Satisfying this condition
If u ^ his in xy plane : u^×v→=i^j^k^u1u20v1v2v3 = w ^
⇒ u 3 =0  
Let v → = v 1   i ^ + v 2 j ^ + v 3 k ^
w^=u2v3i^-u1v3j^+u1v2-u2v1k^ = 1 6 i ^ + 1 6 j ^ + 2 6 k ^
u2v3=16, -u1v3=16 ⇒  u1=u2
If u ^ his in xz plane ⇒ u 2 =0 : u^×v→=i^j^k^u10u3v1v2v3
w^=-v2u3i^-u1v3-u3v1j^+u1v2k^ = 1 6 i ^ + 1 6 j ^ + 2 6 k ^
-v2u3=16, u1v2=26
⇒ 2u3=u1

Asked in: JEE Advanced 2016 (Paper 2)

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