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Question

Find the angle between the lines whose direction cosines are given by the equation 3l+m+5n=0 and 6mn2nl+5lm=0.

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Solution

Given equation 3l+m+5n=0 & 6mn2nl+5lm=0-----(2)
m=(3l+5n)(1)
Sub (1) in (2) 6[(3l+5n)]n2ln+5l[(3l+5n)]=0
(18l30n)n2nl15l225ln=0
18ln30n22ln15l225ln=0
15l245ln30n2=0l2+3ln+2n2=0
l(l+2n)+n(l+2n)=0(l+n)(l+2n)=0
l=n & l=2n
from (1) m=(3l+5n) m=(3l+5n)
l=nm=(3n+5n) l=2nm=(3(2n)+5n)
m=2n m=n
l=n & m=2n l=2n & m=n
l1=n1 m2=n1 l2=n1 m1=n1
(l1,m1,n1)=(1,2,1) (l2,m2,n2)=(2,1,1)
Angle between direction cosines are
cosθ=l1l2+m1m2+n1n2l21+m21+n21l22+m22+n22
=(1)(2)+(2)(1)+1(1)(1)2+(2)2+12(2)2+12+12
=+21+11+4+14+1+1=16.6=16
cosθ=16
θ=cos1(16).


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