Question

If \( |\vec{a}|^2 = 1 \), \( |\vec{b}| = 2 \) and \( \vec{a} \cdot \vec{b} = 2 \), then the value of \( |\vec{a} + \vec{b}| \) is:

• A. \( 9 \)
• B. \( 3 \)
• C. \( -3 \)
• D. \( 2 \)

Hint:

To find the magnitude of the sum of vectors, use the formula \( |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2 \vec{a} \cdot \vec{b} \), and ensure to take the positive square root for the magnitude.

Answer 1

The Correct Option is B

We are given: - \( |\vec{a}|^2 = 1 \), so \( |\vec{a}| = 1 \) (since magnitude is non-negative). - \( |\vec{b}| = 2 \). - \( \vec{a} \cdot \vec{b} = 2 \). We need to find the magnitude of \( |\vec{a} + \vec{b}| \). The magnitude of the sum of two vectors is given by: \[ |\vec{a} + \vec{b}|^2 = (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) \] Expanding the dot product: \[ |\vec{a} + \vec{b}|^2 = \vec{a} \cdot \vec{a} + 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} \] Substituting the given values: - \( \vec{a} \cdot \vec{a} = |\vec{a}|^2 = 1 \), - \( \vec{a} \cdot \vec{b} = 2 \), - \( \vec{b} \cdot \vec{b} = |\vec{b}|^2 = 2^2 = 4 \). So: \[ |\vec{a} + \vec{b}|^2 = 1 + 2 \cdot 2 + 4 = 1 + 4 + 4 = 9 \] Taking the square root (since magnitude is non-negative): \[ |\vec{a} + \vec{b}| = \sqrt{9} = 3 \] \[ \boxed{|\vec{a} + \vec{b}| = 3} \]