Question

Ttwo lines whose d.c's are related by al+bm+cn=0, fmn +gnl +hlm =0 are perpendicular, if

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

Answer 1

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

From the first relation, $n=-\left(\frac{al+bm}{c}\right)$
Put the value of n in second relation,
$fm(-\frac{(al+bm)}{c})+gl(-\frac{(al+bm)}{c})+hlm=0$
or $afml+bf{m}^2+ag{l}^2+bglm-chlm=0$
$ag\frac{l^2}{m^2}+\frac{l}{m}(af+bg-ch)+bf=0....\left(i\right)$
Now if $l_1,m_1,n_1$ and $l_2,m_2,n_2$ be direction cosines of two lines, then from (i)
$\frac{l_1{l}_2}{m_1{m}_2}=\frac{bf}{ag},\left[Sincerootsof\right(i)are\frac{l_1}{m_1},\frac{l_2}{m_2}]$
or $\frac{l_1{l}_2}{\frac{f}{a}}=\frac{m_1{m}_2}{\frac{g}{b}}$
Similarly, elimination of I will yield $\frac{m_1{m}_2}{\frac{g}{b}}=\frac{n_1{n}_2}{\frac{h}{c}}$
$\therefore \frac{l_1{l}_2}{\frac{f}{a}}=\frac{m_1{m}_2}{\frac{g}{b}}=\frac{n_1{n}_2}{\frac{h}{c}}=q$ (Say)
We know that the lines are perpendicular, if
$l_1{l}_2+m_1{m}_2+n_1{n}_2=0$
i.e.,$\left(\frac{f}{a}\right)q+\left(\frac{g}{b}\right)q+\left(\frac{h}{c}\right)q=0or\frac{f}{a}+\frac{g}{b}+\frac{h}{c}=0$.
Note: Student should remember this question as a fact.