Question

Suppose the direction cosines of two lines are given by $al+bm+cn=0$ and $fmn+gln+hlm=0$ where $f,g,h,a,b,c$ are arbitrary constants and $l,m,n$ are direction cosines of the line. On the basis of the above information and for $f=g=h=1$ both lines satisfy the relation:

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

Answer 1

The correct option is D All of the above.

$al+bm+cn=\mathrm{0.....}\left(1\right)$
$\Rightarrow fmn+gln+hlm=\mathrm{0....}\left(2\right)$
From $\left(1\right)\Rightarrow n=-\left(\frac{al+bm}{c}\right)$
Putting in $\left(2\right),$ we get
$\Rightarrow (fm+gl)[-\left(\frac{al+bm}{c}\right)]+hlm=0$
$\Rightarrow ag{\left(\frac{l}{m}\right)}^2+(af+bg-ch)\frac{l}{m}+bf=0....\left(1\right)$
if $f=g=h=1$
$\therefore a{\left(\frac{l}{m}\right)}^2+(a+b-c)\left(\frac{l}{m}\right)+b=0$

From $\left(1\right)\Rightarrow m=-\left(\frac{al+cn}{b}\right)$
Putting in $\left(2\right),$ we get
If $f=g=h=1$
$c{\left(\frac{n}{l}\right)}^2+(c+a-b)\frac{n}{l}+a=0$

From $\left(1\right)\Rightarrow l=-\left(\frac{cn+bm}{a}\right)$
Putting in $\left(2\right),$ we get
if $f=g=h=1$
$b{\left(\frac{m}{n}\right)}^2+(b+c-a)\left(\frac{m}{n}\right)+c=0$