Question

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

• A. $\frac{9}{16}$
• B. $\frac{3}{4}$
• C. $\frac{3}{2}$
• D. $\frac{4}{3}$

Answer 1

The correct option is B $\frac{3}{4}$

Given that $\left|\overrightarrow{a}\right|=3,\left|\overrightarrow{b}\right|=4$ and $\overrightarrow{a}+\lambda \overrightarrow{b}$ is perpendicualr to $\overrightarrow{a}-\lambda \overrightarrow{b}$
$(\overrightarrow{a}+\lambda \overrightarrow{b}).(\overrightarrow{a}-\lambda \overrightarrow{b})=0$
$\Rightarrow \overrightarrow{a}.\overrightarrow{a}-\overrightarrow{a}.\overrightarrow{b}\lambda +\lambda \overrightarrow{b}.\overrightarrow{a}-{\lambda }^2\overrightarrow{b}.\overrightarrow{b}=0$
$\Rightarrow {\left|\overrightarrow{a}\right|}^2-{\lambda }^2{\left|\overrightarrow{b}\right|}^2=0$
$\Rightarrow {\lambda }^2=\frac{{\left|\overrightarrow{a}\right|}^2}{{\left|\overrightarrow{b}\right|}^2}$
$\Rightarrow {\lambda }^2=\frac{9}{16}$
$\Rightarrow \lambda =\pm \frac{3}{4}$