Q. 16 B5.0( 1 Vote )

Find the angle between the lines whose direction cosines are given by the equations:

2l–m+2n=0 and mn+nl+lm=0

Answer :

Given relations are:


mn+nl+lm=0 ……(1)


2l–m+2n=0


m=2l+2n ……(2)


Substituting (2) in (1) we get,


(2l+2n)n+nl+l(2l+2n)=0


2ln+2n2+nl+2l2+2ln=0


2n2+5ln+2l2=0


2n2+4ln+ln+2l2=0


2n(n+2l)+l(n+2l)=0


(2n+l)(n+2l)=0


2n+l=0 or n+2l=0


l=–2n or 2l=–n ……(3)


Substituting the values of(3) in (2) we get,


For the 1st line:


m = 2(–2n)+2n


m=–4n+2n


m=–2n


The direction ratios for the 1st line is (–2n,–2n,n)


For the 2nd line:


m=–n+2n


m=n


The direction ratios for the 2nd line is


We know that the angle between the lines with direction ratios proportional to (a1,b1,c1) and (a2,b2,c2) is given by:



Using the above formula we calculate the angle between the lines.


Let be the angle between the two lines given in the problem.







the angle between two lines is .


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