Question

If $\bar{a}$ and $\bar{b}$ are unit vectors and $\theta$ is the angle between them, then $\bar{a} + \bar{b}$ is a unit vector when $\theta$ is

• A. $\frac{\pi}{3}$
• B. $\frac{2\pi}{3}$
• C. $\frac{\pi}{2}$
• D. $\frac{\pi}{4}$

Hint:

If two unit vectors form an equilateral triangle with their sum, the angle between the vectors must be $120^\circ$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We are given that $\bar{a}$ and $\bar{b}$ are unit vectors, which means their magnitudes are 1.
We need to find the angle $\theta$ such that their sum $\bar{a} + \bar{b}$ is also a unit vector.
Step 2: Key Formula or Approach:
The magnitude squared of the sum of two vectors is given by:
\[ |\bar{a} + \bar{b}|^2 = |\bar{a}|^2 + |\bar{b}|^2 + 2\bar{a} \cdot \bar{b} \]
Recall that the dot product $\bar{a} \cdot \bar{b} = |\bar{a}| |\bar{b}| \cos \theta$.
Step 3: Detailed Explanation:
Given that $\bar{a}$ and $\bar{b}$ are unit vectors, so $|\bar{a}| = 1$ and $|\bar{b}| = 1$.
It is also given that $\bar{a} + \bar{b}$ is a unit vector, so $|\bar{a} + \bar{b}| = 1$.
Squaring both sides of this equation:
\[ |\bar{a} + \bar{b}|^2 = 1^2 = 1 \]
Expand the left side using vector properties:
\[ |\bar{a}|^2 + |\bar{b}|^2 + 2(\bar{a} \cdot \bar{b}) = 1 \]
Substitute the dot product formula:
\[ |\bar{a}|^2 + |\bar{b}|^2 + 2|\bar{a}||\bar{b}|\cos \theta = 1 \]
Substitute the known magnitudes ($|\bar{a}|=1$, $|\bar{b}|=1$):
\[ 1^2 + 1^2 + 2(1)(1)\cos \theta = 1 \]
\[ 1 + 1 + 2\cos \theta = 1 \]
\[ 2 + 2\cos \theta = 1 \]
Subtract 2 from both sides:
\[ 2\cos \theta = 1 - 2 \]
\[ 2\cos \theta = -1 \]
\[ \cos \theta = -\frac{1}{2} \]
The principal value of $\theta$ for which $\cos \theta = -1/2$ is in the second quadrant.
\[ \theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \]
Step 4: Final Answer:
The angle $\theta$ is $\frac{2\pi}{3}$.