📚 Exercise 22G: Vectors - Lines and Planes Mastery | 练习22G：掌握向量–直线与平面
Welcome to this focused revision session on Exercise 22G, which covers the vector equations of lines and planes, intersections, angles, and distances – key topics in IB Mathematics Analysis and Approaches (AA) HL, frequently appearing in Paper 2. Mastering these concepts will not only help you tackle textbook exercises but also build the confidence to solve exam problems efficiently. We will break down every essential technique and connect them through clear examples.
欢迎来到本次针对练习22G的集中复习，内容涵盖直线与平面的向量方程、交点、夹角和距离——这些是IB数学分析与方法（AA）HL的关键主题，常出现在试卷二中。掌握这些概念不仅能帮你解决课本习题，还能让你高效应对考试难题。我们将逐一拆解每种重要技巧，并用清晰的例子把它们串联起来。
1. Understanding Vector Representation | 理解向量表示
In three-dimensional space, any point A with coordinates (x₀, y₀, z₀) can be described by a position vector a ion vectors are free vectors that show the orientation of a line or plane. For instance, a vector d line. All vector quantities in IB questions can be expressed in column or component form, and you must be comfortable converting between them.
在三维空间中，任意点A坐标为 (x₀, y₀, z₀) 可从原点用位置向量 a = (x₀, y₀, z₀) 表示。方向向量是自由向量，用于指示直线或平面的方向。例如，向量 d = (1, -2, 3) 与给定直线平行。在IB考题中所有向量都可以用分量或列形式表示，你必须能够熟练地在各种形式之间转换。
2. Vector Equation of a Line | 直线的向量方程
A line in 3D is determined by a fixed point A on the line and a direction vector d. The vector equation has the form r = a + λd, where λ is a scalar parameter. As λ varies, the tip of r traces every point on the line. If d = (a, b, c) and a is the position vector of (x₀, y₀, z₀), then any point on the line is given by (x₀ + λa, y₀ + λb, z₀ + λc).
三维空间中的直线由一个定点A和一个方向向量d确定。向量方程的形式为 r = a + λd，其中 λ 为标量参数。随着 λ 变化，向量 r 的末端描出直线上的所有点。若 d = (a, b, c) 且 a 是点 (x₀, y₀, z₀) 的位置向量，则直线上任一点可写为 (x₀ + λa, y₀ + λb, z₀ + λc)。
3. Parametric and Symmetric Forms | 参数式与对称式
etric form must be written accordingly, for example, x = x₀ with the remaining equalities.
由向量形式我们可立即写出参数方程：x = x₀ + λa, y = y₀ + λb, z = z₀ + λc。如果 a, b, c 都不为零，则可消去 λ 得到对称式（笛卡尔形式）：(x - x₀)/a = (y - y₀)/b = (z - z₀)/c。当某个分量为零时，直线平行于坐标平面，对称式必须相应地写为例如 x = x₀ 并保留其余等式。
4. Vector Equation of a Plane | 平面的向量方程
A plane is determined by a point A and two non-parallel direction vectors u and v. Its vector equation is r = a + su + tv, where s and t are scalar parameters. Any point on the plane can be reached by starting at A and moving along combinations of u and v. This parametric form is particularly useful when finding intersections or when constructing a plane from three non-collinear points.
平面由一个定点A和两个不平行的方向向量u、v确定。其向量方程为 r = a + su + tv，其中 s 和 t 是标量参数。平面上任何点都可以从A出发沿着 u 与 v 的线性组合得到。这个参数形式在求交线或通过三个不共线点构造平面时特别有用。
5. Scalar Product Form of a Plane | 平面的点积形式
The scalar product (or Cartesian) form of a plane uses a normal vector n perpendicular to the plane. If a is the position vector of a known point on the plane, then for any point r on the plane, r · n = a · n = d. In coordinates, if n = (p, q, r), we get px + qy + rz = d. To find n from the parametric form, we can take n = u × v, the cross product of the two direction vectors.
平面的点积形式（笛卡尔形式）使用垂直于该平面的法向量n。若 a 是平面上已知点的位置向量，则对于平面上任意点 r，有 r · n = a · n = d。在坐标下，设 n = (p, q, r)，即得到 px + qy + rz = d。从参数形式求法向量时，可取 n = u × v，即两个方向向量的叉积。
6. Intersection of Two Lines | 两直线的交点
To find the intersection of two lines, set their vector equations equal: a + λd₁ = b + μd₂. Equate the components to obtain three equations in λ and μ. Solve two equations and check the third for consistency. If consistent, you have a unique intersection point. If the direction vectors are parallel (one is a scalar multiple of the other), the lines are either coincident or parallel and distinct. If not parallel and no solution exists, the lines are skew.
