Question

If the line $\frac{x-1}{l}=\frac{y-2}{m}=\frac{z+1}{n}$ passes through the point (–1, 0, 1), then its direction cosines l, m, n are _______________.

Answer 1

$$\begin{gathered} \mathrm{Given},\mathrm{line}\frac{x-1}{l}=\frac{y-2}{m}=\frac{z+1}{n}\mathrm{passes}\mathrm{through}\left(-1,0,1\right) \\ \mathrm{say}\frac{x-1}{l}=\frac{y-2}{m}=\frac{z+1}{n}=\lambda \\ \mathrm{since}\left(-1,0,1\right)\mathrm{lies}\mathrm{on}\frac{x-1}{l}=\frac{y-2}{m}=\frac{z+1}{n}=\lambda \\ \mathrm{i}.\mathrm{e}.\frac{-1-1}{l}=\frac{0-2}{m}=\frac{1+1}{n}=\lambda \\ \mathrm{i}.\mathrm{e}.l=\frac{-2}{\lambda },m=\frac{-2}{\lambda },n=\frac{2}{\lambda } \\ \mathrm{Let}\lambda =\frac{2}{\mathrm{\mu }} \\ \mathrm{i}.\mathrm{e}.l=-\mathrm{\mu },m=-\mathrm{\mu },n=\mathrm{\mu } \\ \mathrm{Since}{l}^2+m^2+n^2=1 \\ \mathrm{i}.\mathrm{e}.{\left(-\mathrm{\mu }\right)}^2+{\left(-\mathrm{\mu }\right)}^2+{\mathrm{\mu }}^2=1 \\ \mathrm{i}.\mathrm{e}.{\mathrm{\mu }}^2+{\mathrm{\mu }}^2+{\mathrm{\mu }}^2=1 \\ \mathrm{i}.\mathrm{e}.3{\mathrm{\mu }}^2=1 \\ \mathrm{i}.\mathrm{e}.\mathrm{\mu }=\pm \frac{1}{\sqrt{3}} \\ \therefore l=\frac{-1}{\sqrt{3}},m=\frac{-1}{\sqrt{3}},n=\frac{1}{\sqrt{3}} \end{gathered}$$