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in 3D Coordinate Geometry by (29.1k points)
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Find the angle between the lines whose direction cosines are given by the equations:

l + 2m + 3n = 0 and 3lm – 4ln + mn = 0

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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

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

For the 1st line:

The Direction Ratios for the 1st line is (-(3 + 2\(\sqrt{2}\))n, \(\sqrt{2}\)n, n).

For the 2nd line:

The Direction Ratios for the 2nd line is ((2\(\sqrt{2}\) - 3)n, -\(\sqrt{2}\)n, n).

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 \(90^\circ.\)

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