Let be a regular tetrahedron with edge length . Let be the centroid of triangle , and let be the median line in the base plane. Find the distance from vertex to line .

Solid Geometry Solution – Problem 18: Find the distance from vertex to line

Question

Let $ABCD$ be a regular tetrahedron with edge length $2$ . Let $G$ be the centroid of triangle $ABC$ , and let $\ell$ be the median line $AG$ in the base plane. Find the distance from vertex $D$ to line $\ell$ .

Step-by-step solution

Place the base $\triangle ABC$ in $z=0$ : $A(0,0,0),\ B(2,0,0),\ C(1,\sqrt3,0).$ Then the centroid is $G\left(\frac{0+2+1}{3},\frac{0+0+\sqrt3}{3},0\right)=\left(1,\frac{\sqrt3}{3},0\right).$ In a regular tetrahedron, $D$ lies on the line perpendicular to plane $ABC$ through $G$ . Write $D\left(1,\frac{\sqrt3}{3},h\right).$ Use $DA=2$ : $DA^2=1^2+\left(\frac{\sqrt3}{3}\right)^2+h^2=1+\frac13+h^2=\frac43+h^2=4,$ so $h^2=\frac{8}{3}\Rightarrow h=\frac{2\sqrt6}{3}.$ Since $G\in AG=\ell$ and $DG\perp$ plane $ABC$ , we have $DG\perp \ell$ . Therefore the distance from $D$ to $\ell$ equals $DG=h=\frac{2\sqrt6}{3}.$

Final answer

The distance from $D$ to line $AG$ is $\dfrac{2\sqrt6}{3}$ .

Marking scheme

1. Checkpoints (max 7 pts total)

Coordinate setup (2 pts): Place an equilateral $\triangle ABC$ of side $2$ in $z=0$ and compute $G$ .
Solve for height (3 pts): Set $D=(1,\frac{\sqrt3}{3},h)$ and use $DA=2$ to obtain $h=\frac{2\sqrt6}{3}$ .
Distance conclusion (2 pts): Use $DG\perp$ plane $ABC$ and $G\in AG$ to conclude distance $=DG$ .

2. Zero-credit items

Assuming the height $h$ without deriving it from $DA=2$ .
Treating the point-to-line distance as the point-to-plane distance without justification.

3. Deductions

Centroid error (-1): incorrect $G$ coordinate.
Square-root error (-1): incorrect extraction of $h$ from $h^2=\frac{8}{3}$ .

Solid Geometry – Problem 18: Find the distance from vertex to line