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 the given lines will be perpendicular if:

• A. $\frac{f}{a}-\frac{g}{b}+\frac{h}{c}=0$
• B. $\frac{f}{a}+\frac{g}{b}-\frac{h}{c}=0$
• C. $-\frac{f}{a}+\frac{g}{b}+\frac{h}{c}=0$
• D. $\frac{f}{a}+\frac{g}{b}+\frac{h}{c}=0$

Answer 1

The correct option is D $\frac{f}{a}+\frac{g}{b}+\frac{h}{c}=0$

$al+bm+cn=\mathrm{0.....}\left(1\right)$
$\Rightarrow fmn+gln+hlm=\mathrm{0....}\left(2\right)$
$\Rightarrow n=-\left(\frac{al+bm}{c}\right)$
Eliminating $n$ in the above equation,
$\Rightarrow (fm+gl)[-\left(\frac{al+bm}{c}\right)]+hlm=0$
$\Rightarrow ag{l}^2+(bg+af-ch)lm+bf{m}^2=0$
$\therefore \frac{l_1{l}_2}{m_1{m}_2}=\frac{bf}{ag}$
Similarly, $\frac{m_1{m}_2}{n_1{n}_2}=\frac{cg}{bh}$
$\Rightarrow \frac{l_1{l}_2}{\frac{f}{a}}=\frac{m_1{m}_2}{\frac{g}{b}}=\frac{n_1{n}_2}{\frac{h}{c}}$
But for lines to be perpendicular, $l_1{l}_2+m_1{m}_2+n_1{n}_2=0\Rightarrow \frac{f}{a}+\frac{g}{b}+\frac{h}{c}=0$