Answer to Question #114941 in Analytic Geometry for ANJU JAYACHANDRAN

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 not = 0. Show that if L1 ⊥ L2, then f/a +g/b +h/c = 0.

Answer 1

We have given that $L_1 \& L_2$ are equation of line whose direction cosines are given as $(l_1,m_1,n_1) \& (l_2,m_2,n_2)$ respectively obtained from solution of two given equation as follows,

$$al +bm+cn = 0 \hspace{1cm} (1)\\ f mn+gnl +hlm = 0 \hspace{1cm}(2)$$

where $abc \neq 0$ ,also given that $L_1 \bot L_2 \implies l_1l_2+m_1m_2+n_1n_2=0 \: (\star)$ .

Let's eliminate one of the variable from $(1)$ ,say $l$ ,thus $l=-\frac{bm+cn}{a}$ and substitute to $(2)$ we get,

$$fmn+gn(-\frac{bm+cn}{a})+hm(-\frac{bm+cn}{a})=0\\ \implies hbm^2+(hc+gb-af)mn+gcn^2=0\\ \implies hby^2+(hc+gb-af)y+gc=0 \hspace{1cm}(\clubs)$$

where

$$y=\frac{m}{n}$$

Clearly, equation $(\clubs)$ is quadratic in $y$ ,hence by relation between roots and coefficients(Vieta's Theorem) we get,

$$y_1+y_2=-\frac{(hc+gb-af)}{hb} \hspace{1cm} (3)\\ y_1y_2=\frac{gc}{hb}\hspace{1cm} (4)$$

Hence,from equation $(4)$ ,

$$\frac{m_1}{n_1}\frac{m_2}{n_2}=\frac{gc}{hb}\\ \implies \frac{m_1m_2}{\frac{g}{b}}=\frac{n_1n_2}{\frac{h}{c}} \hspace{1cm}(5)$$

In the same manner if we eliminate other two variables $n ,m$ (say $n$ ) and do the same calculation as above we get,

$$\frac{m_1m_2}{\frac{g}{b}}=\frac{l_1l_2}{\frac{f}{a}} \hspace{1cm}(6)$$

Now from equation $5$ and $6$ we finally get

$$\frac{m_1m_2}{\frac{g}{b}}=\frac{l_1l_2}{\frac{f}{a}} =\frac{n_1n_2}{\frac{h}{c}} \hspace{1cm}(7)$$

Thus from equation $7$ if we consider $\frac{m_1m_2}{\frac{g}{b}}=\frac{l_1l_2}{\frac{f}{a}} =\frac{n_1n_2}{\frac{h}{c}}=p$ ,clearly $p \neq 0$ ,hence,

$$l_1l_2+m_1m_2+n_1n_2=p\bigg(\frac{g}{b}+\frac{f}{a}+\frac{h}{c}\bigg )$$

but from equation $(\star)$ we get,

$$l_1l_2+m_1m_2+n_1n_2=p\bigg(\frac{g}{b}+\frac{f}{a}+\frac{h}{c}\bigg )=0\\ \implies p\bigg(\frac{g}{b}+\frac{f}{a}+\frac{h}{c}\bigg )=0\\ \implies \bigg(\frac{g}{b}+\frac{f}{a}+\frac{h}{c}\bigg )=0 \hspace{1cm}(\because p\neq0)\\$$

Therefore we are done.