Let $\vec{a}$ and $\vec{b}$ be two unit vectors such that the angle between them is $\frac{\pi}{3}$. If $\lambda \vec{a}+2 \vec{b}$ and $3 \vec{a}-\lambda \vec{b}$ are perpendicular to each other, then the number of values of $\lambda$ in $[-1,3]$ is :

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Aicharya · Accepted Answer

$\begin{aligned} & \hat{a} \cdot \hat{b}=\frac{1}{2} \ & 	ext { Now }(\lambda \hat{a}+2 \hat{b}) \cdot(3 \hat{a}-\lambda \hat{b})=0 \ & 3 \lambda \hat{a} \cdot \hat{a}-\lambda^2 \hat{a} \cdot \hat{b}+6 \hat{a} \cdot \hat{b}-2 \lambda \hat{b} \cdot \hat{b}=0 \ & 3 \lambda-\frac{\lambda^2}{2}+3-2 \lambda=0 \ & \lambda^2-2 \lambda-6=0 \ & \lambda=1 \pm \sqrt{7} \ & \Rightarrow 	ext { number of values }=0\end{aligned}$ ,

Let $\vec{a}$ and $\vec{b}$ be two unit vectors such that the angle between them is $\frac{\pi}{3}$. If…

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