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Question Number 118161 by gopikrishnan last updated on 15/Oct/20

$$if\overrightarrow{p}=5\hat{i}+\lambda \hat{j}-3\hat{k}and\hat{q}=\hat{i}+3\hat{j}-5\hat{k}thenfindthevalueof\lambda sothat\overrightarrow{p}+\overrightarrow{q}and\overrightarrow{p}-\overrightarrow{q}areperpendicularvectors$$

Answered by Dwaipayan Shikari last updated on 15/Oct/20

$$\overrightarrow{p}+\overrightarrow{q}=6\hat{i}+(3+\lambda )\hat{j}-8\hat{k}$$$$\overrightarrow{p}-\overrightarrow{q}=4\hat{i}+(\lambda -3)\hat{j}+2\hat{k}$$$$(\overrightarrow{p}+\overrightarrow{q}).(\overrightarrow{p}-\overrightarrow{q})=\mid \overrightarrow{p}+\overrightarrow{q}\mid .\mid \overrightarrow{p}-\overrightarrow{q}\mid cos\frac{\pi }{2}$$$$(24+{\lambda }^2-9-16)=0$$$${\lambda }^2=1$$$$\lambda =\pm 1$$

Answered by Rio Michael last updated on 15/Oct/20

$$\overrightarrow{p}=5\hat{i}+\lambda \hat{j}-3\hat{k};\overrightarrow{q}=\hat{i}+3\hat{j}-5\hat{k}$$$$\overrightarrow{p}+\overrightarrow{q}=\left(\begin{array}{c}5\\ \lambda \\ -3\end{array}\right)+\left(\begin{array}{c}1\\ 3\\ -5\end{array}\right)=\left(\begin{array}{c}6\\ 3+\lambda \\ -8\end{array}\right)$$$$\overrightarrow{p}-\overrightarrow{q}=\left(\begin{array}{c}5\\ \lambda \\ -3\end{array}\right)-\left(\begin{array}{c}1\\ 3\\ -5\end{array}\right)=\left(\begin{array}{c}4\\ \lambda -3\\ 2\end{array}\right)$$$$\mathrm{for}\mathrm{perpendicular}\mathrm{vectors},(\overrightarrow{p}+\overrightarrow{q}).(\overrightarrow{p}-\overrightarrow{q})=0$$$$\Rightarrow 24+(3+\lambda )(\lambda -3)-16=0$$$$\Rightarrow {\lambda }^2-9-16+24=0$$$$\Rightarrow \lambda =\pm 1$$

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