Difference between revisions of "1996 AHSME Problems/Problem 28"
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<math> \text{(A)}\ 1.6\qquad\text{(B)}\ 1.9\qquad\text{(C)}\ 2.1\qquad\text{(D)}\ 2.7\qquad\text{(E)}\ 2.9 </math> | <math> \text{(A)}\ 1.6\qquad\text{(B)}\ 1.9\qquad\text{(C)}\ 2.1\qquad\text{(D)}\ 2.7\qquad\text{(E)}\ 2.9 </math> | ||
+ | |||
+ | ==Solution== | ||
+ | |||
+ | 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: | ||
+ | |||
+ | <cmath>\frac{x}{4} + \frac{y}{4} + \frac{z}{3} = 1</cmath> | ||
+ | |||
+ | <cmath>3x + 3y + 4z = 12</cmath> | ||
+ | |||
+ | <cmath>3x + 3y + 4z - 12 = 0</cmath> | ||
+ | |||
+ | 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: | ||
+ | |||
+ | <cmath>\frac{Aa + Bb + Cc + D}{\sqrt{A^2 + B^2 + C^2}}</cmath> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | <cmath>\frac{12}{\sqrt{9 + 9 + 16}}</cmath> | ||
+ | |||
+ | <cmath>\frac{12}{\sqrt{34}}</cmath> | ||
+ | |||
+ | 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 plane. In the 2D case, both <math>c</math> and <math>C</math> are set equal to <math>0</math>. | ||
==See also== | ==See also== | ||
{{AHSME box|year=1996|num-b=27|num-a=29}} | {{AHSME box|year=1996|num-b=27|num-a=29}} |
Revision as of 17:54, 20 August 2011
Problem
On a rectangular parallelepiped, vertices , , and are adjacent to vertex . The perpendicular distance from to the plane containing , , and is closest to
Solution
By placing the cube in a coordinate system such that is at the origin, , , and , we find that the equation of plane is:
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 plane. In the 2D case, both and are set equal to .
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 |