Question

If $\bar{\mathrm{e}}_1$, $\bar{\mathrm{e}}_2$ and $\bar{\mathrm{e}}_1+\bar{\mathrm{e}}_2$ are unit vectors, then the angle between $\bar{\mathrm{e}}_1$ and $\bar{\mathrm{e}}_2$ is

• A. $150^\circ$
• B. $120^\circ$
• C. $90^\circ$
• D. $135^\circ$

Hint:

This configuration creates a perfectly enclosed equilateral vector triangle! If two adjacent sides have length 1 and the closing resultant side also has a length of 1, the internal angle is $60^\circ$. The angle between the outward-pointing vector directions is the supplementary exterior angle: $180^\circ - 60^\circ = 120^\circ$!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given that two individual vectors $\bar{\mathrm{e}}_1$, $\bar{\mathrm{e}}_2$, and their vector sum $\bar{\mathrm{e}}_1+\bar{\mathrm{e}}_2$ are all unit vectors. We need to find the angle $\theta$ between the two original vectors.

Step 2: Key Formula or Approach:
The magnitude of the sum of two vectors is given by the vector dot product expansion identity: $$ |\bar{\mathrm{e}}_1 + \bar{\mathrm{e}}_2|^2 = |\bar{\mathrm{e}}_1|^2 + |\bar{\mathrm{e}}_2|^2 + 2|\bar{\mathrm{e}}_1||\bar{\mathrm{e}}_2|\cos\theta $$

Step 3: Detailed Explanation:
Since $\bar{\mathrm{e}}_1$, $\bar{\mathrm{e}}_2$, and $\bar{\mathrm{e}}_1+\bar{\mathrm{e}}_2$ are all unit vectors, their magnitudes are all exactly equal to 1: $$ |\bar{\mathrm{e}}_1| = 1, \quad |\bar{\mathrm{e}}_2| = 1, \quad |\bar{\mathrm{e}}_1 + \bar{\mathrm{e}}_2| = 1 $$ Substitute these magnitude values directly into our identity formula: $$ (1)^2 = (1)^2 + (1)^2 + 2(1)(1)\cos\theta $$ $$ 1 = 1 + 1 + 2\cos\theta $$ $$ 1 = 2 + 2\cos\theta $$ Subtract 2 from both sides to isolate the trigonometric term: $$ -1 = 2\cos\theta \implies \cos\theta = -\frac{1}{2} $$ The principal angle value where the cosine function equals $-0.5$ is: $$ \theta = \cos^{-1}\left(-\frac{1}{2}\right) = 180^\circ - 60^\circ = 120^\circ $$

Step 4: Final Answer:
The angle between the two unit vectors is $120^\circ$, which matches option (B).