To find the cosine of the acute angle between the lines whose direction cosines satisfy the given equations, let's solve the problem step-by-step.
The given equations relating the direction cosines \(l\), \(m\), and \(n\) are:
- \(4l + m - n = 0\)
- \(2mn + 10nl + 3lm = 0\)
First, solve the first equation for one of the variables, say \(m\):
\(m = n - 4l\)
Substitute this expression for \(m\) in the second equation:
\(2(n - 4l)n + 10nl + 3l(n - 4l) = 0\)
Simplify the equation:
\(2n^2 - 8ln + 10nl + 3ln - 12l^2 = 0\)
Combine the terms:
\(2n^2 + 2ln - 12l^2 = 0\)
Divide the entire equation by 2:
\(n^2 + ln - 6l^2 = 0\)
Consider this as a quadratic in \(n\). Solving the quadratic \(n^2 + ln - 6l^2 = 0\) using the quadratic formula:
The quadratic formula is given by:
\(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
For our quadratic, \(a = 1\), \(b = l\), and \(c = -6l^2\).
\(n = \frac{-l \pm \sqrt{l^2 + 24l^2}}{2}\)
Solving within the square root:
\(n = \frac{-l \pm \sqrt{25l^2}}{2} = \frac{-l \pm 5l}{2}\)
This gives us two solutions:
\(n = 2l\) or \(n = -3l\)
Since the direction cosines should satisfy \(l^2 + m^2 + n^2 = 1\), substitute \(n = 2l\) and \(m = n - 4l\):
For \(n = 2l\):
\(m = 2l - 4l = -2l\)
Thus:
\(l^2 + (-2l)^2 + (2l)^2 = 1\)
\(l^2 + 4l^2 + 4l^2 = 1 \Rightarrow 9l^2 = 1 \Rightarrow l = \pm \frac{1}{3}\)
So, \(n = \pm \frac{2}{3}\) and \(m = \mp \frac{2}{3}\). Therefore, one set of direction cosines is:
\(l = \frac{1}{3}, m = -\frac{2}{3}, n = \frac{2}{3}\)
For \(n = -3l\):
\(m = -3l - 4l = -7l\)
Thus:
\(l^2 + (-7l)^2 + (-3l)^2 = 1\)
\(l^2 + 49l^2 + 9l^2 = 1 \Rightarrow 59l^2 = 1 \Rightarrow l = \pm \frac{1}{\sqrt{59}}\)
Then \(n = \mp \frac{3}{\sqrt{59}}\) and \(m = \mp \frac{7}{\sqrt{59}}\). Therefore, the second set of direction cosines is:
\(l = \frac{1}{\sqrt{59}}, m = -\frac{7}{\sqrt{59}}, n = -\frac{3}{\sqrt{59}}\)
The cosine of the angle \(\theta\) between the lines with direction cosines \( (l_1, m_1, n_1) \) and \( (l_2, m_2, n_2) \) is given by:
\(\cos \theta = l_1 l_2 + m_1 m_2 + n_1 n_2\)
Substituting the values from one set as above \( l = \frac{1}{3}, m = -\frac{2}{3}, n = \frac{2}{3}\) and another set \( l = \frac{1}{\sqrt{59}}, m = -\frac{7}{\sqrt{59}}, n = -\frac{3}{\sqrt{59}} \):
\(\cos \theta = \left(\frac{1}{3}\right)\left(\frac{1}{\sqrt{59}}\right) + \left(-\frac{2}{3}\right)\left(-\frac{7}{\sqrt{59}}\right) + \left(\frac{2}{3}\right)\left(-\frac{3}{\sqrt{59}}\right)\)
\(\cos \theta = \frac{1}{3\sqrt{59}} + \frac{14}{3\sqrt{59}} - \frac{6}{3\sqrt{59}} = \frac{9}{3\sqrt{59}} = \frac{3}{\sqrt{59}}\)
This matches the correct option given:
\(\cos \theta = \frac{10}{\sqrt{38}}\).
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