Question

Let $|a| = 2, |b| = 3$, then maximum value of $3|3a + 2b| + 4|3a - 2b|$ is :

• A. 50
• B. 60
• C. 70
• D. 80

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We need to find the maximum value of a linear combination of vector magnitudes. Let \( \theta \) be the angle between vectors \( \vec{a} \) and \( \vec{b} \).
Step 2: Key Formula or Approach:
\( |x\vec{a} \pm y\vec{b}| = \sqrt{x^2a^2 + y^2b^2 \pm 2xy |\vec{a}||\vec{b}| \cos \theta} \).
Here \( |3\vec{a}| = 3(2) = 6 \) and \( |2\vec{b}| = 2(3) = 6 \).
Step 3: Detailed Explanation:
\( |3\vec{a} + 2\vec{b}| = \sqrt{6^2 + 6^2 + 2(6)(6) \cos \theta} = \sqrt{72(1 + \cos \theta)} = \sqrt{144 \cos^2(\theta/2)} = 12 \cos(\theta/2) \).
Similarly, \( |3\vec{a} - 2\vec{b}| = \sqrt{6^2 + 6^2 - 2(6)(6) \cos \theta} = \sqrt{72(1 - \cos \theta)} = 12 \sin(\theta/2) \).
Let \( \alpha = \theta/2 \). The expression is \( E = 3(12 \cos \alpha) + 4(12 \sin \alpha) = 36 \cos \alpha + 48 \sin \alpha \).
The maximum value of \( A \cos \alpha + B \sin \alpha \) is \( \sqrt{A^2 + B^2} \).
\( E_{max} = \sqrt{36^2 + 48^2} = \sqrt{12^2(3^2 + 4^2)} = 12 \times 5 = 60 \).
Step 4: Final Answer:
The maximum value of the given expression is 60.