Question

The direction cosines of two lines are related by $l+m+n=0$ and $a{l}^2+b{m}^2+c{n}^2=0$. The lines are parallel if

• A. $a+b+c=0$
• B. $a^{-1}+b^{-1}+c^{-1}=0$
• C. $a=b=c$
• D. $Noneofthese$

Answer 1

The correct option is A $a+b+c=0$

Let $l_1{m}_1{n}_1$ and $l_2{m}_2{n}_2$ be the direction cosines.

$l+m+m=0\Rightarrow l=-m-na{l}^2+b{m}^2+c{n}^2=0a(m^2+n^2+2mn)+b{m}^2+c{n}^2=0(a+b)m^2+(a+c)n^2+2amn=0$

$(a+b){\left(\frac{m}{n}\right)}^2+2a\left(\frac{m}{n}\right)+(a+c)=0$ -Quadratic in $\frac{m}{n}$

$\therefore$ product of roots$=\frac{m_1}{n_1}\times \frac{m_2}{n_2}=\frac{a+c}{a+b}$

Similarly,

$m=-l-na{l}^2+b(l^2+n^2+2ln)+c{n}^2=0(a+b)l^2+(b+c)n^2+2bln=0$

$(a+b){\left(\frac{l}{n}\right)}^2+2b\left(\frac{l}{n}\right)+(b+c)=0-$ Quadratic in $\frac{l}{n}$

Product of roots $=\frac{l_1}{n_1}\times \frac{l_2}{n_2}=\frac{b+c}{a+b}$

Since the two lines are perpendicular,

$\therefore l_1{l}_2+m_1{m}_2+n_1{n}_2=cos90l_1{l}_2+m_1{m}_2+n_1{n}_2=0\frac{l_1{l}_2}{n_1{n}_2}+\frac{m_1{m}_2}{n_1{n}_2}+1=\frac{0}{n_1{n}_2}\frac{a+c}{a+b}+\frac{b+c}{a+b}+1=02(a+b+c)=0a+b+c=0$