Ultra-detailed Study Notes on Vectors and Basic Vector Operations

Ultra-detailed study notes on vectors and basic vector operations 1. The core essence and learning significance of vectors Vectors are the bridge connecting geometry and algebra. Their dual properties of "having both magnitude and direction" make them a powerful tool for solving problems such as geometric proofs, physical modeling, and coordinate operations. This note is based on the textbook knowledge points, supplemented by in-depth analysis, error-prone point discrimination, and extended exercises, helping learners build a complete vector knowledge system from "understanding concepts" to "flexible application". The key points of learning are: Distinguish the essential differences between vectors and quantities; Master the geometric meanings and algebraic rules of operations; Skilled use of theorems to solve core problems such as collinearity and decomposition. ② ① ③ Chapter 1 Core Concepts of Vectors (Laying a Solid Foundation) 1.1 Definition and Representation of Vectors Definition: A quantity that has both magnitude and direction (different from a scalar quantity that only has magnitude, such as length, temperature). Representation method: → → → Letter representation: a, b, c, ⋯ (an arrow needs to be added when handwritten, bold in printed form); → → − − Representation by a directed line segment: (starting point is A , ending point is B ), where the direction of AB is from A to B , and the magnitude is − the length of the line segment AB (i.e., the modulus ). Key reminder: The essence of a vector is "direction + magnitude", which has nothing to do with the starting point position (free vector), and only depends on its relative position. For example, a translated vector is equal to the original vector. 1.2 Five Special Vectors (High-frequency Test Points + Error-prone Discrimination) Vector Type Definition Core Properties Warning of Error-prone Points Zero Vector Vector with length 0 Arbitrary direction; ② Parallel to any vector; ③ → → → + = a ; ④ → The arrow cannot be omitted (to distinguish from the number 0); Unit vector Vector with a modulus of 1 unit length → −0 = 0 ② If → → a ∥ b → → → → → → and b ∥ c , when b = 0 , a and c are not necessarily parallel ∣∣∣ ①Module → e = 1 ; ②The unit vector in the same direction as any non-zero vector → a is → a → a The directions of unit vectors are not necessarily the same, so the statement "all unit vectors are equal" is incorrect Parallel vectors (collinear vectors) Non-zero vectors with the same or opposite directions ①Can be translated to the same straight line; ②The zero vector is parallel to any vector; ③Denoted as ①Difference from parallel lines: Parallel lines do not include coincidence, while vector parallelism includes collinearity (coincidence); ②The necessary → → a ∥ b → → and sufficient condition for non-zero vectors to be collinear is that there exists a real number λ such that b = λa Equal vectors Vectors with equal lengths and the same direction ① → → → → a = b ⇔ a = b and the same direction; ② Can coincide after translation; ③ Has transitivity. magnitudes or just the same direction does not constitute equal vectors. For example, equal vectors → a and → −a →→ → → → → (a = b, b = c ⇒ a = c) Just having equal have equal magnitudes but opposite directions and are not Opposite vectors Vectors with equal lengths and opposite directions ① The opposite vector of → a → is −a ; ② → → → a + (−a) = 0; ③ → → → → a = −b ⇔ b = −a The opposite vector of the zero vector is still the zero vector; opposite vectors must be parallel, but parallel vectors are not necessarily opposite vectors 1.3 Concept Discrimination Special Exercise (with Answers) True or False Questions (Mark √ for correct, × for wrong and give reasons) The zero vector has no direction ( ) Explanation: × , the direction of the zero vector is arbitrary, not without direction. If → a = → b → → , then a = b () Analysis: For × , equal modulus only satisfies the size condition, and the directions may be different (such as opposite vectors). → → → → → → If a ∥ b and b ∥ c , then a ∥ c () → → → → Analysis: × , when b = 0 , a and c may not be collinear (the zero vector is parallel to any vector, but any two vectors are not necessarily parallel). A vector is a directed line segment () Analysis: × , a directed line segment is the representation form of a vector. A vector is an abstract "direction + magnitude" and can exist independently of the directed line segment. If two equal vectors share the same starting point, then their endpoints must be the same. ( ) Analysis: Equal vectors have the same direction and magnitude. When their starting points are the same, their endpoints must coincide. Chapter 2 Basic Operations of Vectors (Operation Rules + Geometric Meanings + Question Types) 2.1 Vector Addition (Triangle Rule + Parallelogram Rule) 1. Definition → → The operation of finding the sum of two vectors, denoted as a + b . 