Question

Find the direction cosines of the lines, connected by the relations: l + m + n = 0 and 2lm + 2ln – mn =0.

Answer 1

Given relations are:

⇒ 2lm + 2ln– mn = 0 ……(1)

⇒ l + m + n = 0

⇒ l = (– m – n) ……(2)

Substituting (2) in (1) we get,

⇒ 2(– m –n)m + 2(– m –n)n – mn = 0

⇒ 2(– m² – mn) + 2(– mn – n²) – mn = 0

⇒ – 2m² – 2mn – 2mn – 2n² – mn = 0

⇒ – 2m² – 5mn – 2n² = 0

⇒ 2m² + 5mn + 2n² = 0

⇒ 2m² + 4mn + mn + 2n² = 0

⇒ 2m(m + 2n) + n(m + 2n) = 0

⇒ (2m + n)(m + 2n) = 0

⇒ 2m + n = 0 or m + 2n = 0

⇒ 2m = –n or m = –2n

⇒ m = \(\frac{-n}{2}\) or m = -2n .......(3)

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

For 1^st line:

The direction ratios for the first line is \(\left(\frac{-n}{2},\frac{-n}{2},n \right).\)

Let us assume l₁, m₁, n₁ be the direction cosines of 1^st line.

We know that for a line of direction ratios r₁, r₂, r₃ and having direction cosines l, m, n has the following property.

Using the above formulas we get,

The Direction cosines for the 1^st line is \(\left(\frac{-1}{\sqrt{6}},\frac{-1}{\sqrt{6}},\sqrt{\frac{2}{3}}\right)\)

For 2^nd line:

⇒ l = – (– 2n) – n

⇒ l = 2n – n

⇒ l = n

The direction ratios for the second line is (n, -2n, n).

Let us assume l₂, m₂, n₂ be the direction cosines of 1^st line.

We know that for a line of direction ratios r₁, r₂, r₃ and having direction cosines l, m, n has the following property.

Using the above formulas we get,

The Direction Cosines for the 2^nd line is \(\left(\frac{1}{\sqrt{6}},\frac{-2}{\sqrt{6}},\frac{1}{\sqrt{6}}\right).\)