JEE Main 2026 Apr 2 Shift 1, Mathematics Q14: Scalar (or Dot) Product of Two Vectors

Let $\theta$ be the angle between $\vec{a}$ and $\vec{b}$. Then

$|3\vec{a}+2\vec{b}{|}^2=9|\vec{a}{|}^2+4|\vec{b}{|}^2+12\vec{a}\cdot \vec{b}=36+36+72\cos \theta =72(1+\cos \theta )$

$|3\vec{a}-2\vec{b}{|}^2=72(1-\cos \theta )$

Using the half-angle forms:

$|3\vec{a}+2\vec{b}|=12\left|\cos \frac{\theta }{2}\right|,|3\vec{a}-2\vec{b}|=12\left|\sin \frac{\theta }{2}\right|$

So $E=36\cos (\theta /2)+48\sin (\theta /2)$. The maximum of $A\cos \varphi +B\sin \varphi$ is $\sqrt{A^2+B^2}$:

$E_{\max }=\sqrt{36^2+48^2}=\sqrt{1296+2304}=\sqrt{3600}=60$