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

Answer

Given that
$\vec{\text{p}}=5\hat{\text{i}}+\lambda\hat{\text{j}}-3\hat{\text{k}}$
and $\vec{\text{q}}=\hat{\text{i}}+3\hat{\text{j}}-5\hat{\text{k}},$
$\vec{\text{p}}+\vec{\text{q}}=\big(5\hat{\text{i}}+\lambda\hat{\text{j}}-3\hat{\text{k}}\big)+(\hat{\text{i}}+3\hat{\text{j}}-5\hat{\text{k}})\\=6\hat{\text{i}}+(\lambda+3)\hat{\text{j}}-8\hat{\text{k}}$
$\vec{\text{p}}-\vec{\text{q}}=\big(5\hat{\text{i}}+\lambda\hat{\text{j}}-3\hat{\text{k}}\big)-(\hat{\text{i}}+3\vec{\text{j}}-5\hat{\text{k}})\\=4\hat{\text{i}}+(\lambda-3)\hat{\text{j}}+2\hat{\text{k}}$
Given that $\vec{\text{p}}+\vec{\text{q}}$ is orthogonal to $\vec{\text{p}}-\vec{\text{q}}.$
$\Rightarrow\big(\vec{\text{p}}+\vec{\text{q}}\big).\big(\vec{\text{p}}-\vec{\text{q}}\big)=0$
$\Rightarrow\Big[6\hat{\text{i}}+(\lambda+3)\hat{\text{j}}-8\hat{\text{k}}\Big].\Big[4\hat{\text{i}}+(\lambda-3)\hat{\text{j}}+2\hat{\text{k}}\Big]=0$
$\Rightarrow24+\lambda^2-9-16=0$
$\Rightarrow\lambda^2=1$
$\therefore\lambda=\pm1$

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