Question

Suppose direction cosines of two lines are given by $ul+vm+wn=0$ and $a{l}^2+b{m}^2+c{n}^2=0$ where $u,v,w,a,b,c$ are arbitrary constants and $1,m,n$ are direction cosines of the line. For $u=v=w=1$ both lines satisfies the relation:

• A. $(b+c){\left(\frac{n}{l}\right)}^2+2b\left(\frac{n}{l}\right)+(a+b)=0$
• B. $(c+a){\left(\frac{l}{m}\right)}^2+2c\left(\frac{l}{m}\right)+(b+c)=0$
• C. $(a+b){\left(\frac{m}{n}\right)}^2+2a\left(\frac{m}{n}\right)+(c+a)=0$
• D. All of the above

Answer 1

Answer: (D)

All of the above

Given that direction cosines of the two lines are given by $l+m+n=0$ ------$\left(1\right)$
and $a{l}^2+b{m}^2+c{n}^2=0$ ------$\left(2\right)$
Eliminating $n$ from $\left(1\right)$ and $\left(2\right)$ gives
$a{l}^2+b{m}^2+c(l+m{)}^2=0$
$\Rightarrow (a+c)l^2+2clm+(b+c)m^2=0$
Dividing above equation with $m^2$ gives
$(a+c){\left(\frac{l}{m}\right)}^2+2c\left(\frac{l}{m}\right)+(b+c)=0$
Similarly, eliminating $l$ and $m$ from $\left(1\right)$ and $\left(2\right)$ respectively gives
$(b+c){\left(\frac{n}{l}\right)}^2+2b\left(\frac{n}{l}\right)+(a+b)=0$ and
$(a+b){\left(\frac{m}{n}\right)}^2+2a\left(\frac{l}{m}\right)+(c+a)=0$
Hence, option D.