Question

For any two vectors \( \mathbf{A} \) and \( \mathbf{B} \) if \( \mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos \theta \), the magnitude of \( (\mathbf{A} + \mathbf{B}) \) is (given \( \tan \theta = 1 \), \( \cos \theta = \frac{1}{\sqrt{2}} \))

• A. \( \sqrt{A^2 + B^2 + \sqrt{2} AB} \)
• B. \( \sqrt{A^2 + B^2 + \frac{AB}{\sqrt{2}}} \)
• C. \( A + B \)
• D. \( \sqrt{A^2 + B^2} \)

Hint:

When calculating the magnitude of the sum of two vectors, use the formula \( |\mathbf{A} + \mathbf{B}| = \sqrt{A^2 + B^2 + 2AB \cos \theta} \) and apply the known values of \( \cos \theta \).

Answer 1

The Correct Option is A

Step 1: Using the formula for the magnitude of the sum of two vectors.
The magnitude of the sum of two vectors \( \mathbf{A} \) and \( \mathbf{B} \) is given by: \[ |\mathbf{A} + \mathbf{B}| = \sqrt{A^2 + B^2 + 2AB \cos \theta} \] Step 2: Substituting the given values.
We are given \( \cos \theta = \frac{1}{\sqrt{2}} \), so: \[ |\mathbf{A} + \mathbf{B}| = \sqrt{A^2 + B^2 + 2AB \times \frac{1}{\sqrt{2}}} \] \[ |\mathbf{A} + \mathbf{B}| = \sqrt{A^2 + B^2 + \sqrt{2} AB} \] Thus, the correct answer is (A) \( \sqrt{A^2 + B^2 + \sqrt{2} AB} \).