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

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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