If \( \vec{a} \) and \( \vec{b} \) are two unit vectors inclined at an angle \( \theta \), then the value of \( |\vec{a}-\vec{b}| \) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 2\sin\displaystyle\frac{\theta}{2} \)
Since \( |\vec{a}|=|\vec{b}|=1 \) and \( \vec{a}\cdot\vec{b}=\cos\theta \):
\( \displaystyle |\vec{a}-\vec{b}|^2=|\vec{a}|^2+|\vec{b}|^2-2\,\vec{a}\cdot\vec{b} \)
\( \displaystyle =2-2\cos\theta \)
Using the half-angle identity \( 1-\cos\theta=2\sin^2\frac{\theta}{2} \):
\( \displaystyle |\vec{a}-\vec{b}|^2=4\sin^2\frac{\theta}{2} \)
As \( 0\le\theta\le\pi \), \( \sin\frac{\theta}{2}\ge 0 \), so
\( \displaystyle |\vec{a}-\vec{b}|=2\sin\frac{\theta}{2} \)
Tip: A matching result is worth remembering: for unit vectors, \( |\vec{a}+\vec{b}|=2\cos\frac{\theta}{2} \) and \( |\vec{a}-\vec{b}|=2\sin\frac{\theta}{2} \). A quick check: \( \theta=0 \) must give \( |\vec{a}-\vec{b}|=0 \).