Line and Plane Relationships in 3D | 直线与平面的位置关系

📚 Line and Plane Relationships in 3D | 直线与平面的位置关系

In IB Mathematics, understanding the relative positions of a line and a plane in three-dimensional space is essential for solving problems involving intersections, angles, and distances. This article breaks down the key concepts step by step, with both English and Chinese explanations.

在IB数学中，理解三维空间中直线与平面的相对位置，对于解决涉及交点、角度和距离的问题至关重要。本文将一步步拆解关键概念，并提供中英文双语解析。

1. Defining a Line and a Plane | 直线与平面的基本定义

A line in 3D is usually represented by a point \( \mathbf{a} \) on the line and a direction vector \( \mathbf{d} \), giving the parametric form: \( \mathbf{r} = \mathbf{a} + \lambda \mathbf{d} \). A plane is defined by a point \( \mathbf{p} \) on the plane and a normal vector \( \mathbf{n} \), with equation \( \mathbf{r} \cdot \mathbf{n} = \mathbf{p} \cdot \mathbf{n} \).

三维空间中的直线通常由一个点 \( \mathbf{a} \) 和一个方向向量 \( \mathbf{d} \) 表示，参数形式为 \( \mathbf{r} = \mathbf{a} + \lambda \mathbf{d} \)。平面则由其上一点 \( \mathbf{p} \) 和法向量 \( \mathbf{n} \) 定义，方程为 \( \mathbf{r} \cdot \mathbf{n} = \mathbf{p} \cdot \mathbf{n} \)。

Line: \( \mathbf{r} = \mathbf{a} + \lambda \mathbf{d} \)
Plane: \( \mathbf{r} \cdot \mathbf{n} = k \) (where \( k = \mathbf{p} \cdot \mathbf{n} \))

Here \( \lambda \) is a scalar parameter, and \( \mathbf{d} \) and \( \mathbf{n} \) are non-zero vectors.

其中 \( \lambda \) 为标量参数，\( \mathbf{d} \) 与 \( \mathbf{n} \) 均为非零向量。

2. The Three Possible Positions | 三种位置关系

Given a line and a plane in space, they can only be in one of three relative positions:

给定空间中的一条直线与一个平面，它们之间只可能存在三种相对位置：

The line lies entirely in the plane.
The line is parallel to the plane (no intersection).
The line intersects the plane at a single point.

直线完全在平面内。
直线与平面平行（无交点）。
直线与平面相交于唯一一点。

These can be distinguished by the dot product \( \mathbf{d} \cdot \mathbf{n} \) and by checking whether a point on the line satisfies the plane equation.

这些可以通过点积 \( \mathbf{d} \cdot \mathbf{n} \) 以及检查直线上的点是否满足平面方程来区分。

3. The Line Lies in the Plane | 直线在平面内

If the direction vector \( \mathbf{d} \) is perpendicular to the normal vector \( \mathbf{n} \) (i.e. \( \mathbf{d} \cdot \mathbf{n} = 0 \)), then the line is parallel to the plane. To check whether it lies inside the plane, substitute a known point on the line into the plane equation. If the equation holds, the line is contained in the plane.

如果方向向量 \( \mathbf{d} \) 与法向量 \( \mathbf{n} \) 垂直（即 \( \mathbf{d} \cdot \mathbf{n} = 0 \)），则直线与平面平行或共面。要判断直线是否在平面内，可将直线上的已知点代入平面方程。若方程成立，则直线在平面内。

Condition: \( \mathbf{d} \cdot \mathbf{n} = 0 \) and \( \mathbf{a} \cdot \mathbf{n} = k \)

条件：\( \mathbf{d} \cdot \mathbf{n} = 0 \) 且 \( \mathbf{a} \cdot \mathbf{n} = k \)

For example, the line \( \mathbf{r} = (1,2,0) + \lambda (1,0,1) \) and the plane \( z = 1 \) share no common points because the line’s z-coordinate is always 0, so it is parallel but not in the plane.

例如，直线 \( \mathbf{r} = (1,2,0) + \lambda (1,0,1) \) 与平面 \( z = 1 \) 无公共点，因为直线上 z 坐标恒为 0，因此它平行但不位于平面内。

4. Line Parallel to the Plane | 直线与平面平行

When \( \mathbf{d} \cdot \mathbf{n} = 0 \) but a point on the line does not satisfy the plane equation, the line is parallel to the plane and does not intersect it.

