# If $\overrightarrow{a}=\hat{i}-\hat{j}+7\hat{k}$ and $\overrightarrow{b}=5\hat{i}-\hat{j}+\lambda\hat{k}$,then find the value of $\lambda$,so that $\overrightarrow{a}+\overrightarrow{b}$ and $\overrightarrow{a}-\overrightarrow{b}$ are perpendicular vectors.

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• Since $\overrightarrow{a}+\overrightarrow{b}\; and\; \overrightarrow{a}-\overrightarrow{b}$ are prerendicular, $(\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a}-\overrightarrow{b})=0$
Given $\overrightarrow{a}+\overrightarrow{b}$ and $\overrightarrow{a}- \overrightarrow{b}$ are perpendicular and $\overrightarrow{a}=\overrightarrow{i}-\overrightarrow{j}+7\overrightarrow{k}$ and $\overrightarrow{b}=5\overrightarrow{i}-\overrightarrow{j}+\lambda \overrightarrow{k}$
$\Rightarrow \overrightarrow{a}+\overrightarrow{b}= 6\overrightarrow{i}-2 \overrightarrow{j}+ \overrightarrow{k} (7+\lambda)$
$\Rightarrow \overrightarrow{a}-\overrightarrow{b}=-4\overrightarrow{i}+\overrightarrow{k}(7-\lambda)$
Since $\overrightarrow{a}+\overrightarrow{b}\; and\; \overrightarrow{a}-\overrightarrow{b}$ are prerendicular, $(\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a}-\overrightarrow{b})=0$
$\Rightarrow \bigg[6\overrightarrow{i}-2\overrightarrow{j}+\overrightarrow{k}(7+\lambda)\bigg].\bigg[-4\overrightarrow{i}+\overrightarrow{k}(7-\lambda)\bigg]=0$
$\Rightarrow -24+(7+\lambda).(7-\lambda)=0$
$\Rightarrow 7^2-\lambda^2=24$
$\Rightarrow 49-24=\lambda^2$
$\Rightarrow \lambda=\pm 5$
edited Mar 22, 2013