Question

Let $\vec{a}$ and $\vec{b}$ be two unit vectors such that $\vec{a}+\vec{b}$ is also a 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 $\vec{b}=\frac{2\pi}{3}$
(D) Angle between $\vec{a}$ and $\vec{b}=\frac{\pi}{3}$

• A. (B), (C) and (D) only
• B. (A) and (C) only
• C. (B) and (C) only
• D. (A), (C) and (D) only

Hint:

If magnitude of sum is given, directly use cosine formula to find angle.

Answer 1

The Correct Option is C

Concept: Use identity for magnitude: \[ |\vec{a}+\vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a}\cdot\vec{b} \]

Step 1: {Given condition.}
\[ |\vec{a}|=1,\quad |\vec{b}|=1,\quad |\vec{a}+\vec{b}|=1 \]

Step 2: {Apply formula.}
\[ 1 = 1+1+2\cos\theta \] \[ 1 = 2 + 2\cos\theta \Rightarrow 2\cos\theta = -1 \Rightarrow \cos\theta = -\frac{1}{2} \]

Step 3: {Find angle.}
\[ \theta = \frac{2\pi}{3} \]

Step 4: {Check options.}
\[ |\vec{a}-\vec{b}|^2 = 2 - 2\cos\theta = 2 - 2(-1/2)=3 \] \[ |\vec{a}-\vec{b}|=\sqrt{3} \] So: \[ (B), (C) \text{ are correct} \] (A) is false and (D) is false.