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If $\overrightarrow{p}=5\hat{i}+\lambda\hat{j}-37\hat{k}$ and $\overrightarrow{q}=\hat{i}+3\hat{j}-5\hat{k}$,then find the value of $\lambda$,so that $\overrightarrow{p}+\overrightarrow{q}$ and $\overrightarrow{p}-\overrightarrow{q}$ are perpendicular vectors.

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Given $\overrightarrow{p}=5\hat{i}+\lambda\hat{j}-37\hat{k}$
$\overrightarrow{q}=\hat{i}+3\hat{j}-5\hat{k}$
$\overrightarrow{p}+\overrightarrow{q}=5\hat{i}+\lambda\hat{j}-3\hat{k}+\hat{i}+3\hat{j}-5\hat{k}$
$\qquad\;\;\;=6\hat{i}+\hat{j}(3+\lambda)-8\hat k$
$\overrightarrow{p}-\overrightarrow{q}=5\hat{i}+\lambda\hat{j}-3\hat{k}-(\hat{i}+3\hat{j}-5\hat{k})$
$\qquad\;\;\;=4\hat{i}+\hat{j}(\lambda-3)+2\hat k$
$\overrightarrow{p}+\overrightarrow{q}$ and $\overrightarrow{p}-\overrightarrow{q}$ are perpendicular vectors.
$\overrightarrow{p}+\overrightarrow{q}$.$\overrightarrow{p}-\overrightarrow{q}$=0
$\Rightarrow |\overrightarrow{p}|^2-|\overrightarrow{q}|^2$=0
$p^2=q^2$
(i .e)$(6\hat{i}+\hat{j}(3+\lambda)^2+64)^2=4\hat{i}+j(\lambda-3)+2\hat{k})^2$
$(\sqrt{36+(3+\lambda)^2+64})^2=(\sqrt{16+(\lambda-3)^2+4})^2$
$36+3+\lambda^2+64=16+4+(\lambda-3)^2$
$100+9+6\lambda+\lambda^2=20+\lambda^2-6\lambda+9$
80+12$\lambda$=0
$\lambda=\frac{-80}{12}=\frac{-20}{3}$
Hence $\lambda=\frac{-20}{3}$
answered Mar 25, 2013 by sreemathi.v
 

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