Question

The lines whose direction cosines are given by the relations $al+bm+cn=0$ and $mn+nl+lm=0$ are

• A. perpendicular if $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$
• B. perpendicular if $\sqrt{a}+\sqrt{b}+\sqrt{c}=0$
• C. parallel if $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$
• D. parallel to $a+b+c=0$

Answer 1

Answer: (A)

Given relations
$al+bm+cn=0.....(i)$
and $mn+nl+lm=0.....(ii)$
From relations Eqs. (i) and (ii),
$\left(\frac{al+bm}{-c}\right)(l+m)+lm=0$
[by eliminating ' $n^{\prime }$ ]
$\Rightarrow a{l}^2+(a+b-c)lm+b{m}^2=0$
$\Rightarrow a{\left(\frac{l}{m}\right)}^2+(a+b-c)\left(\frac{l}{m}\right)+b=0\dots (iii)$
Let roots of Eq. (iii) is $\frac{l_1}{m_1}$ and $\frac{l_2}{m_2}$,
$\therefore \frac{l_1{l}_2}{m_1{m}_2}=\frac{b}{a}\dots (iv)$
Similarly, $\frac{l_1{l}_2}{n_1{n}_2}=\frac{c}{a}\dots (v)$
So, $\frac{l_1{l}_2}{\frac{1}{a}}=\frac{m_1{m}_2}{\frac{1}{b}}=\frac{n_1{n}_2}{\frac{l}{c}}[$ from Eqs. (iv) and (v) ]
If lines whose direction cosines given by the relation Eqs. (i) and (ii) are perpendicular if $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$