Question

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

2l + 2m – n = 0 and mn + ln + lm = 0

Answer 1

Given relations are:

⇒ mn + ln + lm = 0 ……(1)

⇒ 2l + 2m – n = 0

⇒ n = 2l + 2m ……(2)

Substituting (2) in (1) we get,

⇒ m(2l + 2m) + l(2l + 2m) + lm = 0

⇒ 2lm + 2m² + 2l² + 2lm + lm = 0

⇒ 2m² + 5lm + 2l² = 0

⇒ 2m² + 4lm + lm + 2l² = 0

⇒ 2m(m + 2l) + l(m + 2l) = 0

⇒ (2m + l)(m + 2l) = 0

⇒ 2m + l = 0 or m + 2l = 0

⇒ 2m = –l or 2l = –m ……(3)

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

For the 1^st line:

⇒ n = 2l – l

⇒ n = l

The Direction Ratios for the first line is \((1, -\frac{1}{2}, 1)\)

For the 2^nd line:

⇒ n = –m + 2m

⇒ n = m

The Direction Ratios for the second line is \(\left(\frac{-m}{2}, m, m \right)\)

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

∴ the angle between two lines is \(\frac{\pi}{2}\) or \(90^\circ.\)