In a rectangular box with , , , find the distance from vertex to the face diagonal on the opposite face .

The distance from to the diagonal is .

Solid Geometry Solution – Problem 20: find the distance from vertex to the face diagonal on the opposite face

Question

In a rectangular box $ABCD-A_1B_1C_1D_1$ with $AB=2$ , $BC=2$ , $AA_1=1$ , find the distance from vertex $A$ to the face diagonal $B_1D_1$ on the opposite face $A_1B_1C_1D_1$ .

Step-by-step solution

Set up coordinates: $A(0,0,0),\ B_1(2,0,1),\ D_1(0,2,1).$ The line $B_1D_1$ has direction $\vec v=\overrightarrow{B_1D_1}=(-2,2,0).$ Use the point-to-line distance formula in $\mathbb R^3$ : $d=\frac{|(\overrightarrow{B_1A})\times\vec v|}{|\vec v|},$ where $\overrightarrow{B_1A}=A-B_1=(-2,0,-1)$ . Compute $(\overrightarrow{B_1A})\times\vec v=(-2,0,-1)\times(-2,2,0)=(2,2,-4),$ so $|(\overrightarrow{B_1A})\times\vec v|=\sqrt{2^2+2^2+(-4)^2}=2\sqrt6,$ and $|\vec v|=\sqrt{(-2)^2+2^2}=2\sqrt2.$ Therefore $d=\frac{2\sqrt6}{2\sqrt2}=\sqrt3.$

Final answer

The distance from $A$ to the diagonal $B_1D_1$ is $\sqrt3$ .

Marking scheme

1. Checkpoints (max 7 pts total)

Coordinate setup (2 pts): Correctly assign coordinates for the $2\times2\times1$ box and the line $B_1D_1$ .
Distance formula (3 pts): Use $d=\frac{|(P-P_0)\times v|}{|v|}$ with correct vectors.
Computation (2 pts): Compute cross product and norms to obtain $\sqrt3$ .

2. Zero-credit items

Using a planar distance in the top face without handling the 3D position of $A$ .
Providing $\sqrt3$ with no vector calculation.

3. Deductions

Vector direction error (-1): wrong direction vector for $B_1D_1$ .
Cross-product arithmetic error (-1): incorrect $(2,2,-4)$ or its norm.

Solid Geometry – Problem 20: find the distance from vertex to the face diagonal on the opposite face