Let \( \vec{a} \) and \( \vec{b} \) are two unit vectors such that \( \vec{a}+\vec{b} \) is also unit vector, then which of the following are TRUE?
(A) \( |\vec{a}-\vec{b}|=0 \)
(B) \( |\vec{a}-\vec{b}|=\sqrt{3} \)
(C) Angle between \( \vec{a} \) and \( \vec{b}=\displaystyle\frac{2\pi}{3} \)
(D) Angle between \( \vec{a} \) and \( \vec{a}+\vec{b}=\displaystyle\frac{\pi}{3} \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (B), (C) and (D) only
Let \( \theta \) be the angle between \( \vec{a} \) and \( \vec{b} \). Since all three vectors are unit vectors,
\( |\vec{a}+\vec{b}|^2=1+1+2\cos\theta=1 \)
\( \displaystyle \cos\theta=-\frac{1}{2} \)
\( \displaystyle \theta=\frac{2\pi}{3} \)
Then
\( |\vec{a}-\vec{b}|^2=1+1-2\cos\theta=2+1=3 \)
\( |\vec{a}-\vec{b}|=\sqrt{3} \)
If \( \alpha \) is the angle between \( \vec{a} \) and \( \vec{a}+\vec{b} \):
\( \displaystyle \cos\alpha=\frac{\vec{a}\cdot(\vec{a}+\vec{b})}{|\vec{a}|\,|\vec{a}+\vec{b}|}=\frac{1-\frac{1}{2}}{1\times 1}=\frac{1}{2} \)
\( \displaystyle \alpha=\frac{\pi}{3} \)
So \( |\vec{a}-\vec{b}|=\sqrt{3} \), the angle between \( \vec{a} \) and \( \vec{b} \) is \( \frac{2\pi}{3} \), and the angle between \( \vec{a} \) and \( \vec{a}+\vec{b} \) is \( \frac{\pi}{3} \).
Checking the statements
(A) \( |\vec{a}-\vec{b}|=0 \) — Incorrect. \( |\vec{a}-\vec{b}| \) is \( \sqrt{3} \).
(B) \( |\vec{a}-\vec{b}|=\sqrt{3} \) — Correct. \( |\vec{a}-\vec{b}|^2=2-2\cos\theta=3 \).
(C) Angle between \( \vec{a} \) and \( \vec{b}=\displaystyle\frac{2\pi}{3} \) — Correct. \( \cos\theta=-\frac{1}{2} \) gives \( \theta=\frac{2\pi}{3} \).
(D) Angle between \( \vec{a} \) and \( \vec{a}+\vec{b}=\displaystyle\frac{\pi}{3} \) — Correct. \( \cos\alpha=\frac{1}{2} \) gives \( \alpha=\frac{\pi}{3} \).