1996 AHSME Problems/Problem 28
Contents
Problem
On a rectangular parallelepiped, vertices , , and are adjacent to vertex . The perpendicular distance from to the plane containing , , and is closest to
Solution 1
By placing the cube in a coordinate system such that is at the origin, , , and , we find that the equation of plane is:
so The equation for the distance of a point to a plane is given by:
Note that the capital letters are coefficients, while the lower case is the point itself. Thus, the distance from the origin (where ) to the plane is given by:
Since , this number should be just a little over , and the correct answer is .
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 line. In the 2D case, both and are set equal to .
Solution 2
Let be the desired distance. Recall that the volume of a pyramid is given by , where is the area of the base and is the height. Consider pyramid . Letting be the base, the volume of is given by , but if we let be the base, the volume is given by . Clearly, these two volumes must be equal, so we get the equation . Thus, to find , we just need to find .
By the Pythagorean Theorem, , , .
The altitude to in triangle has length , so . Then or about . The answer is .
See also
1996 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 27 |
Followed by Problem 29 | |
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