Question

If \(|\vec a|=2\) and \(|\vec b|=3\), then the maximum value of \[ 3\left|\left(\vec a+2\vec b\right)\right| + 4\left|\left(3\vec a-2\vec b\right)\right| \] is:

• A. \(30\)
• B. \(36\)
• C. \(60\)
• D. \(72\)

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We need to find the maximum value of an expression involving magnitudes of linear combinations of vectors. We let \(\theta\) be the angle between \(\vec{a}\) and \(\vec{b}\) and express the magnitudes as functions of \(\theta\).
Step 2: Key Formula or Approach:
1. Magnitude: \(|\vec{u} + \vec{v}| = \sqrt{|\vec{u}|^2 + |\vec{v}|^2 + 2|\vec{u}||\vec{v}|\cos\theta}\).
2. Trigonometric Max: Max of \(A \sin \phi + B \cos \phi = \sqrt{A^2 + B^2}\).
Step 3: Detailed Explanation:
Let \(|\vec{a}| = 2\) and \(|\vec{b}| = 3\). Let the angle between \(\vec{a}\) and \(\vec{b}\) be \(\theta\).
Consider \(|3\vec{a} + 2\vec{b}|\):
\[ |3\vec{a} + 2\vec{b}| = \sqrt{9|\vec{a}|^2 + 4|\vec{b}|^2 + 12|\vec{a}||\vec{b}|\cos\theta} = \sqrt{9(4) + 4(9) + 12(2)(3)\cos\theta} \]
\[ = \sqrt{36 + 36 + 72\cos\theta} = \sqrt{72(1 + \cos\theta)} = \sqrt{144 \cos^2(\theta/2)} = 12|\cos(\theta/2)| \]
Similarly for \(|3\vec{a} - 2\vec{b}|\):
\[ |3\vec{a} - 2\vec{b}| = \sqrt{36 + 36 - 72\cos\theta} = \sqrt{72(1 - \cos\theta)} = \sqrt{144 \sin^2(\theta/2)} = 12|\sin(\theta/2)| \]
The expression is \(E = 3(12|\cos(\theta/2)|) + 4(12|\sin(\theta/2)|)\):
\[ E = 36 |\cos(\theta/2)| + 48 |\sin(\theta/2)| \]
For maximum value, we use the property that \(\max(a \cos \phi + b \sin \phi) = \sqrt{a^2 + b^2}\):
\[ \text{Max } E = \sqrt{36^2 + 48^2} = \sqrt{(12 \times 3)^2 + (12 \times 4)^2} \]
\[ = 12 \sqrt{3^2 + 4^2} = 12 \times 5 = 60 \]
Step 4: Final Answer:
The maximum value is 60.