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Question Number 132327 by liberty last updated on 13/Feb/21

$$\mathrm{If}\mathrm{the}\mathrm{line}\{\begin{array}{l}\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}+1}{3}=\frac{\mathrm{z}-1}{4}\\ \frac{\mathrm{x}-3}{1}=\frac{\mathrm{y}-\mathrm{k}}{2}=\frac{\mathrm{z}}{1}\end{array}$$$$\mathrm{intersect}.\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{k}\mathrm{is}$$

Answered by Ar Brandon last updated on 13/Feb/21

$${\mathrm{L}}_1:\mathrm{i}-\mathrm{j}+\mathrm{k}+\lambda (2\mathrm{i}+3\mathrm{j}+4\mathrm{k})=(1+2\lambda )\mathrm{i}+(3\lambda -1)\mathrm{j}+(4\lambda +1)\mathrm{k}$$$${\mathrm{L}}_2:3\mathrm{i}+\alpha \mathrm{j}+\mu (\mathrm{i}+2\mathrm{j}+\mathrm{k})=(3+\mu )\mathrm{i}+(\alpha +2\mu )\mathrm{j}+\mathrm{k}$$$$\{\begin{array}{l}1+2\lambda =3+\mu \\ 3\lambda -1=\alpha +2\mu \\ 4\lambda +1=1\end{array}\Rightarrow \{\begin{array}{l}\lambda =0\\ \mu =-2\\ \alpha =3\end{array}$$$$\mathrm{k}=\alpha =3\mathrm{and}\mathrm{i}-\mathrm{j}+\mathrm{k}\mathrm{is}\mathrm{point}\mathrm{of}\mathrm{intersection}$$

Answered by EDWIN88 last updated on 15/Feb/21

$$\mathrm{let}\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}+1}{3}=\frac{\mathrm{z}-1}{4}=\lambda \to \{\begin{array}{l}\mathrm{x}=2\lambda +1\\ \mathrm{y}=3\lambda -1\\ \mathrm{z}=4\lambda +1\end{array}$$$$\mathrm{let}\frac{\mathrm{x}-3}{1}=\frac{\mathrm{y}-\mathrm{k}}{2}=\frac{\mathrm{z}}{1}=\vartheta \to \{\begin{array}{l}\mathrm{x}=\vartheta +3\\ \mathrm{y}=2\vartheta +\mathrm{k}\\ \mathrm{z}=\vartheta \end{array}$$$$\mathrm{the}\mathrm{line}\mathrm{are}\mathrm{intersect}\mathrm{the}\mathrm{we}\mathrm{get}$$$$\{\begin{array}{l}2\lambda +1=\vartheta +3\\ 3\lambda -1=2\vartheta +\mathrm{k}\\ 4\lambda +1=\vartheta \end{array};\to \{\begin{array}{l}\vartheta =-5\\ 2\lambda =3;\lambda =-\frac{3}{2}\end{array}$$$$\mathrm{then}\mathrm{k}=3\lambda -2\vartheta -1=-\frac{9}{2}+10-1=-\frac{9}{2}+9=\frac{9}{2}$$$$$$

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