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

Step-by-step solution

1. Write each line in vector form:
2. For $L_1$: point $P_1(1,-2,3)$ and direction $\mathbf{a}=⟨2,-1,4⟩$.
3. For $L_2$: point $P_2(-3,1,-2)$ and direction $\mathbf{b}=⟨-3,2,1⟩$.
4. Compute the vector joining the two points: ${\mathbf{P}}_2{\mathbf{P}}_1=P_2-P_1=(-4,3,-5)$.
5. Find the cross product $\mathbf{a}\times \mathbf{b}$:
6. Using the determinant, $\mathbf{a}\times \mathbf{b}=⟨(-1)\cdot 1-4\cdot 2,-(2\cdot 1-4\cdot (-3)),2\cdot 2-(-1)\cdot (-3)⟩=⟨-9,-14,1⟩$.
7. Calculate its magnitude: $|\mathbf{a}\times \mathbf{b}|=\sqrt{(-9{)}^2+(-14{)}^2+1^2}=\sqrt{81+196+1}=\sqrt{278}$.
8. Take the scalar triple product $|{\mathbf{P}}_2{\mathbf{P}}_1\cdot (\mathbf{a}\times \mathbf{b})|$:
9. Dot product: $(-4)(-9)+3(-14)+(-5)(1)=36-42-5=-11$, absolute value $11$.
10. Apply the distance formula for skew lines: $d=\frac{|{\mathbf{P}}_2{\mathbf{P}}_1\cdot (\mathbf{a}\times \mathbf{b})|}{|\mathbf{a}\times \mathbf{b}|}=\frac{11}{\sqrt{278}}$.
11. State the final distance.

Concept tested: Distance between two skew lines

Published 14 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked