If the unit vectors a and b are inclined at an angle 2θ such that $|\vec a-\vec b|<1$ and $0 ≤θ≤π$, then lies in the interval
Answer & explanation
Correct answer: option 1
We have,
$|\vec a-\vec b|^2=|\vec a|^2+|\vec b|^2-2(\vec a.\vec b)$
$⇒|\vec a-\vec b|^2=|\vec a|^2+|\vec b|^2-2|\vec a||\vec b|\cos 2θ$
$⇒|\vec a-\vec b|^2=2-2\cos 2θ$ $[∵|\vec a|=|\vec b|=1]$
$⇒|\vec a-\vec b|^2=4\sin^2θ$
$⇒|\vec a-\vec b|=2|\sin θ|$
Now,
$|\vec a-\vec b|<1$
$⇒2|\sin θ|<1⇒|\sin θ|<\frac{1}{2}⇒θ∈[0, π/6)∪(5π/6, π]$