Question

If the d.e's $(l,m,n)$ of two lines are connected by the relations $l+m+n=0,2lm-mn+2nl=0$ then the d.e's of the two lines are

• A. $(\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{-2}{\sqrt{6}}),(\frac{1}{\sqrt{6}},\frac{-2}{\sqrt{6}},\frac{1}{\sqrt{6}})$
• B. $(\frac{1}{14},\frac{2}{14},\frac{3}{14}),(\frac{1}{26},\frac{3}{26},\frac{4}{26})$
• C. $(\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}}),(\frac{1}{\sqrt{26}},\frac{3}{\sqrt{26}},\frac{4}{\sqrt{26}})$
• D. $(\frac{1}{\sqrt{16}},\frac{-1}{\sqrt{6}},\frac{-2}{\sqrt{6}}),(\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}},\frac{1}{\sqrt{6}})$

Answer 1

The correct option is A $(\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{-2}{\sqrt{6}}),(\frac{1}{\sqrt{6}},\frac{-2}{\sqrt{6}},\frac{1}{\sqrt{6}})$

The given relations are $l+m+n=0$ and $2lm+2ln-mn=0$

$l+m+n=0$

Putting $l=-(m+n)$ in $2lm+2ln-mn=0$

$-2(m+n)m-2(m+n)n-mn=0\Rightarrow {2m}^2+2mn+2mn+2n^2+mn=0\Rightarrow {2m}^2+5mn+2n^2=0\Rightarrow 2m(m+2n)+n(m+2n)=0\Rightarrow (2m+n)(m+2n)=0\Rightarrow n=-2m,n=\frac{-m}{2}$

When $n=-2ml=-(m-2m)=m$

When $n=\frac{-m}{2}l=-(m-\frac{m}{2})=\frac{-m}{2}$

$\therefore$ directions ratios of the lines proportional to $m,m,-2m$ and $\frac{-m}{2},m,\frac{-m}{2}$ are-

$1,1,2$ and $\frac{-1}{2},1,\frac{-1}{2}$

$1,1,-2$ and $-1,2,-1$

Direction cosines of the first line are

$\pm \frac{1}{\sqrt{{\left(1\right)}^2+{\left(1\right)}^2+{(-2)}^2}},\pm \frac{1}{\sqrt{{\left(1\right)}^2+{\left(1\right)}^2+{(-2)}^2}},\pm \frac{-2}{\sqrt{{\left(1\right)}^2+{\left(1\right)}^2+{(-2)}^2}}=\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{-2}{\sqrt{6}}$

Direction cosines of second line are-

$\pm \frac{1}{\sqrt{{(-1)}^2+{\left(2\right)}^2+{(-1)}^2}},\pm \frac{-2}{\sqrt{{(-1)}^2+{\left(2\right)}^2+{(-1)}^2}},\pm \frac{1}{\sqrt{{(-1)}^2+{\left(2\right)}^2+{(-1)}^2}}=\frac{1}{\sqrt{6}},\frac{-2}{\sqrt{6}},\frac{1}{\sqrt{6}}$

Hence, this is the answer.