2. Operation Rules (Core Key Points) Rule Applicable conditions Drawing steps Geometric meaning Triangle rule Any two vectors (no requirement for starting point) ① Translate the starting point of → → point of b is b → to the ending point of a ; → → a + b ② The vector from the starting point of → a to the ending (end-to-end connection, from starting point to ending point) → → → − − − Such as AB + BC = AC Parallelogram Law Two vectors have the same starting point ①Construct a parallelogram with → → a, b as the adjacent sides; ②The diagonal vector with the same starting point as → → a, b − − → → − → → → a + b OA + OB = OC is (same starting point, diagonal), such as (O is the vertex of the parallelogram) 3. Operational properties → → → → Commutative law: a + b = b + a ; → → → → → → → → Associative law: (a + b) + c = a + (b + c) ; → → → Properties of the zero vector: 0 + a = a + 0 = a , → → → Properties of opposite vectors: a + (−a) = 0 . 2.2 Vector subtraction (Inverse operation of addition) 1. Definition → → → → → → The vector a plus the opposite vector of b , denoted as a − b = a + (−b) . 2. Rules of Operation (Key Breakthrough) Rule Steps of Drawing Core Formula Triangle Rule (Common Starting Point) ① Place → → a, b with a common starting point; ② The vector from the end point of → b → → → to the end point of a is a − b The starting points are the same. The end point of the subtracted vector → is the end point of the vector being subtracted. ∣ Parallelogram Law (common starting point) ①Construct a parallelogram with → a → and b as adjacent sides; → ②The diagonal vector from the starting point of → b to the ending point of a (the non-common starting point diagonal) has the same starting point. For the other diagonal, the subtracting vector points to the vector being subtracted. 3. Important Conclusions − − → → − → AB = OB − OA The modulus of (ending point minus starting point, the basis of coordinate operations); → → a − b → → → → → → a − b ≤ ||a |+|b ∣ satisfies: ||a |−|b || ≤ (Triangle Inequality). 2.3 Product of Real Numbers and Vectors (Scalar Multiplication) 1. Definition → → The product of the real number λ and the vector a is a vector, denoted as λa , which has both magnitude and direction. 2. Provisions for Magnitude and Direction Length: → → λa = |λ| ⋅ a → (the product of the absolute value of λ and the modulus of a ); Direction: → → When λ > 0 , the direction of λa is the same as that of a ; → → When λ < 0 , the direction of λa = 0 is arbitrary. → → When λ = 0 , the direction of λa = 0 is arbitrary. 3. Operational properties (three major laws) → → Commutative law: λa = aλ ; → → Associative law: λ (μa) = (λμ)a ; → → → → → → → Distributive law: (λ + μ)a = λa + μa, λ (a + b) = λa + λb . 4. Core applications: Collinear vector theorem (high-frequency examination point) → → → → Theorem: The vector b is collinear with the non-zero vector a ⇔ there exists and only exists a real number λ such that b = λa . ① → → ② → Error-prone reminder: a must be a non-zero vector (if a = 0 , then λ is not unique); This theorem is the core basis for judging the collinearity of three points and vector parallelism. 2.4 Special Examples of Operations (Including Problem-solving Ideas) Example 1: Geometric Operations of Vector Addition and Subtraction → → → − − − → − − → → → → → ABCD AB = a, AD = b AC, BD, BC a, b In the parallelogram , given , express in terms of . Problem-solving idea: Use the parallelogram rule and triangle rule, combined with vector relationships for transformation. Solution: − − − → → → → → AC = AB + AD = a + b (Parallelogram Law, common starting point and diagonal); → → → → − − − − − − → → → → BD = AD − AB = b − a BD = AD − AB (Triangle Law, , when subtracting vectors with a common starting point, the end point of the subtracted vector points to the end point of the subtracting vector); − → − → → BC = AD = b (The opposite sides of a parallelogram are equal and in the same direction.) Example 2: Scalar Multiplication Operation and Collinearity Problem − − − → → → → → → → → → → → e1 , e2 AB = 2e1 + ke2 , CB = e1 + 3e2 , CD = 2e1 − e2 It is known that are non - zero vectors that are not collinear, , if A, B, Points D are collinear. Find the value of k . − → − − → → Solution idea: If three points are collinear, ⇔ the corresponding vectors are collinear, that is, AB = λBD . First, find BD , and then use the collinearity theorem to set up an equation. Solution: → → − − − − → → → → → → → → BD : BD = CD − CB = (2e1 − e2 ) − (e1 + 3e2 ) = e1 − 4e2 Find ; 、 、 → − − → → → → → A B D AB = λBD 2e1 + ke2 = λ (e1 − 4e2 ) Since → → 1 , e2 Because e are collinear, so , that is ; are not collinear, we get that the coefficients are corresponding equal: { 2 = λ , and the solution is λ = 2, k = −8 . k = −4λ Example 3: Comprehensive Operations and the Properties of the Centroid − − − → → → → Given that G is the centroid of △ABC , prove that GA + GB + GC = 0 . − − → → Idea of solving the problem: Use the parallelogram rule to transform GB + GC , combined with the property of the centroid (the distance from the centroid to the vertex is twice the distance to the midpoint of the opposite side). Proof: → → → → → − − − − − Take GB, GC as adjacent sides to construct parallelogram GBEC , then the diagonal is GE = GB + GC , and G is the intersection point of → → → → − − − − the parallelogram diagonals. Therefore, GD = 1 2 GE (D is the midpoint of BC), that is, GE = 2GD ; − − − − − − → → → → → → Since G is the centroid of △ABC , so AG = 2GD , and AG is in the opposite direction of GD , that is GA = −2GD ; − − − − − → → → → → → Substitute to get: GA + GB + GC = −2GD + 2GD = 0 . 2.5 Operational Expansion Exercises (Progressive Layers) Basic Questions − − − − − − → → → → → → Simplify: AB + BC + CA; OA − OB + AB . − → → Answer: 0; OA . → →→ → → → → → → Given a = 3e, b = −2e (where e is a non-zero vector), determine whether a and b are collinear. If they are collinear, find λ such that a = λb . Answer: Collinear, λ = − 3 2 . Advanced problem As shown in the figure, in parallelogram ABCD, the diagonals OD and AB intersect at point C. There is a point M on BC such that BC = 3BM, − − − − → → → → − → → → →→ N CD = 3CN OA = a, OB = b OM , ON , M N a, b and there is a point on CD such that . Let , express in terms of . − − → → − → → → → → → → Answer: OM = 5 6 b + 1 6 a; ON = 2 3 a + 2 3 b; M N = 1 2 a − 1 6 b . Difficult problem − − − − − → → → → → Let OA, OB be non - collinear, and point P lies on the line AB . Prove that OP = λOA + μOB and λ + μ = 1 (λ, μ ∈ R) . − − → → Solution idea: Use the collinear relationship of vectors on the line to express AP as a real - multiple of AB , and then transform it into a linear − − → → combination of OA, OB through vector operations. Proof: − − − − → → → → Since point P lies on the line AB , AP and AB are collinear. There exists a real number t ∈ R such that AP = tAB ; − − − − − − → → → → → → By the vector subtraction rule, AP = OP − OA, AB = OB − OA ; − − − − → → → → Substituting gives: OP − OA = t (OB − OA) ; − − − → → → After rearrangement, we get: OP = (1 − t)OA + tOB ; − → − − → → Let λ = 1 − t, μ = t , then OP = λOA + μOB , and λ + μ = (1 − t) + t = 1 , proven. Chapter 3 The Fundamental Theorem of Plane Vectors (Core of Vector Decomposition) 3.1 Theorem Content → → − − → e1 , e2 a If are two non-collinear vectors in the same plane, then for any vector in this plane, − − → → → λ1, λ2 a = λ1e1 + λ2e2 there exists one and only one pair of real numbers such that . − − → → {e1 , e2 } Core concept: is called a set of bases for all vectors in the plane. 3.2 Key Points of Theorems Two Core Conditions for Bases: − − → → e1 , e2 Non-collinearity: must not be collinear (if collinear, it can only represent vectors on the same straight line and cannot cover the entire plane); Non-zero property: The base vectors must be non-zero vectors (zero vectors are collinear with any vector and do not meet the non-collinearity requirement). − − − − → → → → → {e1 , e2 } a = λ1e1 + λ2e2 Uniqueness of decomposition: For a given basis → a is unique, that is, λ 1, λ2 , the decomposition formula of any vector is uniquely determined. − − − − − − → → → → → → λ1e1 + λ2e2 = μ1e1 + μ2e2 e1 , e2 Unique application: If , and λ = μ and λ = μ (the coefficients correspond equally). 1 1 2 are non - collinear, then 2 The essential meaning of the theorem: It transforms any complex vector in the plane into a "linear combination" of two non - collinear vectors, achieving the "standardized decomposition" of vectors, and laying a theoretical foundation for →→ subsequent coordinate operations (with i, j as the standard basis). 