Find the shortest distance between the two skew lines $L_1:\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-3}{3}$ and $L_2:\frac{x+2}{1}=\frac{y-4}{2}=\frac{z+1}{-2}.$

Step-by-step solution

1. Write direction vectors: ${\mathbf{d}}_1=(2,-1,3)$ from $L_1$, ${\mathbf{d}}_2=(1,2,-2)$ from $L_2$.
2. Select a point on each line: for $L_1$ when the parameter is $0$, $A(1,-1,3)$; for $L_2$ when the parameter is $0$, $B(-2,4,-1)$.
3. Form the vector joining the two points: $\mathbf{AB}=B-A=(-3,5,-4)$.
4. Compute the cross product ${\mathbf{d}}_1\times {\mathbf{d}}_2$:
5. ${\mathbf{d}}_1\times {\mathbf{d}}_2=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 2& -1& 3\\ 1& 2& -2\end{array}\right|=(-4,7,5)$.
6. Find the scalar triple product $|\mathbf{AB}\cdot ({\mathbf{d}}_1\times {\mathbf{d}}_2)|$:
7. $\mathbf{AB}\cdot (-4,7,5)=(-3)(-4)+5\cdot 7+(-4)\cdot 5=12+35-20=27$.
8. Compute the magnitude of the cross product: $|{\mathbf{d}}_1\times {\mathbf{d}}_2|=\sqrt{(-4{)}^2+7^2+5^2}=\sqrt{90}=3\sqrt{10}$.
9. Apply the distance formula $d=\frac{|\mathbf{AB}\cdot ({\mathbf{d}}_1\times {\mathbf{d}}_2)|}{|{\mathbf{d}}_1\times {\mathbf{d}}_2|}=\frac{27}{3\sqrt{10}}=\frac{9}{\sqrt{10}}=\frac{9\sqrt{10}}{10}$.

Concept tested: Shortest distance between skew lines

Published 4 September 2026 · How this solution was prepared and checked