JEE Main 2024 — 8 April, Shift 1 — Question 18

$\vec{a}=\alpha \hat{i}+6 \hat{j}-3 \hat{k}$

$\overrightarrow{\mathrm{b}}=\hat{\mathrm{t}}-2 \hat{\mathrm{j}}-2 \alpha \hat{\mathrm{k}}$

so $\vec{a} \cdot \vec{b}<0, \forall t \in R$

$\alpha t^{2}-12+6 \alpha t<0$

$\alpha \mathrm{t}^{2}+6 \alpha \mathrm{t}-12<0, \forall \mathrm{t} \in \mathrm{R}$

$\alpha<0$, and $\mathrm{D}<0$

$36 \alpha^{2}+48 \alpha<0$

$12 \alpha(3 \alpha+4)<0$

$\frac{-4}{3}<\alpha<0$

also for $\mathrm{a}=0, \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}<0$

hence a $\alpha \in\left(\frac{-4}{3}, 0\right]$