3.3 Basis Selection and Vector Decomposition Questions (High-frequency Test Points) Question Type 1: Judgment of Basis Problem-solving idea: To determine whether a set of vectors is a basis, only two points need to be verified: ① Non-zero vectors; ② Non-collinear. Example 4: Among the following groups of vectors, which can be used as a basis in the plane? () →→ → − − e1 = 0, e2 = (1, −2) A. − − → → e1 = (1, 2), e2 = (2, 4) B. − − → → e1 = (1, −2), e2 = (2, 1) C. − − → → e1 = (3, 5), e2 = (6, 10) D. Analysis: → − e1 A: is a zero vector and is collinear with any vector, so it is excluded; − − → → e2 = 2e1 B: , the two vectors are collinear, excluded; → C: Assume e 1 → D: e 2 → = 2e1 → = λe2 , then { 1 = 2λ , no solution, so they are not collinear and can be selected; −2 = λ , the two vectors are collinear, excluded; Answer: C. Type 2: Given the basis to represent unknown vectors Problem-solving idea: Use vector addition, subtraction, and scalar multiplication operations, combined with the midpoint and dividing point relationships in the figure, to transform the unknown vector into a linear combination of the basis. → → → − − − → → → → △ABC D BC AB = e1 , AC = e2 AD e1 , e2 Example 5: In , is the midpoint of side , , express in terms of . Solution: → → − − → − By the triangle rule of vector addition, AD = AB + BD ; → − − → Since D is the midpoint of BC , then BD = 1 2 BC ; → → → − − − − → → → → → 1 BC = AC − AB = e2 − e1 BD = (e2 − e1 ) Also , so ; 2 − → → → → → → 1 1 1 AD = e1 + (e2 − e1 ) = e1 + e2 Substituting gives: 2 2 . 2 − − − → → → Conclusion: The vector formula for the median of a triangle is AD = 1 2 (AB + AC) (which can be directly memorized and applied). − − → → − → → → Example 6: In parallelogram ABCD , E is the midpoint of CD , F is the midpoint of AE , express BF in terms of AB = a, AD = b . Solution: − − − − − − → → → → → → − − → → → → → 1 1 1 1 AE : AE = AD + DE DE = CD = AB = a AE = b + a First, express . Since E is the midpoint of CD, 2 2 2 , so 2 ; − − → → → → → → 1 1 1 1 1 F AE AF = AE = (b + a) = a + b is the midpoint of , so 2 2 2 4 2 ; − − → → → − → → → → → 1 1 3 1 BF = BA + AF = −a + ( a + b) = − a + b By vector subtraction, 4 2 4 2 . 3.4 Extended Exercises on the Fundamental Theorem of Plane Vectors Basic Questions → → − → → → →→ → → → → → → → e1 , e2 a = 3e1 − 2e2 , b = −e1 + 4e2 a + b, a − 2b e1 , e2 Let be non - collinear vectors, , find expressed in terms of . Solution: → → → → → → − → → a + b = (3e1 − 2e2 ) + (−e1 + 4e2 ) = 2e1 + 2e2 ; → → → → → → → → → → → → a − 2b = (3e1 − 2e2 ) − 2 (−e1 + 4e2 ) = 3e1 − 2e2 + 2e1 − 8e2 = 5e1 − 10e2 → → → Answer: a + b = 2e 1 Advanced problem → → → → → + 2e2 ; a − 2b = 5e1 − 10e2 . 。 − − − − − → → → → → In △ABC , AN = 1 3 AC , P is a point on BN. If AP 2 = λAB + 9 AC , find the value of λ . Solution: − − − → → → − − − − → → → → Let BP = tBN (t ∈ R) , then AP ; = AB + BP = AB + tBN − − − − − − − − − − → → → → → → → → → → − → Because of BN = AN − AB = 1 3 AC − AB , so AP = AB + t ( 1 3 AC − AB) = (1 − t)AB + t 3 AC ; − − − → → → Compare with the known AP So λ = 1 − t = 1 − Answer: λ = 1 3 2 3 = 1 3 = λAB + 2 9 AC , we get t 3 2 = 9 , and solve to get t = 2 3 ; . . Difficult problem → − − − − − − → → → →→ → → → → e1 , e2 a = e1 + 2e2 , b = 2e1 + ke2 a, b k Given that are not collinear, , if can be used as a basis, find the range of values of . Solution: → → The condition for being able to be used as a basis is that a, b are not collinear; → → − − − − → → → → → → a, b λ b = λa 2e1 + ke2 = λ (e1 + 2e2 ) Suppose are collinear, then there exists a real number such that , that is, ; 2 = λ − − → → e1 , e2 { k = 4 k = 2λ Since are not collinear, we have , and solving it gives ; → → Therefore, when a, b are not collinear, k ≠ 4 . Answer: k ∈ (−∞, 4) ∪ (4, +∞) . Chapter 4 Coordinate Operations of Plane Vectors (Core of Algebraization) 4.1 Coordinate Representation of Plane Vectors 1. Coordinate Definition →→ In the plane rectangular coordinate system, take the unit vectors i, j in the positive directions of the x - axis and the y - axis (referred to as the standard basis). For any vector → → a , according to the fundamental theorem of plane vectors, there exists a → → → unique pair of real numbers (x, y) such that a = xi + yj . Then (x, y) is called the coordinate of the vector a , denoted as → a = (x, y) . → → → Special vector coordinates: 0 = (0, 0); i = (1, 0); j = (0, 1) . 