JEE Main 2025 Jan 22 Shift 2, Mathematics Q15: Scalar (or Dot) Product of Two Vectors

Since $\vec{a},\vec{b}$ are unit vectors and angle is $\frac{\pi }{3}$: $\vec{a}\cdot \vec{b}=\cos \frac{\pi }{3}=\frac{1}{2}$, with $|\vec{a}{|}^2=|\vec{b}{|}^2=1$.

Perpendicularity: $(\lambda \vec{a}+2\vec{b})\cdot (3\vec{a}-\lambda \vec{b})=0$.

$=3\lambda |\vec{a}{|}^2-{\lambda }^2\vec{a}\cdot \vec{b}+6\vec{a}\cdot \vec{b}-2\lambda |\vec{b}{|}^2$

$=3\lambda -\frac{{\lambda }^2}{2}+3-2\lambda =-\frac{{\lambda }^2}{2}+\lambda +3=0$.

Multiplying by $-2$: ${\lambda }^2-2\lambda -6=0\Rightarrow \lambda =1\pm \sqrt{7}$.

$\sqrt{7}\approx 2.646$, so $\lambda \approx 3.646$ or $\lambda \approx -1.646$. Neither lies in $[-1,3]$.

$\therefore$ Number of values of $\lambda$ in $[-1,3]$ is $0$.