The direction cosines l, m, n of two lines satisfy the relations 2l+2…

The correct answer is Option (1) → \(\displaystyle\frac{\pi}{2}\)

From the linear relation:

    \(n = 2l+2m\)

Put this in the quadratic relation:

    \(lm+m(2l+2m)+l(2l+2m) = 0\)

    \(2l^2+5lm+2m^2 = 0\)

    \((2l+m)(l+2m) = 0\)

First line, \(m = -2l\):

    \(n = 2l-4l = -2l\), direction ratios \(1, -2, -2\)

Second line, \(l = -2m\):

    \(n = -4m+2m = -2m\), direction ratios \(-2, 1, -2\)

Testing for perpendicular lines:

    \((1)(-2)+(-2)(1)+(-2)(-2) = -2-2+4 = 0\)

So the lines are perpendicular and the angle is \(\frac{\pi}{2}\).