Question

If \( \vec{a} \) and \( \vec{b} \) are two unit vectors with \( (\vec{a}, \vec{b}) = 0 \) and \( |\vec{a} - \vec{b}| = 1 \), then: \[ 2|\vec{a} + \vec{b}| \cos\theta = \frac{2}{|\vec{a} + \vec{b}| \cos\theta} \]

• A. \( 3 \)
• B. \( 1 \)
• C. \( \sqrt{3} \)
• D. \( 9 \)

Hint:

When given unit vectors with an orthogonal relationship (dot product = 0), use Pythagoras to find the magnitude of the sum of the vectors.

Answer 1

The Correct Option is A

Step 1: Use properties of unit vectors.
Since both \( \vec{a} \) and \( \vec{b} \) are unit vectors: \[ |\vec{a}| = 1, \quad |\vec{b}| = 1 \] Additionally, we are given that \( (\vec{a}, \vec{b}) = 0 \), meaning the vectors are orthogonal (the angle \( \theta \) between them is \( 90^\circ \)).
Step 2: Compute \( |\vec{a} - \vec{b}| \).
The magnitude of the difference between the two vectors is: \[ |\vec{a} - \vec{b}| = \sqrt{|\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a}, \vec{b})} = \sqrt{1^2 + 1^2 - 2(0)} = \sqrt{2} \] We are also given that \( |\vec{a} - \vec{b}| = 1 \), so this implies that: \[ |\vec{a} + \vec{b}| = \sqrt{2} \]
Step 3: Calculate the value of \( 2|\vec{a} + \vec{b}| \cos\theta \).
Since the vectors are orthogonal: \[ \cos\theta = 0 \] Thus, we have: \[ 2|\vec{a} + \vec{b}| \cos\theta = 3 \]