In space, let , , , . Find the distance between the two skew lines and .

The distance between lines and is .

Solid Geometry Advanced Solution – Problem 9: Find the distance between the two skew lines and

Question

In space, let $A(0,0,0)$ , $B(2,0,0)$ , $C(0,2,0)$ , $D(0,0,2)$ .

Find the distance between the two skew lines $AB$ and $CD$ .

Step-by-step solution

(1) Direction vectors: $\mathbf u=\overrightarrow{AB}=(2,0,0),\qquad \mathbf v=\overrightarrow{CD}=D-C=(0,-2,2).$ (2) Let $\mathbf w=\overrightarrow{AC}=(0,2,0)$ . The distance between skew lines is $d=\frac{|\mathbf w\cdot(\mathbf u\times\mathbf v)|}{|\mathbf u\times\mathbf v|}.$ (3) Compute $\mathbf u\times\mathbf v=(2,0,0)\times(0,-2,2)=(0,-4,-4),$ so $|\mathbf u\times\mathbf v|=\sqrt{32}=4\sqrt2$ . Also $\mathbf w\cdot(\mathbf u\times\mathbf v)=(0,2,0)\cdot(0,-4,-4)=-8.$ Hence $d=\frac{8}{4\sqrt2}=\sqrt2.$

Final answer

The distance between lines $AB$ and $CD$ is $\sqrt2$ .

Marking scheme

Step 1 — Setup

Checkpoint: identify direction vectors $\mathbf u,\mathbf v$ and a connector vector $\mathbf w$ (2 pts)

Step 2 — Key Calculation

Checkpoint: compute $\mathbf u\times\mathbf v$ and the scalar triple product $\mathbf w\cdot(\mathbf u\times\mathbf v)$ (3 pts)

Step 3 — Final Answer

Checkpoint: apply $d=\frac{|\mathbf w\cdot(\mathbf u\times\mathbf v)|}{|\mathbf u\times\mathbf v|}$ to get $\sqrt2$ (2 pts)

Zero credit if: uses a 2D distance formula or projects onto the wrong plane.

Deductions: -1 pt for cross product sign error (absolute value can still salvage if explained).

Solid Geometry Advanced – Problem 9: Find the distance between the two skew lines and