Question

The projection of the line segment joining the points (-1,0, 3) and (2, 5, 1) ont the line whose direction ratios are 6, 2, 3 is?

Answer 1

Dear student,

$$\begin{gathered} \mathrm{The}\mathrm{dr}\text{'}\mathrm{s}\mathrm{of}\mathrm{the}\mathrm{other}\mathrm{line}\mathrm{is}: \\ \mathrm{a}=6;\mathrm{b}=2;\mathrm{c}=3 \\ \mathrm{Now},\mathrm{direction}\mathrm{cosines}\mathrm{of}\mathrm{the}\mathrm{above}\mathrm{line}\mathrm{are}: \\ \mathrm{l}=\frac{\mathrm{a}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}}=\frac{6}{\sqrt{6^2+2^2+3^2}}=\frac{6}{\sqrt{36+4+9}}=\frac{6}{\sqrt{49}}=\frac{6}{7} \\ \mathrm{m}=\frac{\mathrm{b}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}}=\frac{2}{\sqrt{6^2+2^2+3^2}}=\frac{2}{\sqrt{36+4+9}}=\frac{2}{\sqrt{49}}=\frac{2}{7} \\ \mathrm{n}=\frac{\mathrm{a}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}}=\frac{3}{\sqrt{6^2+2^2+3^2}}=\frac{3}{\sqrt{36+4+9}}=\frac{3}{\sqrt{49}}=\frac{3}{7} \\ \mathrm{Now},\mathrm{projection}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{segment}\mathrm{joining}\mathrm{the}\mathrm{points}\left({\mathrm{x}}_1,{\mathrm{y}}_1,{\mathrm{z}}_1\right)\mathrm{and}\left({\mathrm{x}}_2,{\mathrm{y}}_2,{\mathrm{z}}_2\right)\mathrm{on}\mathrm{the}\mathrm{line}\mathrm{having}\mathrm{direction}\mathrm{cosines}\left(\mathrm{l},\mathrm{m},\mathrm{n}\right)\mathrm{is} \\ \left({\mathrm{x}}_2-{\mathrm{x}}_1\right)\mathrm{l}+\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)\mathrm{m}+\left({\mathrm{z}}_2-{\mathrm{z}}_1\right)\mathrm{n} \\ \mathrm{Now},\mathrm{projection}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{segment}\mathrm{joining}\mathrm{the}\mathrm{points}\left(-1,0,3\right)\mathrm{and}\left(2,5,1\right)\mathrm{on}\mathrm{the}\mathrm{line}\mathrm{having}\mathrm{direction}\mathrm{cosines}\left(\frac{6}{7},\frac{2}{7},\frac{3}{7}\right)\mathrm{is} \\ \left(2+1\right)\times \frac{6}{7}+\left(5-0\right)\times \frac{2}{7}+\left(1-3\right)\times \frac{3}{7} \\ =\frac{18}{7}+\frac{10}{7}-\frac{6}{7} \\ =\frac{22}{7} \end{gathered}$$

Regards