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向量xiàngliàng

vector
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更新于 2025-10-29
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Core Concept

A vector is a quantity that has both magnitude and direction. Geometrically, vectors are represented by directed line segments; algebraically, they are represented by coordinates.

Vector Notation

1. Geometric Notation

Directed line segment AB\overrightarrow{AB} where AA is the initial point and BB is the terminal point.

2. Letter Notation

Lowercase letters with arrows: a\vec{a}, b\vec{b}, c\vec{c}

3. Coordinate Notation

2D vector: a=(x,y)\vec{a} = (x, y) or a=(a1,a2)\vec{a} = (a_1, a_2)

3D vector: a=(x,y,z)\vec{a} = (x, y, z) or a=(a1,a2,a3)\vec{a} = (a_1, a_2, a_3)

Basic Concepts

Magnitude (Length)

For vector a=(x,y)\vec{a} = (x, y), magnitude is denoted a|\vec{a}| or a\|\vec{a}\|:

a=x2+y2|\vec{a}| = \sqrt{x^2 + y^2}

Zero Vector

Vector with magnitude 0, denoted 0=(0,0)\vec{0} = (0, 0), with arbitrary direction

Unit Vector

Vector with magnitude 1, denoted e\vec{e}

Unit vector in direction of a\vec{a}: e=aa\vec{e} = \frac{\vec{a}}{|\vec{a}|}

Equal Vectors

Vectors with same direction and magnitude

Opposite Vectors

Vectors with opposite directions but equal magnitudes. Opposite of a\vec{a} is a-\vec{a}

Vector Operations

1. Vector Addition

Geometric Meaning

Parallelogram rule or triangle rule

Coordinate Operation

a+b=(x1,y1)+(x2,y2)=(x1+x2,y1+y2)\vec{a} + \vec{b} = (x_1, y_1) + (x_2, y_2) = (x_1 + x_2, y_1 + y_2)

Properties

  • Commutative: a+b=b+a\vec{a} + \vec{b} = \vec{b} + \vec{a}
  • Associative: (a+b)+c=a+(b+c)(\vec{a} + \vec{b}) + \vec{c} = \vec{a} + (\vec{b} + \vec{c})

2. Vector Subtraction

ab=(x1,y1)(x2,y2)=(x1x2,y1y2)\vec{a} - \vec{b} = (x_1, y_1) - (x_2, y_2) = (x_1 - x_2, y_1 - y_2)

Geometric meaning: Points from terminal of b\vec{b} to terminal of a\vec{a}

3. Scalar Multiplication

ka=k(x,y)=(kx,ky)k\vec{a} = k(x, y) = (kx, ky)

where kk is a real number.

Properties:

  • k>0k > 0: Same direction as a\vec{a}, magnitude kak|\vec{a}|
  • k<0k < 0: Opposite direction to a\vec{a}, magnitude ka|k||\vec{a}|
  • k=0k = 0: Results in zero vector

4. Dot Product (Scalar Product)

ab=abcosθ\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta

where θ\theta is the angle between a\vec{a} and b\vec{b}.

Coordinate Form

ab=(x1,y1)(x2,y2)=x1x2+y1y2\vec{a} \cdot \vec{b} = (x_1, y_1) \cdot (x_2, y_2) = x_1x_2 + y_1y_2

Important Properties

  • ab=0ab\vec{a} \cdot \vec{b} = 0 \Leftrightarrow \vec{a} \perp \vec{b} (vectors perpendicular)
  • aa=a2\vec{a} \cdot \vec{a} = |\vec{a}|^2

Real-World Applications

Application 1: Displacement

Problem: Person walks 3km east then 4km north from origin. Find displacement magnitude and direction.

Solution: Displacement vector: s=(3,4)\vec{s} = (3, 4)

Magnitude: s=32+42=5 km|\vec{s}| = \sqrt{3^2 + 4^2} = 5 \text{ km}

Direction: Angle with east axis tanθ=43,θ53.1°\tan\theta = \frac{4}{3}, \quad \theta \approx 53.1°

Answer: 5km at 53.1° north of east

Application 2: Force Composition

Problem: Forces F1=(3,0)\vec{F_1} = (3, 0) N and F2=(0,4)\vec{F_2} = (0, 4) N act at same point. Find resultant.