2. Coordinate formula for a directed line segment → − If the starting point of the directed line segment AB is A (x 1, y1) and the ending point is B (x 2, y2) , then − → AB = (x2 − x1, y2 − y1) (ending point coordinates - starting point coordinates). → Inference: The magnitude of vector a = (x, y) is → 2 2 a = √x + y 4.2 Rules of Coordinate Operations (Memorize Key Points) → Let a = (x 1, → y1), b = (x2, y2), λ ∈ R Type of Operation , then: (derived from the Pythagorean theorem). Coordinate formula Explanation Addition � + � = (�1 + �2 , �1 + �2 ) Add corresponding coordinates Subtraction � − � = (�1 − �2 , �1 − �2 ) Subtract corresponding coordinates Scalar multiplication �� = (��1 , ��1 ) Multiply each coordinate by a real number Equal → → a = b ⇔ x1 = x2 and y = y The coordinates are exactly the same 1 2 Collinear (parallel) � ∥ � ⇔ �1 �2 − �2 �1 = 0 Cross - multiply to get equality (avoid a zero denominator) Magnitude |�| = �21 + �21 Square root of the sum of the squares of the coordinates Warning about error - prone points Difference between vector coordinates and point coordinates: The point A (x, y) represents a → position, while the vector a = (x, y) represents direction and magnitude, independent of the starting point; → → Collinear coordinate condition: The coordinate condition for a ∥ b is x y = (the latter requires x , y ≠ 0 , and the former is more general). 1 x1 y1 x2 y2 2 2 = x2y1 , not 2 Analysis of Coordinate Operation Questions Example 7: Basic Coordinate Operations → → → →→ → → → Given a = (2, −1), b = (−3, 4) , find the coordinates of a + b, a − b, 3a − 2b and Solution: � + � = (2 + ( − 3), − 1 + 4) = ( − 1,3)； → → a − b = (2 − (−3), −1 − 4) = (5, −5); → → 3a − 2b = 3 (2, −1) − 2 (−3, 4) = (6 + 6, −3 − 8) = (12, −11); → a . ∣∣ → 2 2 a = √ 2 + (−1) = √ 5 → → 。 → → → → Answer: a + b = (−1, 3); a − b = (5, −5); 3a − 2b = (12, −11); → a = √5 . Example 8: Solving the collinear problem using coordinates → → → → Given a = (1, 2), b = (m, 4) , and a ∥ b , find the value of m . Solution idea: Use the collinear coordinate condition x 1y2 → − x2y1 = 0 . → Solution: From a ∥ b , we get 1 × 4 − 2 × m = 0 , and the solution is m = 2 . Answer: m = 2 . Example 9: Combining vector coordinates with geometric figures (judging the shape of a triangle) Given A (1, 2), B (3, 4), C (5, −2) , judge the shape of △ABC . Solution: − − − → → → Find the vector coordinates: AB = (3 − 1, 4 − 2) = (2, 2), AC 4) = (2, −6) ; = (5 − 1, −2 − 2) = (4, −4), BC = (5 − 3, −2− − − − − → → → → Verify the perpendicular relationship: AB ⋅ AC = 2 × 4 + 2 × (−4) = 8 − 8 = 0 , so AB⊥AC ; → → − − Calculate the modulus: AB 2 2 = √ 2 + 2 = 2√ 2, AC 2 2 = √ 4 + (−4) = 4√ 2 , the moduli are not equal; Conclusion: △ABC is a right triangle (non - isosceles). 4.4 Extended Exercises on Coordinate Operations Basic Questions → → Given a = (3, −2), b = (−1, 5) , find: → → (1) a + 2b ; (2) → → a − b → → → ; (3) If c = (x, 1) and a ∥ c , find x . Solution: → → (1) a + 2b = (3, −2) + 2 (−1, 5) = (3 − 2, −2 + 10) = (1, 8) ; （2） → → → → 2 2 a − b = (3 + 1, −2 − 5) = (4, −7), a − b = √ 4 + (−7) = √65 → → (3) From a ∥ c , we get 3 × 1 − (−2)x = 0 , and the solution is x = − Answer: (1) (1, 8) ; (2) √65 ; (3) x = − 3 2 3 2 ； . . Advanced Problem Given A (2, 1), B (0, 3), C (−1, 4) , prove that the three points A 、 、 B C are collinear. Proof: − − → → Find the vector coordinates: AB = (0 − 2, 3 − 1) = (−2, 2), AC = (−1 − 2, 4 − 1) = (−3, 3) ; − − → → Verify the collinearity condition: (−2) × 3 − 2 × (−3) = −6 + 6 = 0 , so AB ∥ AC ; Since A is a common point, the three points A, B, and C are collinear. Difficult problem ∣ ∣∣ → → → − − − Given the point A (1, 0), B (0, 1), C (2, 5) , if AC = λAB + μBC , find the value of λ, μ . Solution: − − − → → → Calculate the vector coordinates: AB = (0 − 1, 1 − 0) = (−1, 1), BC = (2 − 0, 5 − 1) = (2, 4), AC = (2 − 1, 5−) 0) = (1, 5) ; − − − → → → From AC , we get (1, 5) = λ (−1, 1) + μ (2, 4) = (−λ + 2μ, λ + 4μ) ; = λAB + μBC Set up the system of equations: { −λ + 2μ = 1 . Adding the two equations gives 6μ = 6 ⇒ μ = 1 , and substituting it into the first equation gives λ = 1 . λ + 4μ = 5 Answer: λ = 1, μ = 1 . Chapter 5 The Scalar Product of Plane Vectors (Core Application) 5.1 Definition of the Dot Product (Geometric Meaning + Algebraic Expression) 1. Geometric Definition → → → → → Given two non-zero vectors a and b , and the angle between them is θ (0 ≤ θ ≤ π) , then the dot product (inner product) of a and b is: a⋅ → → → b = a b cos θ . Special Cases: → → When a⊥b , θ = 90 ∘ , cos θ = 0 → → , so a ⋅ b = 0 ; → → When a and b are in the same direction, θ = 0 ∘ , cos θ = 1 → → When a and b are in the opposite direction, θ = 180 → → → → When a = b , a ⋅ a = → a 2 (denoted as → a 2 →2 = a → → , so a ⋅ b = → a → → b ; → , so a ⋅ b = − ∘ , cos θ = −1 → a → b ; ). 2. Geometric meaning → → a ⋅ b is equal to the product of → the direction of b ( → a cos θ → a → → and the projection of b in the direction of a ( → b cos θ ), or the product of → b → and the projection of a in ). → → Projection formula: The projection of b in the direction of a is → b cos θ = → → a⋅b → a . 