Let a unit vector u^=xi^+yj^+zk^ make angles π2,π3 and 2π3 with the vectors 12i^+12k^, 12j^+12k^ and 12i^+12j^ respectively. If v→=12i^+12j^+12k^, then |u^–v→|2 is equal to [2024]
(4)
Let a→=12i^+12k^, b→=12j^+12k^ and c→=12i^+12j^
Now, u^·a→=|u^|·|a→|cosθ,
⇒ x2+z2=1·1·cosπ2 ⇒ x+z=0 ... (i)
Again u^·b→=y2+z2=1·1·cosπ3 ⇒ y+z=12 ... (ii)
and u^·c→=x2+y2=1·1·cos2π3 ⇒ x+y=–12 ... (iii)
Solving (i), (ii) and (iii), we get
x=–12, y=0, z=12 ∴ u^=–12i^+12k^
So, |u^–v→|2=|–2i^–12j^|2=2+12=52.