1996 AMC 12 第 28 题

Question

在一个

4\times4\times3

的长方体上，顶点

A

、

B

和

C

都与顶点

D

相邻。从

D

到包含

A

、

B

、

C

的平面的垂直距离最接近

On a

4\times4\times3

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

1.6

1.9

2.1

2.7

2.9

答案：C

知识点：planes point-to-plane distance

难度评级：1990

小提示：

把

D

放在原点，并使三条相邻的棱分别沿三条坐标轴

Place

D

at the origin with the three adjacent edges along coordinate axes

大提示：

写出经过

(4,0,0),(0,4,0),(0,0,3)

的平面的截距式

Write the intercept form of the plane through

(4,0,0),(0,4,0),(0,0,3)

解答：

令

D=(0,0,0)

，并把三个相邻顶点取为

(4,0,0),(0,4,0),(0,0,3)

。它们所在的平面为

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

该平面到原点的距离为

\begin{aligned} \frac1{\sqrt{\begin{gathered} (\frac{1}{4})^2+(\frac{1}{4})^2\\ {}+(\frac{1}{3})^2 \end{gathered}}} &=\frac{12}{\sqrt{34}}\\ &\approx2.06 \end{aligned}\text{，}

最接近

2.1

。所以正确答案是 C。

Put

D=(0,0,0)

and take the adjacent vertices as

(4,0,0),(0,4,0),(0,0,3).

Their plane is

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

Its distance from the origin is

\begin{aligned} \frac1{\sqrt{\begin{gathered} (\frac{1}{4})^2+(\frac{1}{4})^2\\ {}+(\frac{1}{3})^2 \end{gathered}}} &=\frac{12}{\sqrt{34}}\\ &\approx2.06, \end{aligned}

closest to

2.1.

Thus the correct answer is C.

1996 AMC 12 第 28 题

答案：C

难度评级：1990

小提示：

大提示：

解答：

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