First, find the vector equation of the line passing through A and B. The direction vector AB is given by:
\(\text{Direction vector } \overrightarrow{AB} = (3\mathbf{i} + \mathbf{j} + \mathbf{k}) - (\mathbf{i} + 2\mathbf{j} - \mathbf{k}) = 2\mathbf{i} - \mathbf{j} + 2\mathbf{k}\)
The vector equation of the line through A and B is:
\(\mathbf{r} = \mathbf{i} + 2\mathbf{j} - \mathbf{k} + \lambda(2\mathbf{i} - \mathbf{j} + 2\mathbf{k})\)
Equate the components of the general points on both lines:
\((1 + 2\lambda, 2 - \lambda, -1 + 2\lambda) = (2 + \mu, 1 + \mu, 1 + 2\mu)\)
Solving these equations:
1. \(1 + 2\lambda = 2 + \mu\)
2. \(2 - \lambda = 1 + \mu\)
3. \(-1 + 2\lambda = 1 + 2\mu\)
From equation 1: \(\lambda = \frac{1}{2} \mu + \frac{1}{2}\)
From equation 2: \(\lambda = -\mu + 1\)
From equation 3: \(\lambda = \mu + 1\)
Substitute \(\lambda = \mu + 1\) into \(\lambda = \frac{1}{2} \mu + \frac{1}{2}\):
\(\mu + 1 = \frac{1}{2} \mu + \frac{1}{2}\)
\(\mu = -1\)
Substitute \(\mu = -1\) into \(\lambda = \mu + 1\):
\(\lambda = 0\)
Verify with the third equation:
\(-1 + 2(0) = 1 + 2(-1)\)
\(-1 \neq -1\)
Since the equations are not satisfied, the lines do not intersect.
