Question

The straight lines whose direction cosines are given by al + bm + cn = 0, fmn + gnl + hlm = 0 are perpendicular, if

• A.
  
• B.
  
• C.
  
• D.
  

Answer 1

The correct option is A

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