8.2 Coordinate Operations of Vectors

8.2 Coordinate Operations of Vectors

Coordinate operations translate geometric vector concepts into algebraic manipulations. By choosing an orthogonal coordinate system, plane vector operations reduce to arithmetic on component pairs, transforming complex geometric proofs into linear algebraic computations.

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1. Cartesian Coordinate System and Vector Coordinates

1.1 Orthonormal Basis

In a Cartesian coordinate system $Oxy$, let $\vec{i}$ and $\vec{j}$ denote the unit vectors along the positive $x$-axis and positive $y$-axis, respectively:

• $|\vec{i}| = 1$, $|\vec{j}| = 1$, and $\vec{i} \perp \vec{j}$.
• The set ${\vec{i}, \vec{j}}$ forms a standard orthonormal basis for the plane.

1.2 Vector Coordinate Representation

By the Fundamental Theorem of Plane Vectors, any vector $\vec{a}$ in the plane can be uniquely expressed as: $\vec{a} = x\vec{i} + y\vec{j}$ The ordered pair of real numbers $(x, y)$ is called the coordinate representation of vector $\vec{a}$, written as $\vec{a} = (x, y)$.

• $x$ is the $x$-component (horizontal projection).
• $y$ is the $y$-component (vertical projection).
• The origin position vector $\vec{OP}$ corresponding to point $P(x, y)$ has coordinates $\vec{OP} = (x, y)$.

1.3 Segment Vector Coordinates

Given initial point $A(x_1, y_1)$ and terminal point $B(x_2, y_2)$, the vector $\vec{AB}$ is computed as terminal point coordinates minus initial point coordinates: $\vec{AB} = \vec{OB} - \vec{OA} = (x_2, y_2) - (x_1, y_1) = (x_2 - x_1, y_2 - y_1)$

Vector Magnitude in Coordinates

The magnitude of vector $\vec{a} = (x, y)$ is given by the Pythagorean formula: $|\vec{a}| = \sqrt{x^2 + y^2}$ For directed segment $\vec{AB}$ with endpoints $A(x_1, y_1)$ and $B(x_2, y_2)$: $|\vec{AB}| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

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2. Algebraic Operations in Coordinate Form

Let $\vec{a} = (x_1, y_1)$, $\vec{b} = (x_2, y_2)$, and scalar $\lambda \in \mathbb{R}$.

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3. Coordinate Criterion for Parallel (Collinear) Vectors

Theorem (Coordinate Parallelism Criterion): Let $\vec{a} = (x_1, y_1)$ and $\vec{b} = (x_2, y_2)$. The vectors $\vec{a}$ and $\vec{b}$ are parallel ((\vec{a} \parallel \vec{b})) if and only if: $x_1 y_2 - x_2 y_1 = 0$

Derivation:

1. If $\vec{a} = \vec{0} = (0,0)$, then $0 \cdot y_2 - x_2 \cdot 0 = 0$ holds trivially, and $\vec{0} \parallel \vec{b}$ is true.
2. If $\vec{a} \neq \vec{0}$, then $\vec{a} \parallel \vec{b} \iff \exists \lambda \in \mathbb{R}$ such that $\vec{b} = \lambda \vec{a}$.
3. In coordinates: $(x_2, y_2) = (\lambda x_1, \lambda y_1) \implies x_2 = \lambda x_1$ and $y_2 = \lambda y_1$.
4. Eliminating $\lambda$: $x_1 y_2 = x_1 (\lambda y_1) = \lambda x_1 y_1 = (\lambda x_1) y_1 = x_2 y_1 \implies x_1 y_2 - x_2 y_1 = 0$

Memory Trick (Cross Multiplication): Multiply the outer components and inner components: $\text{Outer} - \text{Inner} = x_1 y_2 - x_2 y_1 = 0$.

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4. Division of Line Segments and Centroid Formulas

4.1 Section Formula (Internal/External Division)

Let points $A(x_1, y_1)$ and $B(x_2, y_2)$ define a segment. If point $P(x, y)$ divides the directed segment $AB$ such that $\vec{AP} = \lambda \vec{PB}$ ($,\lambda \neq -1,$): $x = \frac{x_1 + \lambda x_2}{1 + \lambda}, \quad y = \frac{y_1 + \lambda y_2}{1 + \lambda}$

• If $\lambda > 0$, $P$ is an internal division point.
• If $-1 < \lambda < 0$ or $\lambda < -1$, $P$ is an external division point.

4.2 Midpoint Formula

When $\lambda = 1$, $P$ is the midpoint $M(x_M, y_M)$ of $AB$: $x_M = \frac{x_1 + x_2}{2}, \quad y_M = \frac{y_1 + y_2}{2}$ In vector form: $\vec{OM} = \frac{1}{2}(\vec{OA} + \vec{OB})$.

4.3 Centroid Formula of a Triangle

For $\triangle ABC$ with vertices $A(x_A, y_A)$, $B(x_B, y_B)$, $C(x_C, y_C)$, the centroid $G(x_G, y_G)$ satisfies: $x_G = \frac{x_A + x_B + x_C}{3}, \quad y_G = \frac{y_A + y_B + y_C}{3}$ Vector characteristic equation of centroid $G$: $\vec{GA} + \vec{GB} + \vec{GC} = \vec{0} \quad \iff \quad \vec{OG} = \frac{1}{3}(\vec{OA} + \vec{OB} + \vec{OC})$

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5. Worked Gaokao Exam Examples

Worked Example 1: Parallel Vectors and Parameters

Problem: Given vectors $\vec{a} = (1, m)$ and $\vec{b} = (m - 2, 3)$ where $m \in \mathbb{R}$.

1. Find the value(s) of $m$ such that $\vec{a} \parallel \vec{b}$.
2. For each such $m$, determine whether $\vec{a}$ and $\vec{b}$ have the same or opposite direction.

Solution:

1. Apply the coordinate parallelism criterion $x_1 y_2 - x_2 y_1 = 0$: $1 \cdot 3 - m(m - 2) = 0 \implies 3 - m^2 + 2m = 0 \implies m^2 - 2m - 3 = 0$
2. Factor the quadratic equation: $(m - 3)(m + 1) = 0 \implies m = 3 \quad \text{or} \quad m = -1$
3. Case 1: $m = 3$
   • $\vec{a} = (1, 3)$ and $\vec{b} = (3 - 2, 3) = (1, 3)$.
   • Here $\vec{b} = 1 \cdot \vec{a}$, so $\lambda = 1 > 0$. Vectors $\vec{a}$ and $\vec{b}$ are codirectional (same direction).
4. Case 2: $m = -1$
   • $\vec{a} = (1, -1)$ and $\vec{b} = (-1 - 2, 3) = (-3, 3)$.
   • Here $\vec{b} = -3 \cdot \vec{a}$, so $\lambda = -3 < 0$. Vectors $\vec{a}$ and $\vec{b}$ are antidirectional (opposite direction).

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