# Difference between revisions of "1996 AHSME Problems/Problem 28"

## Problem

On a $4\times 4\times 3$ rectangular parallelepiped, vertices $A$, $B$, and $C$ are adjacent to vertex $D$. The perpendicular distance from $D$ to the plane containing $A$, $B$, and $C$ is closest to

$[asy] size(120); import three; currentprojection=orthographic(1, 4/5, 1/3); draw(box(O, (4,4,3))); triple A=(0,4,3), B=(0,0,0) , C=(4,4,0), D=(0,4,0); draw(A--B--C--cycle, linewidth(0.9)); label("A", A, NE); label("B", B, NW); label("C", C, S); label("D", D, E); label("4", (4,2,0), SW); label("4", (2,4,0), SE); label("3", (0, 4, 1.5), E); [/asy]$

$\text{(A)}\ 1.6\qquad\text{(B)}\ 1.9\qquad\text{(C)}\ 2.1\qquad\text{(D)}\ 2.7\qquad\text{(E)}\ 2.9$

## Solution

By placing the cube in a coordinate system such that $D$ is at the origin, $A(0,0,3)$, $B(4,0,0)$, and $C(0,4,0)$, we find that the equation of plane $ABC$ is:

$$\frac{x}{4} + \frac{y}{4} + \frac{z}{3} = 1$$

$$3x + 3y + 4z = 12$$

$$3x + 3y + 4z - 12 = 0$$

The equation for the distance of a point $(a,b,c)$ to a plane $Ax + By + Cz + D = 0$ is given by:

$$\frac{Aa + Bb + Cc + D}{\sqrt{A^2 + B^2 + C^2}}$$

Note that the capital letters are coefficients, while the lower case is the point itself.

Thus, the distance from the origin (where $a=b=c=0$) to the plane is given by:

$$\frac{D}{\sqrt{A^2 + B^2 + C^2}}$$

$$\frac{12}{\sqrt{9 + 9 + 16}}$$

$$\frac{12}{\sqrt{34}}$$

Since $\sqrt{34} < 6$, this number should be just a little over $2$, and the correct answer is $\boxed{C}$.

Note that the equation above for the distance from a point to a plane is a 3D analogue of the 2D case of the distance formula, where you take the distance from a point to a plane. In the 2D case, both $c$ and $C$ are set equal to $0$.