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

l+2m+3n=0 and 3lm–4ln+mn=0

Given relations are:


3lm–4ln+mn=0 ……(1)


l+2m+3n=0


l=–2m–3n ……(2)


Substituting (2) in (1) we get,


3(–2m–3n)m –4(–2m–3n)n +mn =0


3(–2m2–3mn) –4(–2mn–3n2) +mn=0


–6m2–9mn+8mn+12n2+mn=0


12n2–6m2=0


m2–2n2=0




……(3)


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


For the 1st line:




The Direction Ratios for the 1st line is .


For the 2nd line:




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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