当 \( \mathbf{d} \cdot \mathbf{n} = 0 \) 但直线上的点不满足平面方程时，直线与平面平行且不相交。

Condition: \( \mathbf{d} \cdot \mathbf{n} = 0 \) and \( \mathbf{a} \cdot \mathbf{n} \neq k \)

条件：\( \mathbf{d} \cdot \mathbf{n} = 0 \) 且 \( \mathbf{a} \cdot \mathbf{n} \neq k \)

Visually, the line never touches the plane. The distance from the line to the plane is constant.

直观上，直线永不相交于平面。直线到平面的距离为常数。

5. Line Intersects the Plane at a Point | 直线与平面相交于一点

If \( \mathbf{d} \cdot \mathbf{n} \neq 0 \), the line is not parallel to the plane, so it must meet the plane at exactly one point.

若 \( \mathbf{d} \cdot \mathbf{n} \neq 0 \)，则直线不平行于平面，因此必定与平面相交于唯一一点。

To find the intersection point, substitute the line’s parametric equations into the plane equation and solve for \( \lambda \).

求交点时，将直线的参数方程代入平面方程，解出 \( \lambda \)。

Solving: \( (\mathbf{a} + \lambda \mathbf{d}) \cdot \mathbf{n} = k \)

求解：\( (\mathbf{a} + \lambda \mathbf{d}) \cdot \mathbf{n} = k \)

6. Line Perpendicular to the Plane | 直线与平面垂直

A line is perpendicular to a plane if its direction vector is parallel to the plane’s normal vector. In that case, \( \mathbf{d} = t \mathbf{n} \) for some non-zero scalar \( t \).

如果直线的方向向量与平面的法向量平行，则直线与平面垂直。此时存在非零标量 \( t \)，使得 \( \mathbf{d} = t \mathbf{n} \)。

Condition: \( \mathbf{d} \times \mathbf{n} = \mathbf{0} \) (or \( \mathbf{d} = t \mathbf{n} \))

条件：\( \mathbf{d} \times \mathbf{n} = \mathbf{0} \)（或 \( \mathbf{d} = t \mathbf{n} \)）

For example, the line \( \mathbf{r} = (0,0,0) + \lambda (0,0,1) \) is perpendicular to the plane \( z = 0 \), because the direction vector is parallel to the normal vector.

例如，直线 \( \mathbf{r} = (0,0,0) + \lambda (0,0,1) \) 垂直于平面 \( z = 0 \)，因为其方向向量与法向量平行。

7. Finding the Intersection Point | 求交点

Given a line \( \mathbf{r} = \mathbf{a} + \lambda \mathbf{d} \) and a plane \( \mathbf{r} \cdot \mathbf{n} = k \), substitute coordinates into the plane equation:

已知直线 \( \mathbf{r} = \mathbf{a} + \lambda \mathbf{d} \) 与平面 \( \mathbf{r} \cdot \mathbf{n} = k \)，将坐标代入平面方程：

\( (a_x + \lambda d_x)n_x + (a_y + \lambda d_y)n_y + (a_z + \lambda d_z)n_z = k \)

Solve for \( \lambda \), then plug it back into the line equation to get the coordinates of the intersection point.

解出 \( \lambda \) 后，代回直线方程得到交点坐标。

8. Angle Between a Line and a Plane | 直线与平面所成角

The angle \( \theta \) between a line and a plane is defined as the complement of the angle between the line’s direction vector \( \mathbf{d} \) and the plane’s normal vector \( \mathbf{n} \). The formula is:

直线与平面所成角 \( \theta \) 定义为直线方向向量 \( \mathbf{d} \) 与平面法向量 \( \mathbf{n} \) 夹角的余角。公式为：

\( \sin \theta = \frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}||\mathbf{n}|} \)

Here \( 0 \leq \theta \leq \frac{\pi}{2} \). If \( \mathbf{d} \cdot \mathbf{n} = 0 \), then \( \theta = 0 \), meaning the line lies in a plane parallel to the given plane.

