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If $|\mathrm{a}|=3,|\mathrm{~b}|=4$, then a value of $\lambda$ for which $\mathrm{a}+\lambda \mathrm{b}$ is perpendicular to $\mathrm{a}-\lambda \mathrm{b}$ is
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The correct answer is:
$\frac{3}{4}$
Since, $(a+\lambda b) \perp(a-\lambda b)$, then
$\begin{aligned}(a+\lambda b) \cdot(a-\lambda b) &=0 \\ \Rightarrow \quad|a|^{2}-\lambda^{2}|b|^{2} &=0 \Rightarrow\left|a^{2}\right|=\lambda^{2}|b|^{2} \\ \Rightarrow \quad \lambda &=\frac{|a|}{|b|}=\frac{3}{4} \end{aligned}$
$\begin{aligned}(a+\lambda b) \cdot(a-\lambda b) &=0 \\ \Rightarrow \quad|a|^{2}-\lambda^{2}|b|^{2} &=0 \Rightarrow\left|a^{2}\right|=\lambda^{2}|b|^{2} \\ \Rightarrow \quad \lambda &=\frac{|a|}{|b|}=\frac{3}{4} \end{aligned}$
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