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: Simplify the magnitude expressions.
Let $\theta$ be the angle between vectors $\u0000a$ and $\u0000b$.
Then $|3a + 2b|^2 = 9|a|^2 + 4|b|^2 + 12|a||b|\cos\theta$
$|3a + 2b|^2 = 9(4) + 4(9) + 12(2)(3)\cos\theta = 36 + 36 + 72\cos\theta = 72(1 + \cos\theta)$.
Similarly, $|3a - 2b|^2 = 72(1 - \cos\theta)$.


Step 2: Substitute back into the main expression $E$.
$E = 3\sqrt{72(1 + \cos\theta)} + 4\sqrt{72(1 - \cos\theta)}$
Using half-angle identities $1+\cos\theta = 2\cos^2(\theta/2)$ and $1-\cos\theta = 2\sin^2(\theta/2)$:
$E = 3\sqrt{72 \cdot 2\cos^2(\theta/2)} + 4\sqrt{72 \cdot 2\sin^2(\theta/2)}$
$E = 3 \cdot \sqrt{144} |\cos(\theta/2)| + 4 \cdot \sqrt{144} |\sin(\theta/2)|$
$E = 36 |\cos(\theta/2)| + 48 |\sin(\theta/2)|$


Step 3: Find the maximum value.
For an expression of the form $A\cos\alpha + B\sin\alpha$, the maximum value is $\sqrt{A^2 + B^2}$.
$E_{max} = \sqrt{36^2 + 48^2}$
Factor out 12: $E_{max} = 12\sqrt{3^2 + 4^2} = 12 \cdot 5 = 60$.
The answer is 60 (Option 2).