The straight lines whose direction cosines are given by al + bm +cn=0 , fmn + gnl + hlm=0 are perpendicular.Find the required condition.

al+bm+cn=0

=>cn=-(al+bm)

=>n=-(al+bm)/c

Substituting this value of n in eq fmn+gnl+hlm=0

=>-fm(al+bm)/c-gl(al+bm)/c+hlm=0

=>aflm+bfm^2+agl^2+bglm-chlm=0

=>(af+bg-ch)lm+ag(l)^2+bf(m)^2=0

=>ag(l)^2+bf(m)^2+(af+bg-ch)lm=0

=>ag(l/m)^2+(af+bg-ch)(l/m)+bf=0

If the roots of this eq are l1/m1 and l2/m2 then sum of roots l1/m1+l2/m2=-(af+bg-ch)/ag

And product of roots l1l2/m1m2=bf/ag

=>a(l1l2)/f =b(m1m2)/g

=>(l1l2)/(f/a)=(m1m2)/(g/b)

Similarly we can show that( m1m2)/(g/b)=(n1n2)/(h/c)

So we can write (l1l2)/(f/a)=(m1m2)/(g/b)=(n1n2)/(h/c) (=k , say)

Since the lines are perpendicular so l1l2+m1m2+n1n2=0

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

=>(f/a)+(g/b) +(h/c) =0 This is the required condition.

Thanks & Regards

Rinkoo Gupta

AskIITians Faculty