要求两直线的交点，可令它们的向量方程相等：a + λd₁ = b + μd₂。对应分量相等得到三个关于 λ 和 μ 的方程。先解其中两个方程，再将结果代入第三个检验一致性。若一致，则存在唯一交点。如果方向向量平行（即一个为另一个的标量倍），则两直线要么重合，要么平行而不相交。若不平行且无解，则两直线为异面直线（歪斜）。
7. Intersection of a Line and a Plane | 直线与平面的交点
ne is parallel to the plane and never meets it. If it gives an identity (0 = 0), the line lies entirely in the plane.
当直线与平面相交时，将直线的参数式 (x = x₀ + λa, y = y₀ + λb, z = z₀ + λc) 代入平面方程 px + qy + rz = d。这样会得到一个关于 λ 的一元一次方程。解出 λ 后再求交点坐标。若方程化简为矛盾式（如 0 = 5），则直线与平面平行且不相交。若得到恒等式（0 = 0），则直线完全落于平面内。
8. Intersection of Two Planes | 两平面的交线
The intersection of two non-parallel planes is a straight line. To find its equation, you need a point on the line and its direction vector. The direction vector d is perpendicular to both normals, so d = n₁ × n₂. A point can be found by setting one coordinate to a convenient value (often zero) and solving the two plane equations simultaneously. The line can then be expressed in vector form r = a + λd.
两个不平行平面的交线是一条直线。要求其方程，需要线上一点和方向向量。方向向量 d 同时垂直于两平面法向量，因此 d = n₁ × n₂。求点时可令一个坐标取方便值（通常设为零），然后联立两个平面方程求解。之后即可将交线表示为向量形式 r = a + λd。
9. Angle Between Lines and Planes | 直线与平面间的夹角
e acute angle θ between two lines equals the angle between their direction vectors: cosθ = |d₁ · d₂|/(|d₁||d₂|). For the angle between a line and a plane, we use the normal vector n. The angle φ between the line and the normal is given by cosφ = |d · n|/(|d||n| The required acute angle α between the line and the plane satisfies α = 90° - φ, so sinα = |d · n|/(|d||n|).
两直线间的锐角 θ 等于它们方向向量间的夹角：cosθ = |d₁ · d₂|/(|d₁||d₂|)。对于直线与平面间的夹角，需利用法向量 n。设直线方向与法向量夹角为 φ，则 cosφ = |d · n|/(|d||n|)。所求直线与平面之间的锐角 α 满足 α = 90° - φ，因此 sinα = |d · n|/(|d||n|)。
10. Distance from a Point to a Line | 点到直线的距离
The shortest distance from a point P to a line r = a + λd is given by the formula d = |(AP × d)| / |d|, where A is any fixed point on the line and AP = p - a. The cross product captures the area of the parallelogram formed by AP and d, and dividing by the length of d yields the perpendicular height.
点P到直线 r = a + λd 的最短距离公式为 d = |(AP × d)| / |d|，其中A为直线上任一定点，且 AP = p - a。叉积的模等于由 AP 与 d 构成的平行四边形面积，除以 d 的长度即得到垂直距离。
11. Distance from a Point to a Plane | 点到平面的距离
For a plane with equation r · n = d or px + qy + rz = d, the perpendicular distance from a point P(x₁, y₁, z₁) to the plane is given by D = |(AP · n)| / |n|, where A is any point on the plane and AP = p - a. In coordinate form, this simplifies to D = |px₁ + qy₁ + rz₁ - d| / √p² + q² + r². This is an extremely high-frequency exam question.
对于方程为 r · n = d 或 px + qy + rz = d 的平面，点P(x₁, y₁, z₁) 到平面的垂直距离为 D = |(AP · n)| / |n|，其中A是平面上任一点，AP = p - a。在坐标形式下，该公式可简化为 D = |px₁ + qy₁ + rz₁ - d| / √p² + q² + r²。这是一个极其高频的考题。
12. Solving Typical Exam Questions | 解典型考题
IB exam problems often combine several of these concepts. For example, you might be given two lines and asked to determine whether they intersect, are parallel, or are skew; then find the shortest distance between them or the equation of a plane containing one line and parallel to the other. Another common task: given three points, find the equation of the plane, then find the angle it makes with another plane or a line. Practice with clear logical steps – always write down the known vectors, choose a parameter, and apply the appropriate formula.
IB考题常常综合以上多个概念。例如，给定两条直线，要求判断它们是相交、平行还是异面，然后求它们之间的最短距离，或求包含其中一条直线且平行于另一条的平面方程。另一个常见任务是：给定三个点，求平面方程，再求该平面与另一平面或直线的夹角。练习时一定要保持清晰的逻辑步骤——先写下已知向量，选取参数，然后套用正确的公式。
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