(a) To find the vector equation for the line through A and B, we first find the direction vector \(\mathbf{AB}\) by subtracting the position vector of A from B:
\(\mathbf{AB} = (\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}) - (2\mathbf{i} + \mathbf{j} + \mathbf{k}) = -\mathbf{i} - 3\mathbf{j} + \mathbf{k}\).
The vector equation of the line through A and B is:
\(\mathbf{r} = 2\mathbf{i} + \mathbf{j} + \mathbf{k} + \lambda(-\mathbf{i} - 3\mathbf{j} + \mathbf{k})\).
(b) To find the acute angle between the directions of \(AB\) and \(l\), we use the dot product formula:
\(\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|}\).
Direction vector of \(l\) is \(\mathbf{i} - 3\mathbf{j} - 2\mathbf{k}\).
\(\mathbf{AB} \cdot (\mathbf{i} - 3\mathbf{j} - 2\mathbf{k}) = (-1)(1) + (-3)(-3) + (1)(-2) = -1 + 9 - 2 = 6\).
\(\|\mathbf{AB}\| = \sqrt{(-1)^2 + (-3)^2 + (1)^2} = \sqrt{11}\).\)
\(\|\mathbf{l}\| = \sqrt{1^2 + (-3)^2 + (-2)^2} = \sqrt{14}\).\)
\(\cos \theta = \frac{6}{\sqrt{11} \cdot \sqrt{14}}\).
\(\theta = \cos^{-1}\left(\frac{6}{\sqrt{11} \cdot \sqrt{14}}\right) \approx 61.1^\circ\).
(c) To show that the line through A and B does not intersect the line \(l\), express the general points of both lines in component form and equate them:
Line through A and B: \((2 - \lambda, 1 - 3\lambda, 1 + \lambda)\).
Line \(l\): \((1 + \mu, 2 - 3\mu, -3 - 2\mu)\).
Equate components and solve for \(\lambda\) and \(\mu\):
\(2 - \lambda = 1 + \mu\), \(1 - 3\lambda = 2 - 3\mu\), \(1 + \lambda = -3 - 2\mu\).
Solving these gives inconsistent results, confirming the lines do not intersect.
