Question and Answers Forum

All Questions      Topic List

Coordinate Geometry Questions

Previous in All Question      Next in All Question      

Previous in Coordinate Geometry      Next in Coordinate Geometry      

Question Number 28504 by rish@bh last updated on 26/Jan/18

Find the direction cosines of two lines which are connected by relation l+m+n=0 mn−2nl−2lm=0 my solution l=−(m+n) mn+2n(m+n)+2(n+m)n=0 2m^2 +5mn+2n^2 =0 (2m+n)(m+2n)=0 m=−2n, or m=−(1/2)n case 1: m=−2n l=−(m+n)=−n l^2 +m^2 +n^2 =1 6n^2 =1⇒n=±(1/(√6)) (l,m,n)=±(−(1/(√6)),−(2/(√6)),(1/(√6))) case 2:m=−(1/2)n l=−(1/2)n l^2 +m^2 +n^2 =1 (n^2 /4)+(n^2 /4)+n^2 =1⇒n=±(2/(√6)) (l,m,n)=±(−(1/(√6)),−(1/(√6)),(2/(√6))) so i get a total of 4 solution. Books answer is 2 lines ((1/(√6)),(1/(√6)),−(2/(√6))) and ((1/(√6)),−(2/(√6)),(1/(√6))) please help.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{direction}\:\mathrm{cosines}\:\mathrm{of}\:\mathrm{two} \\ $$$$\mathrm{lines}\:\mathrm{which}\:\mathrm{are}\:\mathrm{connected}\:\mathrm{by}\:\mathrm{relation} \\ $$$${l}+{m}+{n}=\mathrm{0} \\ $$$${mn}−\mathrm{2}{nl}−\mathrm{2}{lm}=\mathrm{0} \\ $$$$ \\ $$$$\boldsymbol{\mathrm{my}}\:\boldsymbol{\mathrm{solution}} \\ $$$${l}=−\left({m}+{n}\right) \\ $$$${mn}+\mathrm{2}{n}\left({m}+{n}\right)+\mathrm{2}\left({n}+{m}\right){n}=\mathrm{0} \\ $$$$\mathrm{2}{m}^{\mathrm{2}} +\mathrm{5}{mn}+\mathrm{2}{n}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left(\mathrm{2}{m}+{n}\right)\left({m}+\mathrm{2}{n}\right)=\mathrm{0} \\ $$$${m}=−\mathrm{2}{n},\:{or}\:{m}=−\left(\mathrm{1}/\mathrm{2}\right){n} \\ $$$${case}\:\mathrm{1}:\:{m}=−\mathrm{2}{n} \\ $$$${l}=−\left({m}+{n}\right)=−{n} \\ $$$${l}^{\mathrm{2}} +{m}^{\mathrm{2}} +{n}^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{6}{n}^{\mathrm{2}} =\mathrm{1}\Rightarrow{n}=\pm\frac{\mathrm{1}}{\sqrt{\mathrm{6}}} \\ $$$$\left({l},{m},{n}\right)=\pm\left(−\frac{\mathrm{1}}{\sqrt{\mathrm{6}}},−\frac{\mathrm{2}}{\sqrt{\mathrm{6}}},\frac{\mathrm{1}}{\sqrt{\mathrm{6}}}\right) \\ $$$${case}\:\mathrm{2}:{m}=−\left(\mathrm{1}/\mathrm{2}\right){n} \\ $$$${l}=−\frac{\mathrm{1}}{\mathrm{2}}{n} \\ $$$${l}^{\mathrm{2}} +{m}^{\mathrm{2}} +{n}^{\mathrm{2}} =\mathrm{1} \\ $$$$\frac{{n}^{\mathrm{2}} }{\mathrm{4}}+\frac{{n}^{\mathrm{2}} }{\mathrm{4}}+{n}^{\mathrm{2}} =\mathrm{1}\Rightarrow{n}=\pm\frac{\mathrm{2}}{\sqrt{\mathrm{6}}} \\ $$$$\left({l},{m},{n}\right)=\pm\left(−\frac{\mathrm{1}}{\sqrt{\mathrm{6}}},−\frac{\mathrm{1}}{\sqrt{\mathrm{6}}},\frac{\mathrm{2}}{\sqrt{\mathrm{6}}}\right) \\ $$$$\mathrm{so}\:\mathrm{i}\:\mathrm{get}\:\mathrm{a}\:\mathrm{total}\:\mathrm{of}\:\mathrm{4}\:\mathrm{solution}. \\ $$$$\mathrm{Books}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{2}\:\mathrm{lines} \\ $$$$\left(\frac{\mathrm{1}}{\sqrt{\mathrm{6}}},\frac{\mathrm{1}}{\sqrt{\mathrm{6}}},−\frac{\mathrm{2}}{\sqrt{\mathrm{6}}}\right)\:\mathrm{and}\:\left(\frac{\mathrm{1}}{\sqrt{\mathrm{6}}},−\frac{\mathrm{2}}{\sqrt{\mathrm{6}}},\frac{\mathrm{1}}{\sqrt{\mathrm{6}}}\right) \\ $$$$\mathrm{please}\:\mathrm{help}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com