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

Concept: The magnitude of a vector expression depends on the angle between the vectors. Maximum values generally occur when vectors are parallel or anti-parallel. Let the angle between \(\vec a\) and \(\vec b\) be \(\theta\).
Step 1:Find \( |\vec a+2\vec b| \).} \[ |\vec a+2\vec b|^2 = |\vec a|^2+4|\vec b|^2+4\vec a\cdot\vec b \] \[ =4+36+24\cos\theta \] \[ =40+24\cos\theta \]
Step 2:Find \( |3\vec a-2\vec b| \).} \[ |3\vec a-2\vec b|^2 = 9|\vec a|^2+4|\vec b|^2-12\vec a\cdot\vec b \] \[ =36+36-72\cos\theta \] \[ =72-72\cos\theta \]
Step 3:Consider extreme values of \( \cos\theta \).} When vectors are parallel, \[ \cos\theta=1 \] \[ |\vec a+2\vec b|=\sqrt{64}=8 \] \[ |3\vec a-2\vec b|=0 \] Expression value \[ 3(8)+4(0)=24 \] When vectors are opposite, \[ \cos\theta=-1 \] \[ |\vec a+2\vec b|=\sqrt{16}=4 \] \[ |3\vec a-2\vec b|=\sqrt{144}=12 \] Expression value \[ 3(4)+4(12)=12+48=60 \] Thus the maximum value is \[ \boxed{60} \]