(a) To show that the lines are skew, express the general point of each line in component form:
Line 1: \((1 + 2s, 3 - s, 2 + 3s)\)
Line 2: \((2 + t, 1 - t, 4 + 4t)\)
Equate the components to find \(s\) and \(t\):
\(1 + 2s = 2 + t\)
\(3 - s = 1 - t\)
\(2 + 3s = 4 + 4t\)
Solving these, possible values are \(s = -1, 6, \frac{2}{5}\) and \(t = -3, 4, -\frac{1}{5}\).
Verify that all three component equations are not satisfied simultaneously, showing the lines are not parallel and thus skew.
(b) To find the acute angle between the directions of the two lines, calculate the scalar product of the direction vectors:
Direction vector of Line 1: \(\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}\)
Direction vector of Line 2: \(\begin{pmatrix} 1 \\ -1 \\ 4 \end{pmatrix}\)
Scalar product: \(2 \times 1 + (-1) \times (-1) + 3 \times 4 = 2 + 1 + 12 = 15\)
Calculate the moduli of the direction vectors:
\(\| \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} \| = \sqrt{2^2 + (-1)^2 + 3^2} = \sqrt{14}\)
\(\| \begin{pmatrix} 1 \\ -1 \\ 4 \end{pmatrix} \| = \sqrt{1^2 + (-1)^2 + 4^2} = \sqrt{18}\)
Cosine of the angle \(\theta\):
\(\cos \theta = \frac{15}{\sqrt{14} \times \sqrt{18}}\)
\(\theta = \cos^{-1}\left(\frac{15}{\sqrt{14} \times \sqrt{18}}\right) \approx 19.1^\circ\) or \(0.333\) radians.
