If $\left|\overrightarrow{a}\right|=3,\left|\overrightarrow{b}\right|=4$ then a value of $\lambda$, for which $\overrightarrow{a}+\lambda \overrightarrow{b}$ is perpendicular to $\overrightarrow{a}-\lambda \overrightarrow{b}$ is :

If $\overrightarrow{a}+\lambda \overrightarrow{b}$ is perpendicular to $\overrightarrow{a}-\lambda \overrightarrow{b}$, then

$\begin{array}{l}(\overrightarrow{a}+\lambda \overrightarrow{b})\cdot (\overrightarrow{a}-\lambda \overrightarrow{b})=|\overrightarrow{a}+\lambda \overrightarrow{b}||\overrightarrow{a}-\lambda \overrightarrow{b}|\cdot \cos {90}^{\circ }\\ \Rightarrow (\overrightarrow{a}+\lambda \overrightarrow{b})\cdot (\overrightarrow{a}-\lambda \overrightarrow{b})=0\\ \Rightarrow \overrightarrow{a}\cdot \overrightarrow{a}-\lambda \cdot \overrightarrow{a}\cdot \overrightarrow{b}+\lambda \cdot \overrightarrow{b}\cdot \overrightarrow{a}-{\lambda }^2\cdot \overrightarrow{b}\cdot \overrightarrow{b}=0\\ \Rightarrow a^2-{\lambda }^2{b}^2=0\Rightarrow {\lambda }^2=\frac{a^2}{b^2}\Rightarrow {\lambda }^2=\frac{3^2}{4^2}\\ \Rightarrow \lambda =\frac{3}{4}\end{array}$