Question

$\mathrm{If}\left|\overrightarrow{a}\right|=3,\left|\overrightarrow{b}\right|=4\mathrm{and}\overrightarrow{a}+\lambda \overrightarrow{b}$ is perpendicular to $\overrightarrow{a}-\lambda \overrightarrow{b},\mathrm{then}\lambda =\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_.$

Answer 1

Given:
$\left|\overrightarrow{a}\right|=3,\left|\overrightarrow{b}\right|=4$
$\overrightarrow{a}+\lambda \overrightarrow{b}$ is perpendicular to $\overrightarrow{a}-\lambda \overrightarrow{b}$


$$\begin{gathered} \mathrm{Since},\left(\overrightarrow{a}+\lambda \overrightarrow{b}\right)\mathrm{is}\mathrm{perpendicular}\mathrm{to}\left(\overrightarrow{a}-\lambda \overrightarrow{b}\right) \\ \mathrm{Therefore}, \\ \left(\overrightarrow{a}+\lambda \overrightarrow{b}\right)·\left(\overrightarrow{a}-\lambda \overrightarrow{b}\right)=0 \\ \Rightarrow \overrightarrow{a}·\overrightarrow{a}-\lambda \overrightarrow{b}·\lambda \overrightarrow{b}=0 \\ \Rightarrow {\left|\overrightarrow{a}\right|}^2-{\lambda }^2{\left|\overrightarrow{b}\right|}^2=0 \\ \Rightarrow {\left(3\right)}^2-{\lambda }^2{\left(4\right)}^2=0\left(\because \left|\overrightarrow{a}\right|=3,\left|\overrightarrow{b}\right|=4\right) \\ \Rightarrow 9-16{\lambda }^2=0 \\ \Rightarrow 16{\lambda }^2=9 \\ \Rightarrow {\lambda }^2=\frac{9}{16} \\ \Rightarrow \lambda =\pm \frac{3}{4} \end{gathered}$$

Hence, $\lambda =\underline{\pm \frac{3}{4}}$.