Problem Statement
The lines $L_1$ and $L_2$ have vector equations given by:
$$L_1: \mathbf{r} = \begin{pmatrix} 1 \\ 0 \\ -2 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}, \quad \lambda \in \mathbb{R}$$
$$L_2: \mathbf{r} = \begin{pmatrix} 4 \\ 3 \\ 1 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}, \quad \mu \in \mathbb{R}$$
**(a)** [3 marks] Show that the lines $L_1$ and $L_2$ are not parallel and do not intersect (i.e. they are skew lines).
**(b)** [3 marks] Find a vector perpendicular to both $L_1$ and $L_2$, and hence calculate the shortest distance between the two skew lines.
**(c)** [3 marks] Find the Cartesian equation of the plane $\Pi$ that contains the line $L_1$ and is parallel to the line $L_2$.
Verified Solution & Marking Scheme
Part (a) [M1]
Direction vectors $\mathbf{d}_1 = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$ and $\mathbf{d}_2 = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$. Since no scalar $k$ exists such that $\mathbf{d}_1 = k\mathbf{d}_2$ (ratio $2/1 \ne -1/2$), the lines are not parallel.
Part (a) [M1]
Equate components for intersection: $1 + 2\lambda = 4 + \mu$, $-\lambda = 3 + 2\mu$, $-2 + 3\lambda = 1 - \mu$. From first two: $2\lambda - \mu = 3$ and $\lambda + 2\mu = -3 \implies 5\lambda = 3 \implies \lambda = 3/5, \mu = -9/5$.
Part (a) [R1]
Substitute into third equation: LHS $=-2 + 3(3/5) = -1/5$; RHS $= 1 - (-9/5) = 14/5$. Since $-1/5 \ne 14/5$, the system is inconsistent. Therefore, the lines do not intersect and are skew.
Part (b) [M1]
Compute cross product $\mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2 = \begin{pmatrix} (-1)(-1) - (3)(2) \\ (3)(1) - (2)(-1) \\ (2)(2) - (-1)(1) \end{pmatrix} = \begin{pmatrix} -5 \\ 5 \\ 5 \end{pmatrix} \propto \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}$
Part (b) [M1]
Vector between initial points $\mathbf{a}_2 - \mathbf{a}_1 = \begin{pmatrix} 4 - 1 \\ 3 - 0 \\ 1 - (-2) \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \\ 3 \end{pmatrix}$
Part (b) [A1]
Shortest distance $D = \frac{|(\mathbf{a}_2 - \mathbf{a}_1) \cdot \mathbf{n}|}{|\mathbf{n}|} = \frac{|3(-1) + 3(1) + 3(1)|}{\sqrt{(-1)^2 + 1^2 + 1^2}} = \frac{|3|}{\sqrt{3}} = \sqrt{3} \approx 1.73$
Part (c) [M1]
Normal vector to the plane is $\mathbf{n} = \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}$. Equation is $-x + y + z = d$.
Part (c) [M1]
Substitute point on $L_1$, $(1, 0, -2)$: $-(1) + (0) + (-2) = -3 \implies d = -3$.
Part (c) [A1]
Cartesian equation: $-x + y + z = -3$ (or $x - y - z = 3$).