Let \(\vec{a}\) and \(\vec{b}\) be unit vectors and let \(\theta\) be the angle between them. Which of the following statements are correct?
(A) \(|\vec{a}-\vec{b}| = 2\cos\displaystyle\frac{\theta}{2}\)
(B) \(|\vec{a}+\vec{b}| = 2\cos\displaystyle\frac{\theta}{2}\)
(C) If \(|\vec{a}+\vec{b}| = 1\), then \(\theta = \displaystyle\frac{2\pi}{3}\)
(D) If \(\theta = \displaystyle\frac{\pi}{2}\), then \(\vec{a}\cdot\vec{b} = 1\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (B) and (C) only
Since \(|\vec{a}| = |\vec{b}| = 1\), we have \(\vec{a}\cdot\vec{b} = \cos\theta\). Using \(|\vec{v}|^2 = \vec{v}\cdot\vec{v}\):
\(|\vec{a}+\vec{b}|^2 = 1+1+2\cos\theta = 2(1+\cos\theta)\)
\( \displaystyle = 4\cos^2\frac{\theta}{2} \)
\(|\vec{a}-\vec{b}|^2 = 1+1-2\cos\theta = 2(1-\cos\theta)\)
\( \displaystyle = 4\sin^2\frac{\theta}{2} \)
For \(0 \le \theta \le \pi\), both \(\cos\frac{\theta}{2}\) and \(\sin\frac{\theta}{2}\) are non-negative.
So \(|\vec{a}+\vec{b}| = 2\cos\frac{\theta}{2}\), \(|\vec{a}-\vec{b}| = 2\sin\frac{\theta}{2}\), and \(\vec{a}\cdot\vec{b} = \cos\theta\).
Checking the statements
(A) \(|\vec{a}-\vec{b}| = 2\cos\frac{\theta}{2}\) — Incorrect. \(|\vec{a}-\vec{b}| = 2\sin\frac{\theta}{2}\); for \(\theta = 0\) it is 0, not 2.
(B) \(|\vec{a}+\vec{b}| = 2\cos\frac{\theta}{2}\) — Correct. \(|\vec{a}+\vec{b}|^2 = 2(1+\cos\theta) = 4\cos^2\frac{\theta}{2}\).
(C) If \(|\vec{a}+\vec{b}| = 1\), then \(\theta = \frac{2\pi}{3}\) — Correct. \(2+2\cos\theta = 1\) gives \(\cos\theta = -\frac{1}{2}\).
(D) If \(\theta = \frac{\pi}{2}\), then \(\vec{a}\cdot\vec{b} = 1\) — Incorrect. \(\vec{a}\cdot\vec{b} = \cos\frac{\pi}{2} = 0\).