Question

Let \(\vec{a}\) and \(\vec{b}\) be two unit vectors, and \(\theta\) be the angle between them. If \(\vec{a}-\vec{b}\) is a unit vector, then \(\theta\) is equal to

• A. \(\frac{\pi}{3}\)
• B. \(\frac{\pi}{2}\)
• C. \(\frac{\pi}{4}\)
• D. \(\frac{2\pi}{3}\)
• E. \(\frac{\pi}{6}\)

Hint:

For vector problems, always use identities like \(|\vec{a}-\vec{b}|^2=|\vec{a}|^2+|\vec{b}|^2-2\vec{a}\cdot\vec{b}\). It simplifies calculations quickly.

Answer 1

The Correct Option is A

Step 1: Use the condition that \(\vec{a}\) and \(\vec{b}\) are unit vectors.
Since both \(\vec{a}\) and \(\vec{b}\) are unit vectors, we have:
\[ |\vec{a}|=1 \quad \text{and} \quad |\vec{b}|=1 \]

Step 2: Use the condition given in the question.
It is given that \(\vec{a}-\vec{b}\) is also a unit vector. Therefore:
\[ |\vec{a}-\vec{b}|=1 \]

Step 3: Use the identity for magnitude.
We know that:
\[ |\vec{a}-\vec{b}|^2=|\vec{a}|^2+|\vec{b}|^2-2\vec{a}\cdot\vec{b} \] Substituting the given values:
\[ 1^2=1^2+1^2-2\vec{a}\cdot\vec{b} \]

Step 4: Simplify the equation.
\[ 1=1+1-2\vec{a}\cdot\vec{b} \] \[ 1=2-2\vec{a}\cdot\vec{b} \] \[ 2\vec{a}\cdot\vec{b}=1 \] \[ \vec{a}\cdot\vec{b}=\frac{1}{2} \]

Step 5: Use the dot product formula.
We know that:
\[ \vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos\theta \] Since both are unit vectors:
\[ \vec{a}\cdot\vec{b}=\cos\theta \]

Step 6: Solve for \(\theta\).
\[ \cos\theta=\frac{1}{2} \] Thus,
\[ \theta=\frac{\pi}{3} \]

Step 7: Match with the options.
The value \(\frac{\pi}{3}\) corresponds to option \((1)\). Hence, the correct answer is:
\[ \boxed{\frac{\pi}{3}} \]