Answer to Question #104581 in Analytic Geometry for Deepak

Question

Consider two lines L1 and L2 whose direction cosines l1,m1,n1 and l2,m2,n2 are
given by the equations for l,m,n :
al +bm+cn = 0, f mn+gnl +hlm = 0,
where abc 6= 0. Show that if L1 ⊥ L2, then f/a + g/b +h/c = 0.

Answer 1

As per the given question,

The equation of the lines are

$al+bm+cn=0−−−−−−−(i)$

And

$fmn+gnl+hlm=0−−−−−−−(ii)$

from the equation (i)

$n=-(\dfrac{al+bm}{c})$ and $n=-(\dfrac{al+cn}{b})$

Now, substituting the value of n in the equation (ii)

$fm(\dfrac{−(al+bm)}{c})+gl(\dfrac{−(al+bm)}{c})+hlm=0$

$fm(-\dfrac{al+bm}{c})+gl(-\dfrac{al+bm}{c})+hlm=0$

multiplying both side by c

$-fm(al+bm​)-gl(al+bm​)+chlm=0$

now, multiplying both side by -1

$fm(al+bm)+gl(al+bm)−hlmc=0$

$fmal+fbm^2+agl^2+bglm−hlmc=0$

$agl^2+afml+bfm^2+bmgl−hlmc=0$

Now, both side divided by $m^2$

$ag(\dfrac{l}{m})2+(af+bg−ch)(\dfrac{l}{m})+bf=0−−−−−−−−(iii)$

Similarly, after substituting the value of m in the question (ii) and we will rearrange it accordingly,

$ah(\dfrac{l}{n})^2+(af+gb−hc)(\dfrac{l}{n})+fc=0−−−−−−−−(iv)$

$l_1​,m_1​,n_1​$ and $l_2,m_2,n_2$ are the direction cosine of straight line

roots of the equation (iii)

so, from the equation (iii)

$\dfrac{l_1 l_2}{m_1 m_2}=\dfrac{\dfrac{f}{a}}{\dfrac{g}{b}}=\dfrac{fb}{ag}$ similarly, from equation (iv) $\dfrac{l_1 l_2}{n_1 n_2}=\dfrac{fc}{ah}$

So, we can write it as

$\dfrac{l_1 l_2}{\dfrac{f}{a}}=\dfrac{m_1 m_2}{\dfrac{g}{b}}=\dfrac{n_1 n_2}{\dfrac{h}{c}}=k,$ let it is equal to a constant K

So,

$l_1 l_2+m_1m_2+n_1n_2=0$

$K(\dfrac{f}{a}+\dfrac{g}{b}+\dfrac{h}{c})=0$

Hence, dividing both side by K

$(\dfrac{f}{a}+\dfrac{g}{b}+\dfrac{h}{c})=0$

Hence L1 and L2 are perpendicular.