Solution: F=F1+F2=(3,4) N\vec{F} = \vec{F_1} + \vec{F_2} = (3, 4) \text{ N}

Magnitude: F=32+42=5 N|\vec{F}| = \sqrt{3^2 + 4^2} = 5 \text{ N}

Application 3: Velocity Decomposition

Problem: Plane flies at 200 km/h toward northeast (45° from east). Find east and north components.

Solution: vx=200cos45°=1002141.4 km/hv_x = 200\cos45° = 100\sqrt{2} \approx 141.4 \text{ km/h} vy=200sin45°=1002141.4 km/hv_y = 200\sin45° = 100\sqrt{2} \approx 141.4 \text{ km/h}

CSCA Practice Problems

💡 Note: The following practice problems are designed based on the CSCA exam syllabus and Chinese standardized test formats to help students familiarize themselves with question types and problem-solving approaches.

Example 1: Basic (Difficulty ★★☆☆☆)

Given vector a=(3,4)\vec{a} = (3, 4), find a|\vec{a}|.

Options:

  • A. 3
  • B. 4
  • C. 5
  • D. 7

Solution: a=32+42=25=5|\vec{a}| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5

Answer: C


Example 2: Intermediate (Difficulty ★★★☆☆)

Given a=(2,1)\vec{a} = (2, 1) and b=(1,3)\vec{b} = (1, -3), find a+2b\vec{a} + 2\vec{b}.

Solution:

2b=2(1,3)=(2,6)2\vec{b} = 2(1, -3) = (2, -6) a+2b=(2,1)+(2,6)=(4,5)\vec{a} + 2\vec{b} = (2, 1) + (2, -6) = (4, -5)

Answer: (4,5)(4, -5)


Example 3: Advanced (Difficulty ★★★★☆)

Given a=(1,2)\vec{a} = (1, 2) and b=(2,1)\vec{b} = (2, -1), find:

  1. ab\vec{a} \cdot \vec{b}
  2. Angle θ\theta between a\vec{a} and b\vec{b}

Solution:

(1) Dot product: ab=1(2)+2(1)=0\vec{a} \cdot \vec{b} = 1(2) + 2(-1) = 0

(2) Angle: Since ab=0\vec{a} \cdot \vec{b} = 0: ab\vec{a} \perp \vec{b} θ=90°\theta = 90°

Answers: (1) 0 (2) 90° (perpendicular)

Common Mistakes

❌ Mistake 1: Equal vectors must have same initial point

Correction: Vectors are equal if they have same direction and magnitude, regardless of position. Vectors can be translated.

❌ Mistake 2: Magnitude can be negative

Correction: Magnitude (length) is always non-negative: a0|\vec{a}| \geq 0.

❌ Mistake 3: Dot product result is a vector

Correction: Dot product result is a scalar (number), not a vector.

❌ Mistake 4: Zero vector has no direction

Correction: Zero vector direction is arbitrary, can be parallel to any vector.

❌ Mistake 5: Wrong angle range

Correction: Angle θ\theta between two non-zero vectors ranges from [0°,180°][0°, 180°] or [0,π][0, \pi].

Study Tips

  1. Understand essence: Both magnitude and direction, different from scalars
  2. Master coordinate operations: Addition, subtraction, scalar multiplication, dot product
  3. Grasp geometric meaning: Visualize vector operations in coordinate system
  4. Remember perpendicularity: ab=0ab\vec{a} \cdot \vec{b} = 0 \Leftrightarrow \vec{a} \perp \vec{b}
  5. Apply to real world: Displacement, force, velocity are all vectors

💡 Exam Tip: Vectors are essential in CSCA geometry and physics. Coordinate operations and dot product are frequently tested!

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