3. Warnings about error-prone points The result of the dot product is a real number, not a vector (different from addition, subtraction, and scalar multiplication operations); → → Definition of the included angle θ : Place a, b with the same starting point, and the smallest positive angle formed (range 0 ≤ θ ≤ π) ); → → → → The dot product of the zero vector and any vector is 0 ( 0 ⋅ a = 0 ), but it cannot be deduced that a⊥0 (the direction of the zero vector is arbitrary, and perpendicularity is meaningless). 5.2 Operational properties of the dot product (core formulas) → → → Let a, b, c be a non-zero vector, λ ∈ R , then: → → → → Commutative law: a ⋅ b = b ⋅ a ; → → → → → → Associative law of scalar multiplication: (λa) ⋅ b = λ (a ⋅ b) = a ⋅ (λb) ; → → → → → → → Distributive law: (a + b) ⋅ c = a ⋅ c + b ⋅ c ; Modulus formula: → √→ → a = a ⋅ a ; ∣∣ ∣ Angle formula: cos θ = → → a⋅b 为 的夹角 →→ a, b (θ →→ a b → → ) ; →→ Perpendicular condition: a⊥b ⇔ a ⋅ b = 0 (non-zero vectors); Inequality relation: →→ → → a ⋅ b ≤ a b → → (the equal sign holds if and only if a ∥ b ). Common misunderstandings → → → → → → → → Does not satisfy the associative law: (a ⋅ b) ⋅ c ≠ a ⋅ (b ⋅ c) (the left side is consistent with c , and the right side is consistent with a ); →→ → → → → → → The cancellation law does not hold: If a ⋅ b = a ⋅ c and a ≠ 0 , it cannot be deduced that b = c . 5.3 Coordinate Operations of the Dot Product → Let a = (x → 1, , then: y 1 ), b = (x 2 , y 2 ) →→ Dot product formula: a ⋅ b = x Magnitude formula: 1 x2 + y1 y2 → 2 2 a = √x + y 1 ; 1 Angle formula: cos θ = ; 2 2 2 2 1 1 2 2 √x +y ⋅√x +y → → Vertical condition: a⊥b ⇔ x ; x 1 x 2 +y 1 y 2 1 x2 + y1 y2 = 0 ; → → Projection formula: The projection of b in the direction of a is x 1 x 2 +y 1 y 2 2 2 1 1 √ x +y 5.4 Analysis of Special Types of Dot Product Questions Type 1: Basic Calculations (Geometric Method + Coordinate Method) → → → a = 3, b = 4, a Example 10: Given the included angle θ = 60 between ∘ → → → → → → → and b , find a ⋅ b, (a + 2b) ⋅ (a − b) . Solution (geometric method): →→ ∘ a ⋅ b = 3 × 4 × cos 60 = 12 × 1 = 6; 2 → → → → → →2 →→ 2 (a + 2b) ⋅ (a − b) = a + a ⋅ b − 2b = 9 + 6 − 2 × 16 = −17 →→ → → → 。 → Answer: a ⋅ b = 6; (a + 2b) ⋅ (a − b) = −17 . → → →→ → → → a + b ,a Example 11: Given a = (2, −1), b = (−3, 4) , find the angle θ between a ⋅ b, Solution (coordinate method): →→ a ⋅ b = 2 × (−3) + (−1) × 4 = −10 → → → → a + b = (−1, 3), a + b = √ (−1) → → a = √ 5, b = 5, cos θ = →→ Answer: a ⋅ b = −10; −10 √ 5×5 2 ; + 3 = − 2 2√ 5 5 = √ 10; , so θ = arccos (− → → a + b = √ 10; θ = arccos (− Question type 2: Solving vertical problems 2√ 5 5 ) . 2√ 5 5 ) . → and b . . ∣∣ ∣ → → → → Example 12: Given a = (1, 2), b = (k, −1) , if a⊥b , find the value of k . Solution: From the perpendicular condition x 1x2 , we get 1 × k + 2 × (−1) = 0 ⇒ k = 2 . + y1y2 = 0 Answer: k = 2 . Question Type 3: Angle and Projection Calculation → → → → → → Example 13: Given a = (3, 4), b = (0, −2) , find the angle θ between a and b and the projection of a in the direction of b . Solution: → →→ → a ⋅ b = 3 × 0 + 4 × (−2) = −8, a = 5, b = 2, cos θ = → a → → a⋅b = →→ a b −8 5×2 = − 4 5 , since θ ∈ [0, π] , then θ = arccos (− → 's projection in the direction of b is → → a⋅b → −8 = 5 ) ; → (the negative sign indicates that the projection direction is opposite to the direction of b ). = −4 2 4 b Answer: θ = arccos (− 4 5 ) ; the projection is -4. Question type 4: Comprehensive application of dot product (geometric proof + maximum/minimum problem) Example 14: Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its four sides (i.e., 2 2 2 − 2 − − − → → → → − AC + BD = 2 ( AB + AD ) ) . Proof: − − − → → → − → → → → → → → AB = a, AD = b AC = a + b, BD = b − a Let , then ; 2 2 → − 2 → → → → → → By the property of the dot product, 2 AC 2 = (a + b) = a + 2a ⋅ b + ; b 2 2 → − 2 → → → →→ → BD = (b − a) = b − 2a ⋅ b + a ; 2 2 2 2 2 − − − → 2 → → → − → → Adding the two equations gives: AC + BD → → = 2 a → + 2 b = 2 ( AB + AD ) , which is proven. → → Example 15: Given a = (1, 1), b = (2, −1) , if c = a + tb (t ∈ R) , find the minimum value of Problem-solving idea: Express 2 → c →→ Calculate the coordinates of c : c = (1 + 2t, 1 − t) ; 2 2 → c : → c 2 = (1 + 2t) 2 + (1 − t) 2 = 1 + 4t + 4t The axis of symmetry of the quadratic function y = 5t 2 The minimum value is y min Therefore, → c = √ min Answer: When t = − 1 5 g 5 and the value of t at this time. as a quadratic