Geometry & Trigonometry: Shortest Distance Between Skew Lines

Direction vectors $\vec{d}_1 = \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}$ and $\vec{d}_2 = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}$ are not scalar multiples ($2/1 \ne 1/(-1)$), hence lines are not parallel. Now test for intersection: $\begin{cases} 1 + 2\lambda = 4 + \mu \implies 2\lambda - \mu = 3 \\ 0 + \lambda = 1 - \mu \implies \lambda + \mu = 1 \\ -1 + 3\lambda = 1 + 2\mu \implies 3\lambda - 2\mu = 2 \end{cases}$ From the first two equations, adding gives $3\lambda = 4 \implies \lambda = 4/3$, which gives $\mu = -1/3$. Test in third equation: $3(4/3) - 2(-1/3) = 4 + 2/3 = 14/3 \ne 2$. Since no consistent solution exists, lines do not intersect. Non-parallel and non-intersecting implies $L_1$ and $L_2$ are skew.

A common normal vector $\vec{n}$ is: $\vec{n} = \vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & 1 & 3 \\ 1 & -1 & 2 \end{vmatrix} = \vec{i}(2 - (-3)) - \vec{j}(4 - 3) + \vec{k}(-2 - 1) = \begin{pmatrix} 5 \\ -1 \\ -3 \end{pmatrix}$ Magnitude: $|\vec{n}| = \sqrt{5^2 + (-1)^2 + (-3)^2} = \sqrt{25 + 1 + 9} = \sqrt{35}$. Displacement between points $\vec{a}_2 - \vec{a}_1 = \begin{pmatrix} 4 - 1 \\ 1 - 0 \\ 1 - (-1) \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix}$.

$D = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot \vec{n}|}{|\vec{n}|} = \frac{|3(5) + 1(-1) + 2(-3)|}{\sqrt{35}} = \frac{|15 - 1 - 6|}{\sqrt{35}} = \frac{8}{\sqrt{35}} \approx 1.35$