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+2n^{2}+nl+2l^{2}+2ln=0

⇒ 2n^{2}+5ln+2l^{2}=0

⇒ 2n^{2}+4ln+ln+2l^{2}=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 1^{st} line:

⇒ m = 2(–2n)+2n

⇒ m=–4n+2n

⇒ m=–2n

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

For the 2^{nd} line:

⇒ m=–n+2n

⇒ m=n

The direction ratios for the 2^{nd} line is

We know that the angle between the lines with direction ratios proportional to (a_{1},b_{1},c_{1}) and (a_{2},b_{2},c_{2}) 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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