function of t and use the maximum and minimum values of the quadratic function to solve (avoid taking the square root to simplify the calculation). Solution: Calculate → c = = 5 × (− 3√5 5 1 5 2 ) + 2 × (− 1 5 + 2t + 2 ) + 2 = . , the minimum value of → c is 2 + 1 − 2t + t 3√ 5 5 . is t = − 1 5 − 2 5 2 = 5t b 2a + 2t + 2 = − + 2 = 9 5 ; 2 2×5 ; = − 1 5 ; ∣∣ ∣ 5.5 Scalar Product Extended Exercises (Progressive Layers) Basic Questions Given → →→ = √ 3, a ⋅ b = 3 → a = 2, b → → a⋅b Solution: cos θ = 3 = → → a b = 2×√ 3 → → , find the angle θ between a and b . . Since θ ∈ [0, π] , then θ = 30 (or √3 ∘ 2 π 6 radians). Answer: θ = 30 . ∘ → → →→ → → Given a = (2, 3), b = (−4, 1) , find the projection of a ⋅ b and a in the direction of b . Solution: →→ a ⋅ b = 2 × (−4) + 3 × 1 = −8 + 3 = −5 → b 2 = √ (−4) 2 + 1 = √ 17 ; , the projection is →→ a⋅b → b →→ Answer: a ⋅ b = −5 ; The projection is − 5√ 17 17 = − 5 √ 17 = − . 5√ 17 17 . Advanced problem → → Given that the angle between a and b is 120 → → , a = 1, b = 2 ∘ , find → → a − 2b . Solution: → → a − 2b 2 2 → → = (a − 2b) = → a 2 → →→ − 4a ⋅ b + 4 b 2 ; →→ ∘ a ⋅ b = 1 × 2 × cos 120 = −1; Substituting gives: Answer: → → a − 2b 2 = 1 − 4 × (−1) + 4 × 4 = 1 + 4 + 16 = 21 , so → → a − 2b = √ 21 . . √ 21 → → → → → Given a = (1, k), b = (2, 1) , if a⊥ (a − b) , find the value of k . Solution: → → a − b = (−1, k − 1) ; → → → From the perpendicular condition a ⋅ (a − b) = 0 , we get 1 × (−1) + k × (k − 1) = 0 ; After rearrangement: k Answer: k = 1+√5 2 2 − k − 1 = 0 or k = 1−√5 2 , and the solution is k = 1±√5 2 . . Difficult problem →→ Given that a, b is a non-zero vector and → → → → a = b = a − b → → → , find the angle θ between a and a + b . Solution: Let → a = Expanded to: → b → a = m (m > 0) 2 → →→ − 2a ⋅ b + b , from → → a − b = m 2 2 = m , substituting into , we get → → a − b → → a = b = m 2 2 = m , we get m 2 ; →→ 2 2 − 2a ⋅ b + m = m →→ , and the solution is a . b = 2 m 2 ∣∣ Calculate → → → → a + b : a + b → → a + b = √ 3m 2 = → a 2 → →→ + 2a ⋅ b + b 2 2 = m 2 + 2 × m 2 → → → a⋅(a+b) ; Calculate the included angle: cos θ = Due to θ ∈ [0, π] , so θ = 30 (or ∘ Answer: θ = 30 . ∘ →→ → a a+b π 6 2 + m 2 = 3m 2 → →→ a +a⋅b = m×√ 3m radians). 2 = , so m + 2 m 2 √ 3m2 = √3 2 ; ∣ Chapter 6 Comprehensive Review and Summary of Core Key Points 6.1 Sorting out Core Knowledge Points 1. The Essence and Representation of Vectors Core: It has both magnitude and direction (free vector, independent of the starting point); − → → a AB (x, y) Representation: Letter , directed line segment , coordinate . 2. Special Vectors and Key Relationships → The zero vector o : has an arbitrary direction and is parallel to any vector; → Unit vector: e = → → a → a ; → → → Collinear (parallel): a ∥ b ⇔ b = λa (coordinates: x → → 1y2 →→ Perpendicular: a⊥b ⇔ a ⋅ b = 0 (coordinates: x 1x2 − x2y1 = 0 + y1y2 = 0 ); ). 3. Three major operation systems Operation type Geometric rules → Coordinate formula (a = (x Addition → 1, y1), b = (x2, y2)) Triangle/Parallelogram rule Subtraction � + � = (�1 + �2 , �1 + �2 ) Triangle rule (common starting point) Scalar multiplication Length → → λa = |λ| a Dot product � − � = (�1 − �2 , �1 − �2 ) , direction changes with λ �� = (��1 , ��1 ) � ⋅ � = |�||�|cos� 4. Two core theorems → → → → � ⋅ � = �1 �2 + �1 �2 → → Collinear vector theorem: a ∥ b (a ≠ 0) ⇔ ∃λ ∈ R, b = λa ; → → → → − → a λ1e1 + λ2e2 (e1 , e2 Fundamental theorem of plane vectors: Any vector can be uniquely represented as (not collinear). 6.2 Comprehensive Example Analysis Example 16: Vector and Geometry Comprehensive (Three Points Collinear + Perpendicular) ∣ ∣∣ Given A (−1, 2), B (3, 4), C (5, k) , if the three points A 、 、 B C are collinear, find the value of k ; if AB⊥AC , find the value of k . Solution: → → − − Three points are collinear: AB = (4, 2), AC = (6, k − 2) , from the collinearity condition 4 (k − 2) − 2 × 6 = 0 , we solve to get k = 5 ; − − → → Vertical: From AB ⋅ AC = 4 × 6 + 2 (k − 2) = 0 , solve to get 24 + 2k − 4 = 0 ⇒ k = −10 . Answer: When collinear, k = 5 ; when perpendicular, k = −10 . Example 17: Comprehensive of vector maximum/minimum value and included angle → → → → → Given a = (1, 2), b = (−2, 1), c = a + tb , find: → → → → (1) The angle between a and b ; (2) When t takes what value, c is perpendicular to b ; (3) The minimum value of Solution: Angle: cos θ = → → → →→ a⋅b = →→ a b 1×(−2)+2×1 = 0 √ 5×√ 5 → c . , so θ = 90 ; ∘ Vertical: c = (1 − 2t, 2 + t), c ⋅ b = −2 (1 − 2t) + 1 (2 + t) = 0 ⇒ −2 + 4t + 2 + t = 0 ⇒ t = 0 ; Minimum value: 2 → c 2 = (1 − 2t) Answer: (1) 90 ; (2) t = 0 ; (3), √5 . ∘ 2 + (2 + t) = 5t 2 + 5 , when t = 0 , → c = √5 . min 6.3 Summary of Learning Suggestions and Error-Prone Points 1. Avoidance of Error-Prone Points Confusing the results of vector operations: The dot product is a real number, while addition/subtraction/scalar multiplication are vectors; Confusing the conditions of collinearity and perpendicularity: For collinearity, check if the cross product is 0, and for perpendicularity, check if the dot product is 0; Incorrect definition of the included angle: The starting points must be the same, and the range is [0, π] ; → Ignoring the zero vector: In the collinear theorem, a must be non-zero. The dot product of the zero vector and any vector is 0, but they are not perpendicular. 2. Learning methods Combining numbers and shapes: The geometric meaning of vectors (such as projections, angles, graphical characteristics of parallel/perpendicular) is closely combined with algebraic operations (coordinates, dot product formulas). When solving problems, first draw a sketch for analysis, and then use the formulas for calculation; Formula classification: Organize and compare the geometric expressions and coordinate expressions of collinearity, perpendicularity, modulus length, angle, and projection to avoid confusion; Question type summary: Focus on practicing the four core question types of "basis decomposition", "coordinate operation", "perpendicular/collinear determination", and "application of dot product", and summarize the problem-solving templates; Review of wrong questions: Set up a wrong question notebook for error-prone points (such as zero vectors, angle range, confusion of operation properties), and mark the reasons for errors and the correction ideas. 6.4 Expansion of Vector Application Scenarios As an instrumental knowledge, vectors are widely used in disciplines such as mathematics and physics. The core application scenarios include: Geometric field: Prove positional relationships such as parallelism, perpendicularity, and bisection (such as using the dot product to prove perpendicularity and collinear vectors to prove the collinearity of three points); Calculate lengths, angles, areas (such as the area of a triangle S = 1 2 → a → b sin θ , the area of a parallelogram → → a × b ); ∣∣ Solve problems of graphic transformations such as folding, translation, and rotation. Physics field: Composition and decomposition of forces (parallelogram rule/triangle rule); Composition and decomposition of displacement, velocity, and acceleration (the essence of vector operations is vector calculation); Calculation of work (constant force work W → → = F ⋅ s = → → F s cos θ , which is consistent with the definition of the dot product). 6.5 After-class comprehensive test questions (consolidation and improvement) I. Multiple-choice questions (Basic consolidation) Which of the following propositions is correct? () → → → → A. If a ∥ b , then a = λb (λ ∈ R) → → → → → → → → → B. If a ⋅ b = 0 , then a = 0 or b = 0 C. If → → a = b → → → , then a = b or a = −b → → → → → → D. If a and b are collinear, b and c are collinear, and b ≠ 0 , then a ∥ c → → → → Given a = (1, −2), b = (m, 4) , and a⊥b , then m = () A. 8 B. -8 C. 2 D. -2 III. Answer Questions (Comprehensive Application) Given A (2, 1), B (−1, 3), C (1, t) , if △ABC is a right triangle, find the value of t . → → Given that a and b are non-zero vectors, and （1） （2） → → → a ⋅ (a + 2b) → → a − b ； → → → a = 2 , → b = 1 → → , the included angle between a and b is 60 , find: ∘ ； → (3) If (a − kb)⊥a , find the value of k . Answer Key I. Multiple Choice Questions → → → → D (Explanation: In A, it requires b ≠ 0 ; in B, it may be a⊥b ; in C, the magnitudes are equal but the directions are not necessarily the same or opposite); 2. A (Explanation: 1 × m + (−2) × 4 = 0 ⇒ m = 8 ). III. Solution Questions Solution: − − − − → → → → If ∠A = 90 , then AB ⋅ AC ∘ = 0, AB = (−3, 2), AC = (−1, t − 1) , resulting in (−3) (−1) + 2 (t − 1) = 1 0 ⇒ t = − 2 → → − − → → − − If ∠B = 90 , then BA ⋅ BC ∘ = 0, BA = (3, −2), BC = (2, t − 3) , resulting in 3 × 2 + (−2) (t − 3) = 0 ⇒ t = 6 → → − − → → − − If ∠C ∘ = 90 2 t) = 0 ⇒ t , then CA ⋅ CB = 0, CA = (1, 1 − t), CB = (−2, 3 − t) , resulting in 1 × (−2) + (1 − t)(3− ; − 4t + 1 = 0 ⇒ t = 2 ± √ 3 ∣ In summary, t = − or 6 or 2 + √3 or 2 − √3 . 1 2 Solution: → → → (1) a ⋅ (a + 2b) = (2) → → a − b (3) From 2 = → a 2 → a 2 →→ ∘ + 2a ⋅ b = 4 + 2 × 2 × 1 × cos 60 = 4 + 2 = 6 → →→ − 2a ⋅ b + b → → → (a − kb)⊥a , we get 2 = 4 − 2 + 1 = 3 → → → (a − kb) ⋅ a = 0 , so → → a − b = √3 , that is, → a 2 ; ; →→ − ka ⋅ b = 0, 4 − k × 1 = 0 ⇒ k = 4 .