Answer to Question #114941 in Analytic Geometry for ANJU JAYACHANDRAN

Question #114941
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.
1
Expert's answer
2020-05-18T19:13:58-0400

We have given that L1&L2L_1 \& L_2 are equation of line whose direction cosines are given as (l1,m1,n1)&(l2,m2,n2)(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(1)fmn+gnl+hlm=0(2)al +bm+cn = 0 \hspace{1cm} (1)\\ f mn+gnl +hlm = 0 \hspace{1cm}(2)

where abc0abc \neq 0 ,also given that L1L2    l1l2+m1m2+n1n2=0()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)(1) ,say ll ,thus l=bm+cnal=-\frac{bm+cn}{a} and substitute to (2)(2) we get,


fmn+gn(bm+cna)+hm(bm+cna)=0    hbm2+(hc+gbaf)mn+gcn2=0    hby2+(hc+gbaf)y+gc=0()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=mny=\frac{m}{n}



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


y1+y2=(hc+gbaf)hb(3)y1y2=gchb(4)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)(4) ,

m1n1m2n2=gchb    m1m2gb=n1n2hc(5)\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,mn ,m (say nn ) and do the same calculation as above we get,

m1m2gb=l1l2fa(6)\frac{m_1m_2}{\frac{g}{b}}=\frac{l_1l_2}{\frac{f}{a}} \hspace{1cm}(6)

Now from equation 55 and 66 we finally get


m1m2gb=l1l2fa=n1n2hc(7)\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 77 if we consider m1m2gb=l1l2fa=n1n2hc=p\frac{m_1m_2}{\frac{g}{b}}=\frac{l_1l_2}{\frac{f}{a}} =\frac{n_1n_2}{\frac{h}{c}}=p ,clearly p0p \neq 0 ,hence,


l1l2+m1m2+n1n2=p(gb+fa+hc)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,


l1l2+m1m2+n1n2=p(gb+fa+hc)=0    p(gb+fa+hc)=0    (gb+fa+hc)=0(p0)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.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog