13.3 Spatial Coordinates and Spatial Vector Operations

13.3 Spatial Coordinates and Spatial Vector Operations

Spatial vectors convert geometric problems into algebraic computations. Introduced in the modern Gaokao syllabus, the spatial vector method provides an algorithmic, reliable approach to solving complex 3D geometry questions without relying on difficult synthetic auxiliary constructions.

---

1. Spatial Cartesian Coordinate System

Fix an origin point $O$ in space and three mutually perpendicular directed axes: the x-axis (abscissa), y-axis (ordinate), and z-axis (applicate), forming a right-handed Cartesian coordinate system $O-xyz$.

• Basis Vectors: Let $\boldsymbol{i}, \boldsymbol{j}, \boldsymbol{k}$ be unit vectors along the positive x, y, and z axes, respectively. Then $|\boldsymbol{i}| = |\boldsymbol{j}| = |\boldsymbol{k}| = 1$ and $\boldsymbol{i} \cdot \boldsymbol{j} = \boldsymbol{j} \cdot \boldsymbol{k} = \boldsymbol{k} \cdot \boldsymbol{i} = 0$.
• Point and Position Vector: For any point $P(x, y, z)$ in space, its position vector is: $\vec{OP} = x\boldsymbol{i} + y\boldsymbol{j} + z\boldsymbol{k} = (x, y, z)$
• Vector Between Two Points: Given $P_1(x_1, y_1, z_1)$ and $P_2(x_2, y_2, z_2)$: $\vec{P_1 P_2} = (x_2 - x_1, y_2 - y_1, z_2 - z_1)$
• Distance Formula: $d(P_1, P_2) = |\vec{P_1 P_2}| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$
• Midpoint Formula: The midpoint $M$ of segment $P_1P_2$ has coordinates: $M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right)$

---

2. Component Algebra of Spatial Vectors

Let $\boldsymbol{a} = (x_1, y_1, z_1)$ and $\boldsymbol{b} = (x_2, y_2, z_2)$ be two vectors in space, and $\lambda \in \mathbb{R}$:

1. Vector Addition & Subtraction: $\boldsymbol{a} \pm \boldsymbol{b} = (x_1 \pm x_2, y_1 \pm y_2, z_1 \pm z_2)$
2. Scalar Multiplication: $\lambda \boldsymbol{a} = (\lambda x_1, \lambda y_1, \lambda z_1)$
3. Dot Product (Scalar Product): $\boldsymbol{a} \cdot \boldsymbol{b} = |\boldsymbol{a}||\boldsymbol{b}| \cos \theta = x_1 x_2 + y_1 y_2 + z_1 z_2$
4. Vector Magnitude: $|\boldsymbol{a}| = \sqrt{\boldsymbol{a} \cdot \boldsymbol{a}} = \sqrt{x_1^2 + y_1^2 + z_1^2}$
5. Collinear (Parallel) Condition: $\boldsymbol{a} \parallel \boldsymbol{b} \iff \boldsymbol{a} = \lambda \boldsymbol{b} \iff \frac{x_1}{x_2} = \frac{y_1}{y_2} = \frac{z_1}{z_2} = \lambda \quad (\text{for } \boldsymbol{b} \neq \boldsymbol{0})$
6. Perpendicular Condition: $\boldsymbol{a} \perp \boldsymbol{b} \iff \boldsymbol{a} \cdot \boldsymbol{b} = 0 \iff x_1 x_2 + y_1 y_2 + z_1 z_2 = 0$

---

3. Direction Vectors and Normal Vectors

Direction Vector of a Line (直线的方向向量)

Any non-zero vector $\boldsymbol{v}$ parallel to a straight line $l$ is called a direction vector of $l$. If line $l$ passes through point $A(x_0, y_0, z_0)$, any point $P(x, y, z)$ on $l$ satisfies $\vec{AP} = t \boldsymbol{v}$ for $t \in \mathbb{R}$.

Normal Vector of a Plane (平面的法向量)

Any non-zero vector $\boldsymbol{n}$ perpendicular to a plane $\alpha$ is called a normal vector of $\alpha$. That is, for every vector $\vec{AB}$ in plane $\alpha$: $\boldsymbol{n} \cdot \vec{AB} = 0$

Standard System of Equations Algorithm for Normal Vector $\boldsymbol{n}$:

1. Choose two non-collinear vectors $\boldsymbol{u} = (x_u, y_u, z_u)$ and $\boldsymbol{w} = (x_w, y_w, z_w)$ lying in plane $\alpha$.
2. Set $\boldsymbol{n} = (x, y, z)$.
3. Set up the homogeneous linear system: $\begin{cases} \boldsymbol{n} \cdot \boldsymbol{u} = 0 \\ \boldsymbol{n} \cdot \boldsymbol{w} = 0 \end{cases} \implies \begin{cases} x_u x + y_u y + z_u z = 0 \\ x_w x + y_w y + z_w z = 0 \end{cases}$
4. Assign a convenient non-zero value to one free variable (e.g., $z = 1$ or $x = 1$) and solve for the remaining variables to get a simple integer-component normal vector $\boldsymbol{n}$.

---

4. Classic Gaokao Worked Example

Problem

In a right quadrangular pyramid $P-ABCD$, the base $ABCD$ is a square with side length $2$. The altitude of the pyramid is $PO = 2$, where $O$ is the center of the square base $ABCD$. Point $M$ is the midpoint of lateral edge $PC$.

1. Establish an appropriate spatial coordinate frame.
2. Calculate the coordinates of all vertices and point $M$.
3. Compute a normal vector $\boldsymbol{n}$ for plane $MBD$.

Solution

1. Establish Coordinate Frame: Since base $ABCD$ is a square centered at $O$, and $PO \perp$ base $ABCD$, take center $O$ as the coordinate origin $(0,0,0)$.

   • Let $x$-axis be parallel to $\vec{AB}$ (or along $OK$ where $K$ is midpoint of $BC$).
   • Let $y$-axis be along $OM$ (or parallel to $\vec{AD}$).
   • Let $z$-axis be along directed line segment $OP$.

2. Vertex Coordinates:

   • Base vertices: $A(-1, -1, 0)$, $B(1, -1, 0)$, $C(1, 1, 0)$, $D(-1, 1, 0)$.
   • Apex: $P(0, 0, 2)$.
   • Midpoint $M$ of $PC$: $M = \left( \frac{0+1}{2}, \frac{0+1}{2}, \frac{2+0}{2} \right) = \left( \frac{1}{2}, \frac{1}{2}, 1 \right)$

3. Finding Normal Vector of Plane $MBD$:

   • Identify two vectors in plane $MBD$: $\vec{BD}$ and $\vec{BM}$. $\vec{BD} = D - B = (-1-1, 1-(-1), 0-0) = (-2, 2, 0)$ $\vec{BM} = M - B = \left( \frac{1}{2}-1, \frac{1}{2}-(-1), 1-0 \right) = \left( -\frac{1}{2}, \frac{3}{2}, 1 \right)$
   • Let $\boldsymbol{n} = (x, y, z)$ be the normal vector of plane $MBD$. Set up equations: $\begin{cases} \boldsymbol{n} \cdot \vec{BD} = 0 \\ \boldsymbol{n} \cdot \vec{BM} = 0 \end{cases} \implies \begin{cases} -2x + 2y = 0 \\ -\frac{1}{2}x + \frac{3}{2}y + z = 0 \end{cases}$
   • From $-2x + 2y = 0$, we get $x = y$.
   • Substitute $x = y$ into second equation: $-\frac{1}{2}x + \frac{3}{2}x + z = 0 \implies x + z = 0 \implies z = -x$.
   • Assign $x = 1$. Then $y = 1$ and $z = -1$.
   • Conclusion: A normal vector for plane $MBD$ is $\boldsymbol{n} = (1, 1, -1)$.