A) \[\frac{9}{16}\]
B) \[\frac{3}{4}\]
C) \[\frac{3}{2}\]
D) \[\frac{3}{2}\]
Correct Answer: B
Solution :
Given that, \[|\overrightarrow{a}|=3,\,\,|\overrightarrow{b}|=4\] and \[\overrightarrow{a}+\lambda \,\overrightarrow{b}\] is perpendicular to \[\overrightarrow{a}-\lambda \,\overrightarrow{b}\] \[\therefore \] \[(\overrightarrow{a}+\lambda \,\overrightarrow{b})\,.\,(\overrightarrow{a}-\lambda \overrightarrow{b})=0\] or \[\overrightarrow{a}\,.\,\,\overrightarrow{b}-\overrightarrow{a}\,.\,\,\overrightarrow{b}\,.\,\,\lambda +\lambda \,\overrightarrow{b}\,.\,\,\overrightarrow{a}-{{\lambda }^{2}}\overrightarrow{b}\,\,.\,\,\overrightarrow{b}=0\] \[\Rightarrow \] \[{{(\overrightarrow{a})}^{2}}-{{\lambda }^{2}}{{(\overrightarrow{b})}^{2}}=0\] \[\Rightarrow \] \[{{\lambda }^{2}}=\frac{{{(\overrightarrow{a})}^{2}}}{{{(\overrightarrow{b})}^{2}}}\] or \[\lambda =\frac{|\overrightarrow{a}|}{|\overrightarrow{b}|}=\frac{3}{4}\] [Given]
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