Answer to Question #102906 in Analytic Geometry for ag

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, fmn+gnl +hlm = 0,
where abc is not equal to 0. Show that if L1 ⊥ L2, then f÷a + g÷b + h÷c = 0.

Answer 1

al+bm+cn=0, fmn+gnl+hlm=0,

n=-(al+bm)/c,

fm*[-(al+bm)/c]+gl*[-(al+bm)/c]+hlm=0

fmal+fm²b+gal²+gbml-hlmc=0

gal²+lm(af+bg-ch)+bfm²=0

ga(l/m)²+(l/m)(af+bg-ch)+bf=0

l1/m1 & l2/m2 - solving the equation

l1,m1,n1 &l2,m2,n2 of straight line 

L1=/=L2, because L1_|_L2

By Viet's theorem

l1/m1 * l2/m2=bf/ga,

l1*l2/(f/a)=m1*m2/(g/b).

Similarly, expressing the variables m and l,

l1*l2/(f/a)=m1*m2/(g/b)=n1*n2/(h/c)=const=k

A widely known sign of perpendicular straight lines is

the sum of the corresponding products of the directing cosines is zero.

l1*l2+l1*l2+n1*n2=0

k*f/a + k*g/b +k*h/c=0

k(f/a+g/b+h/c)=0

$k \ne 0$ because $l, m, n \ne 0$ (cos x=0 if x=0 ) & $abc \ne 0$

f/a+g/b+h/c=0