这里 \( 0 \leq \theta \leq \frac{\pi}{2} \)。若 \( \mathbf{d} \cdot \mathbf{n} = 0 \)，则 \( \theta = 0 \)，表示直线平行于给定平面。

9. Distance from a Point to a Plane | 点到平面的距离

For a point \( P \) with position vector \( \mathbf{p} \) and a plane \( \mathbf{r} \cdot \mathbf{n} = k \), the perpendicular distance \( D \) is given by:

对于点 \( P \)（位置向量 \( \mathbf{p} \)）与平面 \( \mathbf{r} \cdot \mathbf{n} = k \)，垂直距离 \( D \) 为：

\( D = \frac{|\mathbf{p} \cdot \mathbf{n} – k|}{|\mathbf{n}|} \)

If the line is parallel to the plane, the distance between the line and the plane equals the distance from any point on the line to the plane.

当直线与平面平行时，直线到平面的距离等于直线上任意一点到平面的距离。

10. Worked Example | 例题精讲

Example: Determine the relationship between the line \( \mathbf{r} = (1,2,3) + \lambda (2,-1,1) \) and the plane \( x + 2y – z = 4 \).

例：判断直线 \( \mathbf{r} = (1,2,3) + \lambda (2,-1,1) \) 与平面 \( x + 2y – z = 4 \) 的位置关系。

Solution: Here \( \mathbf{d} = (2,-1,1) \) and \( \mathbf{n} = (1,2,-1) \). Compute \( \mathbf{d} \cdot \mathbf{n} = 2 \times 1 + (-1) \times 2 + 1 \times (-1) = 2 – 2 – 1 = -1 \neq 0 \). So the line intersects the plane at exactly one point.

解：这里 \( \mathbf{d} = (2,-1,1) \)，\( \mathbf{n} = (1,2,-1) \)。计算 \( \mathbf{d} \cdot \mathbf{n} = 2 \times 1 + (-1) \times 2 + 1 \times (-1) = 2 – 2 – 1 = -1 \neq 0 \)。因此直线与平面相交于唯一一点。

To find the intersection point, substitute \( x = 1 + 2\lambda, y = 2 – \lambda, z = 3 + \lambda \) into \( x + 2y – z = 4 \):

求交点：将 \( x = 1 + 2\lambda, y = 2 – \lambda, z = 3 + \lambda \) 代入 \( x + 2y – z = 4 \)：

\( (1 + 2\lambda) + 2(2 – \lambda) – (3 + \lambda) = 4 \)

Simplifying gives \( 1 + 2\lambda + 4 – 2\lambda – 3 – \lambda = 4 \), so \( 2 – \lambda = 4 \), hence \( \lambda = -2 \).

化简得 \( 1 + 2\lambda + 4 – 2\lambda – 3 – \lambda = 4 \)，即 \( 2 – \lambda = 4 \)，所以 \( \lambda = -2 \)。

The intersection point is \( (1 + 2(-2), 2 – (-2), 3 + (-2)) = (-3, 4, 1) \).

交点坐标为 \( (1 + 2(-2), 2 – (-2), 3 + (-2)) = (-3, 4, 1) \)。

11. Common Mistakes and Tips | 常见错误与技巧

Mistake: Assuming \( \mathbf{d} \cdot \mathbf{n} = 0 \) means the line is in the plane. Always check a point on the line.
错误：认为 \( \mathbf{d} \cdot \mathbf{n} = 0 \) 就意味着直线在平面内。务必检查直线上的一个点。
Mistake: Forgetting to take the absolute value when computing angles. The angle must be between 0 and 90 degrees.
错误：计算角度时忘记取绝对值。直线与平面所成角必须在0到90度之间。
Tip: When solving for intersection, always verify the obtained point satisfies the plane equation.
技巧：求解交点时，务必验证所得点是否满足平面方程。

12. Summary | 总结

The relative position between a line and a plane can be determined by evaluating \( \mathbf{d} \cdot \mathbf{n} \) and checking whether a point on the line lies on the plane. This leads to three cases: contained in the plane, parallel, or intersecting at a single point. Understanding these relationships is fundamental for solving many IB exam questions involving 3D geometry.

直线与平面的位置关系可通过计算 \( \mathbf{d} \cdot \mathbf{n} \) 并检查直线上的点是否在平面内来确定。这产生三种情况：在平面内、平行、或相交于一点。理解这些关系是解决IB考试中许多三维几何问题的关键。

Practise with past exam questions to become fluent in applying these steps.

多做历年真题，熟练运用上述步骤。

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Line and Plane Relationships in 3D | 直线与平面的位置关系