Question

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

Answer 1

$$\begin{gathered} \text{Given that} \\ \overrightarrow{p}=5\hat{i}+\lambda \hat{j}-3\hat{k} \\ and\overrightarrow{q}=\hat{i}+3\hat{j}-5\hat{k} \\ \overrightarrow{p}+\overrightarrow{q}=\left(5\hat{i}+\lambda \hat{j}-3\hat{k}\right)+\left(\hat{i}+3\hat{j}-5\hat{k}\right)=6\hat{i}+\left(\lambda +3\right)\hat{j}-8\hat{k} \\ \overrightarrow{p}-\overrightarrow{q}=\left(5\hat{i}+\lambda \hat{j}-3\hat{k}\right)-\left(\hat{i}+3\hat{j}-5\hat{k}\right)=4\hat{i}+\left(\lambda -3\right)\hat{j}+2\hat{k} \\ \text{Given that}\overrightarrow{p}+\overrightarrow{q}\text{is orthogonal to}\overrightarrow{p}-\overrightarrow{q}. \\ \Rightarrow \left(\overrightarrow{p}+\overrightarrow{q}\right).\left(\overrightarrow{p}-\overrightarrow{q}\right)=0 \\ \Rightarrow \left[6\hat{i}+\left(\lambda +3\right)\hat{j}-8\hat{k}\right].\left[4\hat{i}+\left(\lambda -3\right)\hat{j}+2\hat{k}\right]=0 \\ \Rightarrow 24+{\lambda }^2-9-16=0 \\ \Rightarrow {\lambda }^2=1 \\ \therefore \lambda =\pm 1 \end{gathered}$$