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

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==Problem==
 
==Problem==
  
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On a <math> 4\times 4\times 3 </math> rectangular parallelepiped, vertices <math>A</math>, <math>B</math>, and <math>C</math> are adjacent to vertex <math>D</math>. The perpendicular distance from <math>D</math> to the plane containing
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On a <math> 4\times 4\times 3 </math> rectangular [[parallelepiped]], vertices <math>A</math>, <math>B</math>, and <math>C</math> are adjacent to vertex <math>D</math>. The perpendicular distance from <math>D</math> to the plane containing
 
<math>A</math>, <math>B</math>, and <math>C</math> is closest to  
 
<math>A</math>, <math>B</math>, and <math>C</math> is closest to  
  
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By placing the cube in a coordinate system such that <math>D</math> is at the origin, <math>A(0,0,3)</math>, <math>B(4,0,0)</math>, and <math>C(0,4,0)</math>, we find that the equation of plane <math>ABC</math> is:
 
By placing the cube in a coordinate system such that <math>D</math> is at the origin, <math>A(0,0,3)</math>, <math>B(4,0,0)</math>, and <math>C(0,4,0)</math>, we find that the equation of plane <math>ABC</math> is:
  
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<cmath>\frac{x}{4} + \frac{y}{4} + \frac{z}{3} = 1</cmath>
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<cmath>\frac{x}{4} + \frac{y}{4} + \frac{z}{3} = 1,</cmath> so <math>3x + 3y + 4z - 12 = 0.</math> The equation for the distance of a point <math>(a,b,c)</math> to a plane <math>Ax + By + Cz + D = 0</math> is given by:
  
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<cmath>3x + 3y + 4z = 12</cmath>
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<cmath>\frac{Aa + Bb + Cc + D}{\sqrt{A^2 + B^2 + C^2}}.</cmath>
  
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<cmath>3x + 3y + 4z - 12 = 0</cmath>
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Note that the capital letters are coefficients, while the lower case is the point itself. Thus, the distance from the origin (where <math>a=b=c=0</math>) to the plane is given by:
  
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The equation for the distance of a point <math>(a,b,c)</math> to a plane <math>Ax + By + Cz + D = 0</math> is given by:
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<cmath>\frac{D}{\sqrt{A^2 + B^2 + C^2}} = \frac{12}{\sqrt{9 + 9 + 16}} = \frac{12}{\sqrt{34}}.</cmath>
  
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<cmath>\frac{Aa + Bb + Cc + D}{\sqrt{A^2 + B^2 + C^2}}</cmath>
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Since <math>\sqrt{34} < 6</math>, this number should be just a little over <math>2</math>, and the correct answer is <math>\boxed{\text{(C)}}</math>.
−
 
 
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Note that the capital letters are coefficients, while the lower case is the point itself.
 
−
 
 
−
Thus, the distance from the origin (where <math>a=b=c=0</math>) to the plane is given by:
 
−
 
 
−
<cmath>\frac{D}{\sqrt{A^2 + B^2 + C^2}}</cmath>
 
−
 
 
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<cmath>\frac{12}{\sqrt{9 + 9 + 16}}</cmath>
 
−
 
 
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<cmath>\frac{12}{\sqrt{34}}</cmath>
 
−
 
 
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Since <math>\sqrt{34} < 6</math>, this number should be just a little over <math>2</math>, and the correct answer is <math>\boxed{C}</math>.
 
  
 
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 <math>c</math> and <math>C</math> are set equal to <math>0</math>.
 
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 <math>c</math> and <math>C</math> are set equal to <math>0</math>.
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==See also==
 
==See also==
 
{{AHSME box|year=1996|num-b=27|num-a=29}}
 
{{AHSME box|year=1996|num-b=27|num-a=29}}
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[[Category:Introductory Geometry Problems]]
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[[Category:3D Geometry Problems]]

Revision as of 22:14, 20 August 2011

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:

\[\frac{x}{4} + \frac{y}{4} + \frac{z}{3} = 1,\] so $3x + 3y + 4z - 12 = 0.$ 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 $\boxed{\text{(C)}}$.

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 line. In the 2D case, both $c$ and $C$ are set equal to $0$.

See also

1996 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 27 		Followed by
Problem 29
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions
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