JEE Main 2024 — 4 April, Shift 2 — Question 2

$(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=0, \lambda>0$

$|\overrightarrow{\mathrm{a}}|^{2}-|\overrightarrow{\mathrm{b}}|^{2}=0 \rightarrow 1+\lambda^{2}+9=9+1+4$

$\therefore \lambda=2, \cos \theta=\frac{\vec{a}-\vec{b}}{|\vec{a}| \cdot|\vec{b}|}=\frac{3-\lambda-6}{\sqrt{14} \cdot \sqrt{14}}$

$14 \cos \theta=3-8=-5$

$\therefore(14 \cos \theta)^{2}=25$