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.


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

$Find the direction cosines of two$ $lines which are connected by relation$ $l + m + n = 0$ $m n - 2 n l - 2 l m = 0$ $my solution$ $l = - (m + n)$ $m n + 2 n (m + n) + 2 (n + m) n = 0$ $2 m^{2} + 5 m n + 2 n^{2} = 0$ $(2 m + n) (m + 2 n) = 0$ $m = - 2 n, o r m = - (1 / 2) n$ $c a s e 1 : m = - 2 n$ $l = - (m + n) = - n$ $l^{2} + m^{2} + n^{2} = 1$ $6 n^{2} = 1 \Rightarrow n = \pm \frac{1}{\sqrt{6}}$ $(l, m, n) = \pm (- \frac{1}{\sqrt{6}}, - \frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}})$ $c a s e 2 : m = - (1 / 2) n$ $l = - \frac{1}{2} n$ $l^{2} + m^{2} + n^{2} = 1$ $\frac{n^{2}}{4} + \frac{n^{2}}{4} + n^{2} = 1 \Rightarrow n = \pm \frac{2}{\sqrt{6}}$ $(l, m, n) = \pm (- \frac{1}{\sqrt{6}}, - \frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}})$ $so i get a total of 4 solution .$ $Books answer is 2 lines$ $(\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, - \frac{2}{\sqrt{6}}) and (\frac{1}{\sqrt{6}}, - \frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}})$ $please help .$
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