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 :

<p>We are given two unit vectors $\vec{a}$ and $\vec{b}$ with an angle of $\frac{\pi}{3}$ between them. This means:</p> <p><p>$\vec{a} \cdot \vec{a} = \vec{b} \cdot \vec{b} = 1$ (since they are unit vectors)</p></p> <p><p>$\vec{a} \cdot \vec{b} = \cos\frac{\pi}{3} = \frac{1}{2}$</p></p> <p>We need to find the number of values of $\lambda$ in the interval $[-1, 3]$ for which the vectors $\lambda \vec{a}+2 \vec{b}$ and $3 \vec{a}-\lambda \vec{b}$ are perpendicular, meaning their dot product is zero.</p> <p>Here’s how to solve the problem step by step:</p> <p><p>Set up the perpendicular condition by equating the dot product to zero:</p> <p>$(\lambda \vec{a}+2 \vec{b})\cdot(3\vec{a}-\lambda \vec{b}) = 0.$ </p></p> <p><p>Expand the dot product using the distributive property:</p> <p>$$\begin{align*} (\lambda \vec{a}+2 \vec{b})\cdot(3\vec{a}-\lambda \vec{b}) &= \lambda \cdot 3 (\vec{a}\cdot\vec{a}) - \lambda^2 (\vec{a}\cdot\vec{b}) + 2\cdot 3 (\vec{b}\cdot\vec{a}) - 2\lambda (\vec{b}\cdot\vec{b})\\[1mm] &= 3\lambda - \lambda^2\left(\frac{1}{2}\right) + 6\left(\frac{1}{2}\right) - 2\lambda\\[1mm] &= 3\lambda - \frac{\lambda^2}{2} + 3 - 2\lambda. \end{align*}$$</p></p> <p><p>Simplify the expression:</p> <p>$3\lambda - 2\lambda = \lambda,$</p> <p>so the equation becomes:</p> <p>$\lambda - \frac{\lambda^2}{2} + 3 = 0.$</p></p> <p><p>Multiply the entire equation by 2 to eliminate the fraction:</p> <p>$2\lambda - \lambda^2 + 6 = 0,$</p> <p>which can be rearranged as:</p> <p>$-\lambda^2 + 2\lambda + 6 = 0.$</p></p> <p><p>Multiply both sides by -1 to get a standard quadratic form:</p> <p>$\lambda^2 - 2\lambda - 6 = 0.$</p></p> <p><p>Solve the quadratic equation using the quadratic formula:</p> <p>$$\lambda = \frac{2 \pm \sqrt{(-2)^2 - 4\cdot1\cdot(-6)}}{2} = \frac{2 \pm \sqrt{4+24}}{2} = \frac{2 \pm \sqrt{28}}{2}.$$</p> <p>Notice that:</p> <p>$\sqrt{28} = 2\sqrt{7},$</p> <p>thus:</p> <p>$\lambda = \frac{2 \pm 2\sqrt{7}}{2} = 1 \pm \sqrt{7}.$</p></p> <p><p>Evaluate the solutions numerically:</p></p> <p><p>$\lambda = 1 + \sqrt{7} \approx 1 + 2.6458 \approx 3.6458,$</p></p> <p><p>$\lambda = 1 - \sqrt{7} \approx 1 - 2.6458 \approx -1.6458.$</p></p> <p>Check if the solutions are in the interval $[-1, 3]$:</p> <p><p>$1 + \sqrt{7} \approx 3.6458$ is greater than 3.</p></p> <p><p>$1 - \sqrt{7} \approx -1.6458$ is less than -1.</p></p> <p>Since neither value lies within the interval $[-1, 3]$, there are no valid values of $\lambda$ in that interval.</p> <p>Therefore, the number of values of $\lambda$ in the interval $[-1, 3]$ is:</p> <p>0</p>